PATENT ABSTRACT
Charge-Driven Electrostatic Induction is a method for using modest voltage to induce large density electric charge across a large insulation gap. Large density equal and opposite charges are first created in a high performance capacitor adjacent said insulation gap. One charge is trapped on its electrode and the other charge is relocated further from the gap so the electric field from the trapped charge, with minimum interference, induces equal and opposite charge across the gap and stores large density electric energy in the insulation. With electrode area to gap ratio kept sufficiently large to limit field fringing, Charge-Driven Electrostatic Induction will rival electromagnetic motor performance. In practice it will be superior. Using layered, thin film components will eliminate permanent magnets, coils, ferromagnetic materials and large power current sources. The multi-step process will permit high operating speeds.

PATENT DESCRIPTION
CROSS REFERENCE TO RELATED APPLICATION 
       [0001]    The U.S. patent application claims the priority of U.S. Provisional Application No. 61/458,238 filed on Nov. 19, 2010. 
     
    
       [0002]    The invention is related to an invention shown and described in Provisional Patent, U.S. Ser. No. 61/458,238 entitled “Charge-Driven Electrostatic Induction”, filed in the name of John M. Vranish, the present inventor on Nov. 19, 2010. The above are assigned to the assignee of the present invention. The teachings of this related application is, herein meant to be incorporated by reference. The invention is also related to an invention shown in: Vranish, J. M., Device, System and Method for a Sensing Electric Circuit, U.S. Pat. No. 7,622,907, Nov. 24, 2009. [“Driven Ground”] the rights to which are held by the US Government. 
     
    
     ORIGIN OF THE INVENTION 
       [0003]    The invention was made by John M. Vranish as President of Vranish Innovative Technologies LLC and may be used John M. Vranish and Vranish Innovative Technologies LLC without the payment of any royalties therein or therefore. 
       BACKGROUND OF THE INVENTION 
       [0004]    The Charge-Driven Electrostatic Induction concept began with a need to actuate flexures in a Tape Motor invention. Permanent magnet electromagnetic drives were too large and cumbersome. Electrostatic drives were too weak and required voltages that were too high. John M. Vranish looked to electrets as an alternative to permanent magnets. But, it soon became apparent that devices that behave like electrets could be produced by trapping and isolating electric charge on capacitor electrodes. In the process of investigating alternatives to electrets, it soon became apparent that new capacitive materials enabled exceptional charge density from modest voltage. But, there were still problems in isolating the charge and in directing the electric flux from the charge. Step by step, the Charge-Driven electrostatic concept began to evolve to this point. It will continue to evolve. 
       FIELD OF THE INVENTION 
       [0005]    The invention relates generally to Electrostatic Induction and more particularly to working level voltage, electrostatic applications. The invention relates generally to electret applications as an alternative method. The invention relates generally to electromagnetic induction as an electrostatic alternative for electromagnetic induction applications. The invention relates generally to high voltage applications as a working level voltage alternative and more particularly to step up and step down voltage transformers. The invention relates particularly to electrostatic power generation devices, power transfer devices, motor devices and sensors, both static and quasi-static. 
       DESCRIPTION OF THE PRIOR ART 
       [0006]    Electrostatic Motors, Micromotors, Piezoelectric Travelling Wave Motors and Piezoelectric Inch Worms have, traditionally, performed precision positioning. Charge-Driven Electrostatic Induction, in combination with bending flexures, is presented as an alternative with advantages. (Bending Flexures is presented separately from this patent application.) 
         [0007]    Electric Motors, using electromagnetism, constitutes a body of prior art. Charge-Driven Electrostatic Induction introduces an electrostatic alternative with advantages. 
         [0008]    Electromagnetic Generators also constitute a body of prior art. Charge-Driven Electrostatic Induction introduces an electrostatic alternative with advantages. 
         [0009]    Electret Microphones use elements with permanent polarization to perform functions of converting mechanical oscillating motion to electrical energy and output voltage. Charge-Driven Electrostatic Induction performs the same function without using permanently polarized elements and with the advantage of being able to easily neutralize stray charge. This argument can be extended to energy harvesting and scavenging devices and methods. 
         [0010]    Transformers use coils and electromagnetism to step up or step down voltage as per traditional prior art. Charge-Driven Electrostatic Induction presents an alternative with advantages using multiple stacked capacitors rather than multiple coils. 
         [0011]    Electromagnetic means, analogous to Electromagnetic Motors, has been used to transfer electric power across a joint with an air or vacuum gap between the moving members. Charge-Driven Electrostatic Induction performs the same function with advantages using electrostatics. 
       SUMMARY OF THE INVENTION 
       [0012]    Charge-Driven Electrostatic Induction is a method for using modest voltage to induce large density electric charge across a large insulation gap. Large density equal and opposite charges are first created in a high performance capacitor adjacent said gap. One charge is removed and the electric field of the remaining charge is reflected into the gap where it induces equal and opposite charge on the far side and stores large density electric energy in said gap as per Gauss&#39; Law of Charges and the method of images. With electrode area to gap ratio kept sufficiently large to limit field fringing, Charge-Driven Electrostatic Induction can rival electromagnetic motor performance. In practice, it will be superior. Constructed of layered, thin film components, Charge-Driven Inductance devices will be lighter, more compact and less expensive than their permanent magnet, electromagnetic counter parts. Coils, winding process, ferromagnetic materials, rare earth permanent magnets and large current power sources will be unnecessary and integration of controls and action devices will be more seamless. The multi-step process will permit high operating speeds. The principles behind Charge-Driven Electrostatic Induction are explained and construction of a device using Charge-Driven Electrostatic Induction is illustrated. Applications are presented illustrating use as an electrostatic motor, an electrostatic generator, and an electrostatic device for transferring power across a large air or vacuum gap. Performance enhancing techniques of Electric Field Projection and Charge Compression are introduced. 
     
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         [0013]    A more complete appreciation of the invention and many of its attendant advantages will be readily appreciated as the same becomes better understood by reference to the following detailed description when considered in connection with the accompanying drawings wherein: 
           [0014]      FIG. 1  illustrates the major components of a basic Charge-Driven Electrostatic Induction Device and how they are arranged with respect to each other. 
           [0015]      FIG. 2   a  illustrates how equal and opposite charges are created in close proximity to each other and how one of the charges is trapped in place on its electrode. 
           [0016]      FIG. 2   b  illustrates how the second of the charges is displaced from the trapped charge to provide separation and isolation for said trapped first charge. 
           [0017]      FIG. 2   c  illustrates how the separated equal and opposite charges are trapped in place when the voltage source removed. 
           [0018]      FIG. 3   a.  illustrates the charge arrangement before the grounded conductor is introduced. 
           [0019]      FIG. 3   b.  illustrates the charge arrangement after a grounded conductor is introduced nearby and shows energy stored in the insulation gap between the stack of electrodes and the grounded conductor. 
           [0020]      FIG. 4   a.  illustrates the effects of a nearby grounded conductor on initial charge formation and initial charge arrangement after a first charge is trapped. 
           [0021]      FIG. 4   b.  illustrates the effects of a nearby grounded conductor on charge formation and charge arrangement when a second step in charge formation and charge trapping is performed. 
           [0022]      FIG. 5   a  shows the floating outer electrode case. 
           [0023]      FIG. 5   b  shows the grounded outer electrode case. 
           [0024]      FIG. 6  illustrates the effects of simultaneously charging a first one third of the electrodes in a stack of capacitors. 
           [0025]      FIG. 7  illustrates the effects of simultaneously charging a second one third of the electrodes. 
           [0026]      FIG. 8  illustrates the effects of simultaneously charging the remaining one third of the electrodes. 
           [0027]      FIG. 9  illustrates the energy stored in the charged stack of electrodes. 
           [0028]      FIG. 10   a  illustrates the electric field and charge configuration when electric field is applied in a motor application, where the moving member moves transverse to the electric field. 
           [0029]      FIG. 10   b  illustrates residual effects when the electric field is removed. 
           [0030]      FIG. 11   a  illustrates charge distribution and electric field configuration in a motor application in which the moving member moves along the direction of the field, after the field is applied, but before the moving member has moved. 
           [0031]      FIG. 11   b  illustrates the charge distribution and electric field configuration after limited movement has occurred. 
           [0032]      FIG. 12   a  shows the effects of applying an electric field in a power transfer application across an air/vacuum gap typical of moving joints of machines, motors and generators. 
           [0033]      FIG. 12   b  shows the effects when the electric field is removed. 
           [0034]      FIG. 13   a  illustrates a first position in an apparatus that converts time varying mechanical energy to electrical energy. 
           [0035]      FIG. 13   b  illustrates a second position. A comparison of the two positions and the effect on the electric energy stored in the apparatus insulation gap provides insight into the charge-driven electrostatic energy conversion process. 
           [0036]      FIG. 14  illustrates the electrical ground termination apparatus for receiving the generated electrical power illustrated in  FIG. 13   a  and  FIG. 13   b  and for making it available for external use. 
           [0037]      FIG. 15   a  illustrates a first position in an apparatus that uses passive electronic components in converting time varying mechanical energy to electrical energy. 
           [0038]      FIG. 15   b  illustrates a second position. A comparison of the two positions and the effect on the electric energy stored in the apparatus insulation gap provides insight into the charge-driven electrostatic energy conversion process using passive electronic components. 
           [0039]      FIG. 16  illustrates an electrical ground termination and electrical energy storage system for receiving and storing the generated electrical power illustrated in  FIGS. 14   a  and  14   b  where passive electronic components. 
       
    
    
     DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENT 
     A. Description 
       [0040]    In accordance with the present invention, A Charge-Driven Electrostatic Induction System a Charge-Driven Electrostatic Induction System includes a 1). A Charge Creation and Isolation Capacitor System, 2). An Air or Vacuum Gap adjacent to said Charge Creation and Isolation Capacitor, with Remote Electrical Conductor on the Far Side of the Gap, 3). A Housing. The preferred embodiment of A Charge-Driven Electrostatic Induction System is configured according to  FIG. 1 , with said Charge Creation and Isolation System comprising a stack of parallel electrodes each separated from its neighbors by an ultra-thin, high dielectric constant film. The electrodes are charged in parallel but, the stack of electrodes performs like a set of charged capacitors connected in series. As such, the voltage of each of the individual charged capacitors adds to provide a very high stack voltage capable of performing significant electrostatic induction across large insulation gaps. A novel trapped charge step method enables energy to be stored in a stack of capacitors such that the voltages add in series. This method of storing energy is illustrated in  FIGS. 2   a ,  2   b  and  2   c . The internal adjustments made in said stored energy under the influence of a nearby grounded conductor is shown in  FIGS. 3   a ,  3   b ,  4   a  and  4   b . Effects of grounding or floating the electrode in the stack furthest from said nearby grounded conductor are illustrated in  FIGS. 5   a  and  5   b.    
         [0041]    Current technology permits stacks with electrodes and dielectric films in the hundreds so a method for rapidly charging said electrodes is presented, whereby multiple electrodes can be charged or discharged simultaneously. This method is illustrated in  FIGS. 6 ,  7 ,  8  and the resulting stored energy distribution in the stack is illustrated in  FIG. 9 . 
         [0042]    The invention will now be described in more detail by way of example with reference to the embodiments shown in the accompanying figures. It should be kept in mind that the following described embodiments are only presented by way of example and should not be construed as necessarily limiting the inventive concept to any particular physical configuration. 
       B. Construction 
       [0043]    The construction of a Charge-Driven Electrostatic Induction System according to  FIG. 1  will now be described. 
         [0044]    1). Charge Creation and Isolation System. Said Charge Creation and Isolation System, (according to  FIGS. 6 ,  7  and  8 ) comprising a stack of parallel electrode capacitors wherein a first outer electrode, labeled  2 O 1  is followed by an ultra-thin, high dielectric constant film, labeled C 2 , followed in turn by n Inner Electrodes, labeled  2 I 1 ,  2 I 2 ,  2 I 3 , . . .  2 In, each separated from another by an ultra-thin, high dielectric constant film labeled C 2  terminating in second outer electrode, labeled  2 O 2 . The second outer electrode  2 O 2  is separated from the n&#39;th inner electrode by an ultra-thin, high dielectric constant film, labeled C 2 , on one surface and by a second insulating film on its other surface, labeled C 3 . C 3 , in turn, connects said stack of electrodes to Support Structure, labeled  3 . Film C 3 , need be an electrical insulator but, it need not be thin nor have a high dielectric constant. 
         [0045]    2). Switch System. Said Switch System comprising a voltage source, a system of electrically conducting wires whereby said voltage source is connected to each of the electrodes described in 1). Charge Creation and Isolation System and a set of computer controlled switches (according to  FIGS. 1 ,  6 ,  7  and  8 ) whereby each said voltage source and electrode connection can be opened or closed on command. Said Switch System also comprising a system of electrically conducting wires whereby each said electrode is connected to an electrical ground and a set of computer controlled switches whereby each said electrode and ground connection can be opened or closed on command. Said Switching System component (labeled  3 ) that connects and disconnects said electrodes and said voltage sources is separate and distinct from said Switching System component (labeled  4 ) that connects and disconnects said electrodes and said electrical ground. Said electrodes are each served by both Switching System components. 
         [0046]    The switches in 3 service all n+2 electrodes. Each inner electrode in said stack of electrodes (labeled  2 ) has one switch that connects it to and disconnects it from +V S  (labeled S 1 P, S 2 P, S 3 P, etc.) and has one switch that connects it to and disconnects it from −V S  (labeled S 1 N, S 2   n,  S 3 N, etc.). The outer electrodes are each, independently, serviced by two switches, with one switch (labeled SO 1 P) and one switch (labeled SO 1 N) connecting and disconnecting said first outer electrode (labeled  201 ) to and from +V S  and −V S  respectively and one switch (labeled SO 2 P) and another switch (labeled SO 2 N) connecting and disconnecting said second outer electrode (labeled  202 ) to and from +V S  and −V S  respectively. Said electrodes can be connected in groups so that a group of electrodes can be simultaneously serviced by a single switch. In this arrangement, S 1 P, S 1 N, S 2 P, S 2 N etc, connect groups of electrodes, hard wired together. 
         [0047]    The switches in 4 service all n+2 electrodes. Each inner electrode in said stack of electrodes (labeled  2 ) has one switch that connects it to and disconnects it from ground (labeled S 1 G, S 2 G, S 3 G, etc.). The second outer electrode ( 202 ) has a dedicated switch to and from ground (labeled SO 2 G). A first outer electrode ( 201 ) will also require a dedicated switch (labeled SO 1 G) to and from ground except when the said electrodes are grouped. With grouped electrodes, S 1 G eliminates the need for SO 1 G. 
         [0048]    3). Target Conductor. Said Target Conductor, labeled  1  in  FIG. 1 , is separated from said stack of parallel electrode capacitors by insulation gap C 1 . Said Target Conductor is electrically grounded to said Support Structure ground. 
         [0049]    4). Support Structure. Said Support Structure (labeled  5 ), houses and fixes the components of the Charge-Driven Electrostatic Induction System. It also provides the common electrical ground for all components therein. 
         [0050]    5). Moving Member (labeled  6  in  FIGS. 10   a,    10   b,    11   a  and  11   b ). This applies for motor, generator and sensor applications. 
       C. Operations 
       [0051]    A Charge-Driven Electrostatic Induction System operates by first charging a stack of capacitors to emulate a series connection of multiple capacitors and, then using the stepped up voltage from said charged multi-electrode capacitor to induce significant charge in a target conductor separated by a thick insulator (typically air or vacuum) from said stack of capacitors. Stack charging is done in a series of steps so as to enable relatively small working level voltages to charge the individual capacitors in such that the stored charge and energy inside the stack of capacitors emulates that a series arrangement. That is, if the attack of capacitors were charged in series with a very large voltage the same arrangement of stored charge and energy would result. 
         [0052]    The method by which electrodes of individual capacitors in said stack can be charged so as to create a result similar to series charging with much higher voltage will be illustrated as per  FIGS. 2   a,    2   b  and  2   c.  The method uses electrodes in groups of three, functioning as two back to back capacitors with a shared middle electrode. The first electrode E 1  is connected to V S  through switch S 1  and to electrical ground through switch SG 1 . The second electrode E 2  is connected to V S  through switch S 2  and to ground through switch SG 2 . The third electrode E 3  is connected to V S  through switch S 3  and to ground through switch SG 3 . I In the first step, the first electrode E 1  is connected to a voltage source (say +V S ) and the middle electrode E 2  is grounded. As a result positive charge is induced on said first electrode (E 1 ) and equal negative charge is induced on said middle electrode E 2 . Both electrodes are disconnected and floated with charge trapped on each. The middle electrode E 2  is then connected to +V S  and the third electrode E 3  is connected to ground. Positive charge is induced on said middle electrode E 2  and equal negative charge is induced on said third electrode E 3 . Again said voltage source and said ground are disconnected and the induced charges are trapped on their electrodes. The middle electrode E 2  has both positive and negative charge trapped on it. In this example the negative charge is trapped on the middle electrode E 2  surface nearest the first electrode E 1  and the positive charge is trapped on the middle electrode E 2  surface nearest the third electrode E 3 . Electrical energy is stored in the dielectric films between said electrodes E 1  and E 3 . Electrode E 1  is charged positive and electrode E 3  is charged negative with electrode E 2  charged both positive and negative in emulation of the two capacitors being charged in series with +2V S . The process can be repeated until an entire stack of multiple capacitors is charged as if in series. It is possible to double the amount of electric charge on each of the electrodes by using +V S  as source voltage and −V S  as the termination potential, rather than ground. The same charge steps apply. We will return to this discussion in more detail, later. 
         [0053]    Using stored energy in a series stack of charged capacitors to induce charge across a large insulation gap is distinct from using an equivalent large voltage source. The stack of capacitors transfers electric energy from internal storage to the insulation gap and the fixed charge distribution on the external electrode nearest said gap adjusts accordingly as does the remainder of the charge in said stacked capacitors. An equivalent very large voltage source would have to supply charge according to the basic equation relating charge, voltage and a a stack of multiple capacitors connected in series. These differences and their ramifications will be discussed in more detail, later. 
         [0054]    Speed of operation is maintained even though multiple steps are required by charging a multiple capacitor stack as three sets of parallel capacitors and then trapping the charges so as to leave a stored charge and energy arrangement emulating a series charged capacitor. With this method, a stack of 100 or more capacitors can be charged in three steps and high speed ac operations can be conducted. We will also return to this in more detail. 
       1. Charging Method. 
       [0055]    The Charging Method will be discussed in steps. In the step a), the Charging Method will be discussed for the Isolated Three-Electrode case, with no Target Conductor nearby. In step b), a grounded Target Conductor will be positioned nearby the charged Three-Electrode Capacitor of the first step so the reader can see how the trapped charge on the Three-Electrode Capacitor re-arranges itself to induce charge in said grounded Target Conductor and store electrical energy in the air gap that separates said grounded Target Conductor and said three electrode capacitor. This will illustrate differences in electrostatic induction using trapped charge sources and electrostatic induction using voltage sources in the simplest case. We now begin to add some real world complexity, so in step c) we have said grounded Target Conductor present when we charge said three-electrode capacitor. This illustrates charge formation, electrostatic induction and charge arrangement during expected operational circumstances. 
         [0056]    a). Isolated Three-Electrode Case ( FIGS. 2   a,    2   b,    2   c ). This will enable the reader to see how charge forms in the stack of capacitors with minimum complicating factors. It will clearly illustrate how capacitors can be charged in parallel to emulate a series charge arrangement. When the first electrode E 1  is connected to a voltage source +V S  and the middle electrode E 2  is connected to ground as per  FIG. 2   a,  a positive charge is formed on electrode E 1  and an equal but opposite negative charge is formed on electrode E 2  according to Q=V S /C [1]. The Ground and voltage connections are then opened, leaving positive charge trapped on E 1  and negative charge trapped on E 2 , with electrical energy stored in the dielectric film between E 1  and E 2 . The charge on E 1  are on the electrode surface nearest E 2  and the charge on E 2  is on the electrode surface nearest E 1 . Next, +V S  is connected to electrode E 2  and third electrode E 3  is connected to ground as per  FIG. 2   b,  A charge +Q forms on said the electrode E 2  surface nearest electrode E 3  and a −Q charge forms on the E 3  electrode surface nearest E 2 . This leaves us with positive charge on one surface of E 2  and negative charge on the other surface of E 2  while E 2  is connected to +V S . The negative charge on E 2  is attracted to both the positive charge on E 1  and to voltage source +V S . It stays in place so when the ground connection to E 3  is opened, followed by opening the connection between E 2  and +V S , both positive and negative charges are held in place on E 2 , while positive charge is trapped on E 1  and negative charge is trapped on E 3 . Why doesn&#39;t the negative charge on E 2  get removed by +V S  and what does this imply? The negative charge on E 2  is not removed by +V S  because to do so would move the system away from its minimum available energy state and violate conservation of energy. With the negative charge on E 2  removed, the positive charge on E 1  would be forced to induce charge in the nearest available conductor other than E 2  and the energy expended to do so would be greater than the energy expended to hold negative charge on E 2 , even against +V S . And, what does this imply? It implies the voltage on E 1  has increased to the point where V E1 −+V S =+V S , V E1 =+2V S . 
         [0057]    The resulting charge arrangement is similar to what would occur if the Three-Electrode Capacitor had been charged by +2V S  voltage across electrodes E 1  and E 3  and the energy stored in the dielectric layers between E 1 , E 2  and E 3  also matches the series charge arrangement. The positive charge on E 1  and the negative charge on E 3  match the series charge arrangement. The middle electrode E 2  does also, with a negative charge on its surface nearest E 1  and a positive charge on its surface nearest E 3 . This means we have a +2V S  potential difference between E 1  and E 3 , even though only +V S  has been used to charge the capacitor(s). 
         [0058]    b). Grounded Target Conductor introduced after Charging ( FIGS. 3   a,    3   b ). When a grounded Target Conductor is brought near a Three-Electrode Capacitor charged in said series configuration as per  FIG. 3   b,  said positive charge on electrode E 1  can store energy in C 1  and C 2 , rather than C 2  only. As per conservation of energy, it will pick a combination of C 1  and C 2  which is stores the least amount of energy. The portion of positive charge that is stored in C 1  subtracts from the charge available to store energy in C 2 . The total amount of negative charge on electrode E 3  is reduced accordingly. With said three electrodes E 1 , E 2  and E 3  working in series, the charge on both surfaces of middle electrode E 2  are reduced by an amount equal to the amount of charge diverted to C 1 . The charge reduction on middle electrode E 2  is accomplished by some negative charge on one surface of middle electrode E 2  combining with an equal amount of positive charge on the other surface of E 2  and the remaining excess of negative charge on electrode E 3  dispersing back into ground. 
         [0059]    These adjustments are reflected in eq. (1). 
         [0000]        Q   1   =Q   11   +Q   12   [2] eq. (1)
 
         [0000]    Where Q 1  is the total charge trapped on electrode E 1 , Q 12  is the charge on E 1  that capacitively couples by capacitance with E 3  (with E 2  as an intermediary) and Q 11  is the charge on E 1  that couples by capacitance with Target Conductor  1  across capacitance C 1 . C 2  is the capacitance between electrodes E 1  and E 2  and between E 2  and E 3 . 
         [0060]    The voltage on E 1  is: 
         [0000]        Q   11   /C 1 =Q   12 /0.5 C 2  [3] eq. (2)
 
         [0000]    Which provides information on the charge distribution on E 1  as per: 
         [0000]        Q   11   /Q   12 =2 C 1 /C 2  eq. (3)
 
         [0000]    From equations (1) and (3) we have: 
         [0000]    
       
         
           
             
               
                 
                   
                     
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         [0061]    So we can see that some of Q 1  is diverted from coupling with E 3  to coupling with Target Conductor  1  and we postulate the charge and voltages on E 2  and E 3  must adjust accordingly. The amount of charge available to couple with E 2  and E 3  has been reduced so there is less attractive electrostatic force to hold charge on E 3  and the charge on E 3  disperses back into ground until a new balance is restored at a lower level. E 2  responds by reducing the positive charge on one surface and the negative charge on the other surface by allowing limited charge cancellation (or recombination) consistent with the new balance point. The charge neutrality of E 2  is unchanged. 
         [0062]    If E 3  is disconnected from ground before Target Conductor  1  is introduced, it seems Q 11 =0 because the positive charge on E 1  is balanced by a trapped negative charge on E 3  and by Gauss&#39; Law of Charge, the electric flux is zero outside the closed system E 1 , E 2 , E 3 . When E 3  is grounded, the system seeks the balance just described. Once the balance is reached and Q 11  comes into existence, E 3  can be connected and disconnected to ground with no effect on Q 11 . 
         [0000]    c). Grounded Target Conductor present during Charging. 
         [0063]    When a grounded Target Conductor is present throughout a charging cycle, additional electric energy is stored in C 1  during the charge cycle as per  FIGS. 4   a  and  4   b.  A portion of this additional energy is stored in C 1  while electrode E 1  and electrode E 2  are charged and energy stored in C 2  as per a) and b) above, with the remainder stored in another C 2  when electrodes E 2  and E 3  are charged. The potential of electrode E 1  is raised to +V S  during the first step in charging, while extra charge is added to and trapped on E 1 , with charge on both surfaces of E 1 . The first electrode. The potential of said first electrode is raised to approximately +2V S  during the second step in charging and more of the charge trapped on it links with said grounded Target Structure. No additional charge is added to the first electrode during this second stage. 
         [0064]    The added extra charge on said first electrode is given as: 
         [0000]      Δ Q   11   =V   S   C 1=Δ Q   1   , ΔQ   2 =0  eq. (6)
 
         [0000]    For a total charge trapped on E 1  of: 
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                     8 
                     ) 
                   
                 
               
             
           
         
       
     
         [0065]    We note, as per  FIGS. 5   a  and  5   b,  that grounding or floating electrode E 3  has an effect on the amount of electrostatic charge induced in said grounded Target Conductor. When said third electrode is disconnected from ground before said middle electrode is disconnected from +V S , as per  FIG. 5   a,  an amount of negative charge is trapped on said third electrode that is equal and opposite to the positive charge on said middle electrode. This trapped negative charge acts to hold the positive charge on said middle electrode in place and inhibits charge combination/cancellation in said middle electrode, thereby inhibiting charge induction between said first electrode and grounded Target Conductor. When, thereafter, said third electrode is connected to ground, the negative charge on said third electrode is no longer trapped in place and can be reduced to accommodate the most efficient energy storage arrangement in C 1  and both C 2  dielectric layers. This results in more energy storage in C 1  and more induced charge on said grounded Target Conductor, as per  FIG. 5   b.  This happens because charge induced in a capacitor is held in place by a balance between applied voltage and dispersive forces in the charge. When the charge is trapped before the applied voltage is removed, the dispersive forces are physically opposed by the boundaries of the electrodes. When the electrode is grounded, dispersive forces spread the charge until a new balance is reached where more energy is stored in C 1  and more charge is induced in Target Conductor  1 . 
       2. Stack of Multiple Capacitors 
       [0066]    We note that a Three-Electrode Capacitor, charged to emulate a series capacitor between said first and third electrodes has an approximately 2V S  potential on its outer electrodes while a Two-Electrode Capacitor has V S . We can repeat the pattern by adding a fourth electrode E 4  and third dielectric film to obtain approximately +3V S  between electrodes E 1  and E 4 . In this instance said third electrode would be connected to source voltage V S  and capacitance coupled to a grounded fourth electrode. The resulting added positive charge could be trapped on said third electrode and we would have +3V S  between said electrodes E1 and E4 with net positive charge on said electrode E 1 , net negative charge on said electrode E 4  and self-cancelling positive and negative charges on electrodes E 2  and E 3 . This process can be continued until one hundred or more capacitors are added to the stack. 
         [0067]    For the n capacitors stacked in series, we estimate the effective series capacitance of the stack as: 
         [0000]        C   ST   =C   2   /n   [4] eq. (9)
 
         [0000]    Q 1  trapped on electrode E 1  in a stack of n capacitors in series has two parallel capacitance paths to ground, C 1  and C ST =C 2 /n. 
       So: 
       [0068]    
       
         
           
             
               
                 
                   
                     
                       Q 
                       11 
                     
                     
                       Q 
                       ST 
                     
                   
                   = 
                   
                     
                       
                         
                           C 
                           1 
                         
                         
                           C 
                           ST 
                         
                       
                        
                       
                           
                       
                        
                       or 
                        
                       
                           
                       
                        
                       
                         Q 
                         ST 
                       
                     
                     = 
                     
                       
                         Q 
                         11 
                       
                        
                       
                         
                           C 
                           ST 
                         
                         
                           C 
                           1 
                         
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     
                       9 
                        
                       a 
                     
                     ) 
                   
                 
               
             
           
         
       
     
         [0000]    This leads to 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       Q 
                       11 
                     
                     + 
                     
                       
                         Q 
                         11 
                       
                        
                       
                         
                           C 
                           ST 
                         
                         
                           C 
                           1 
                         
                       
                     
                   
                   = 
                   
                     
                       
                         Q 
                         1 
                       
                        
                       
                           
                       
                        
                       or 
                        
                       
                           
                       
                        
                       
                         
                           Q 
                           11 
                         
                          
                         
                           ( 
                           
                             1 
                             + 
                             
                               
                                 C 
                                 ST 
                               
                               
                                 C 
                                 1 
                               
                             
                           
                           ) 
                         
                       
                     
                     = 
                     
                       Q 
                       1 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     
                       9 
                        
                       b 
                     
                     ) 
                   
                 
               
             
           
         
       
     
         [0000]    Which simplifies to: 
         [0000]        Q   1   C   1   =Q   11 ( C   1   +C   ST )  eq. (9c)
 
       Resulting in: 
       [0069]    
       
         
           
             
               
                 
                   
                     
                       
                         
                           Q 
                           11 
                         
                         = 
                         
                           
                             Q 
                             1 
                           
                            
                           
                             ( 
                             
                               
                                 C 
                                  
                                 
                                     
                                 
                                  
                                 1 
                               
                               
                                 
                                   C 
                                    
                                   
                                       
                                   
                                    
                                   1 
                                 
                                 + 
                                 
                                   C 
                                   ST 
                                 
                               
                             
                             ) 
                           
                         
                       
                     
                   
                   
                     
                       
                         = 
                         
                           
                             Q 
                             1 
                           
                            
                           
                             ( 
                             
                               
                                 nC 
                                  
                                 
                                     
                                 
                                  
                                 1 
                               
                               
                                 
                                   nC 
                                    
                                   
                                       
                                   
                                    
                                   1 
                                 
                                 + 
                                 
                                   C 
                                    
                                   
                                       
                                   
                                    
                                   2 
                                 
                               
                             
                             ) 
                           
                         
                       
                     
                   
                   
                     
                       
                         = 
                         
                           
                             V 
                             S 
                           
                            
                           C 
                            
                           
                               
                           
                            
                           2 
                            
                           
                             ( 
                             
                               
                                 nC 
                                  
                                 
                                     
                                 
                                  
                                 1 
                               
                               
                                 
                                   nC 
                                    
                                   
                                       
                                   
                                    
                                   1 
                                 
                                 + 
                                 
                                   C 
                                    
                                   
                                       
                                   
                                    
                                   2 
                                 
                               
                             
                             ) 
                           
                         
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     10 
                     ) 
                   
                 
               
             
           
         
       
     
         [0000]    Without using stacked capacitors we could expect an induced charge of: 
         [0000]      Q 10 =V S C1  eq. (11)
 
         [0000]    Dividing eq. 10 by eq. 11 we find: 
         [0000]    
       
         
           
             
               
                 
                   Gain 
                   = 
                   
                     
                       
                         Q 
                         11 
                       
                       
                         Q 
                         10 
                       
                     
                     = 
                     
                       
                         
                           
                             V 
                             S 
                           
                            
                           C 
                            
                           
                               
                           
                            
                           2 
                            
                           
                               
                           
                            
                           nC 
                            
                           
                               
                           
                            
                           1 
                         
                         
                           
                             V 
                             S 
                           
                            
                           C 
                            
                           
                               
                           
                            
                           1 
                            
                           
                             ( 
                             
                               
                                 nC 
                                  
                                 
                                     
                                 
                                  
                                 1 
                               
                               + 
                               
                                 C 
                                  
                                 
                                     
                                 
                                  
                                 2 
                               
                             
                             ) 
                           
                         
                       
                       = 
                       
                         
                           nC 
                            
                           
                               
                           
                            
                           2 
                         
                         
                           
                             nC 
                              
                             
                                 
                             
                              
                             1 
                           
                           + 
                           
                             C 
                              
                             
                                 
                             
                              
                             2 
                           
                         
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     12 
                     ) 
                   
                 
               
             
           
         
       
     
       3. Charge-Driven vs Voltage-Driven Electrostatic Induction. 
       [0070]    Charge-Driven Electrostatic Induction has operating characteristics that differ from Voltage-Driven Electrostatic Induction, particularly when capacitors are used to supply the charge-drive. In some respects a capacitance-based Charge-Drive is analogous to a current source. There is a fixed amount of current available in a current drive and there is a fixed amount of trapped charge available in capacitance-based Charge-Drive. But, capacitance-based Charge-Drive has a unique problem in separating the charge. Charging a capacitor can yield equal and opposite charges in close proximity to each other. Even if the charges are large, if they are in close proximity to each other, their electric fields tend to cancel when we try to perform charge induction across a large insulation gap. Achieving charge separation and isolation is as important as achieving large charges. Electret devices use one method of achieving charge separation so charge-drive can be employed. This Invention uses a method to separate charge by the length of a stack of capacitors as its method. The method used in this invention can completely remove and return charge or can change polarity on command, while electret devices have a fixed polarity. 
       4. Speed of Operation (FIGS. 6,  7 ,  8  and  9 ). 
       [0071]    To obtain proper charge separation using low voltage sources requires capacitive stacks with electrodes numbering in the hundreds. If we charge them one electrode at a time, responding to high frequency signals becomes problematic. We seek a means by which we can charge several at a time but, still obtain a series charge arrangement in the stack. We choose to organize the electrodes in groups of three, with a first, middle and third electrode in each group as illustrated in  FIGS. 2   a,    2   b  and  2   c.  We stack the groups on top of each other and connect all the first electrodes to a first common switching circuit, all the second electrodes to a second common switching circuit and all third electrodes to a third common switching circuit. The outer electrodes each have their own, independent switching circuit. A common voltage source(s) powers the entire stack. The two step charge sequence for a three electrode capacitor is applied as described in  1   a ),  1   b ),  1   c ), above, except that when the first step is performed, n/3 first electrodes and n/3 second electrodes are charged simultaneously. When the second step is performed, n/3 second electrodes and n/3 third electrodes are charged simultaneously. In this manner, a hundred or more electrodes can be charged in three steps and with high speed switching, a multi-layer stack can track high speed signals up to ⅓ the frequency of the switches. 
         [0072]    We now detail how this charging system will work. In the first step, all first set electrodes ( 2 O 1 ,  2 I 3 ,  2 I 6 ) are connected to source voltage +V S  and all second set electrodes ( 2 I 1 ,  2 I 4 ,  2 I 7 ) are connected to −V S  as per  FIG. 6 . All third set electrodes ( 2 I 2 ,  2 I 5 ,  2 I 8 ) are left floating. As a secondary effect, we see additional cross talk charge on the bottom surface of each second electrode. This charge is reduced by the separation caused by floating third set electrodes. We, then, trap the charge on all first and second electrodes. We trap the charge on the second set of electrodes fractionally before trapping charge on the first set of electrodes. In the second step, as per  FIG. 7 , we connect all second set electrodes to +V S  and all third set electrodes to −V S . All first set electrodes are left floating with trapped charge in place. Positive charge is induced on all second set electrodes and equal and opposite negative charge is induced in all third set electrodes. The parasitic negative charge on all second electrodes is eliminated. No new parasitic charges are induced on the third electrodes because the only available electrodes are floating and unable to acquire or remove charge. The charge on the third set of electrodes is trapped fractionally before trapping the charge on trapping charge on the second set of electrodes. In the third step as per  FIG. 8 , all third set electrodes are connected to +V S  and all first set electrodes, with the exclusion of  2 O 1  and the inclusion of  2 O 2  are connected to −V S . Again, the charge on the first set of electrodes is trapped fractionally before the charge on the third set of electronics is trapped. The charging detail described above shows 8 internal electrodes and two external electrodes but, it applies for many more so long as they can be connected in three groups plus an independent  2 O 1  and  2 O 2 . 
       D. Expected Prototype Performance 
       [0073]    From eq. (13) above we can determine the effective voltage that can be applied to induce electrostatic charge in the Target Conductor and electric energy in the air/vacuum gap C 1 . 
         [0000]    
       
         
           
             
               
                 
                   Gain 
                   = 
                   
                     
                       
                         Q 
                         11 
                       
                       
                         Q 
                         10 
                       
                     
                     = 
                     
                       
                         
                           
                             V 
                             S 
                           
                            
                           C 
                            
                           
                               
                           
                            
                           2 
                            
                           
                               
                           
                            
                           nC 
                            
                           
                               
                           
                            
                           1 
                         
                         
                           
                             V 
                             S 
                           
                            
                           C 
                            
                           
                               
                           
                            
                           1 
                            
                           
                             ( 
                             
                               
                                 nC 
                                  
                                 
                                     
                                 
                                  
                                 1 
                               
                               + 
                               
                                 C 
                                  
                                 
                                     
                                 
                                  
                                 2 
                               
                             
                             ) 
                           
                         
                       
                       = 
                       
                         
                           nC 
                            
                           
                               
                           
                            
                           2 
                         
                         
                           
                             nC 
                              
                             
                                 
                             
                              
                             1 
                           
                           + 
                           
                             C 
                              
                             
                                 
                             
                              
                             2 
                           
                         
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     13 
                     ) 
                   
                 
               
             
           
         
       
     
         [0074]    Performance is measured as increased voltage across an insulation gap. We will assume the gap to be air or vacuum for our performance estimates. 
         [0075]    We choose 3M embedded capacitance material C1011 [5] for our dielectric material between electrodes. This material is 0.00043 in thick with dielectric constant of 20. It has a dielectric strength of 3300 volt/mil and is tested to over 100 volts DC. This calculates to 1419 volts dielectric strength for our 0.43 mil thick layers. We assume operating voltages of 400 volts (+/−200 v using push pull operation). The dielectric layer is coated by copper 0.0015 in thick. The actual thickness of a capacitor is 0.00043+0.0015×2=0.00343 in. Of this, only 0.00043 in is used for separating the positive charges, which is critical to electrostatic induction. We expect we can reduce the copper thickness to 0.0005 in without any adverse effects, especially where multi-layer construction is employed as in our case. 
         [0076]    For our case, we wish to penetrate an air/vacuum gap of 0.030 in. (typical for motor or noncontact energy transfer between moving joints). This means C 2 /C 1 =20(0.030)/0.00043=1395.3488372093 
         [0000]    This makes our Gain 
         [0000]    
       
         
           
             
               
                 
                   Gain 
                   = 
                   
                     
                       n 
                        
                       
                           
                       
                        
                       1395.3488372093 
                     
                     
                       n 
                       + 
                       1395.3488372093 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     14 
                     ) 
                   
                 
               
             
           
         
       
     
         [0077]    We want n as large as possible. We try 200=n. This provides a gain of 174.927113702624 to 1. Using multiple layers means increasing device thickness so we must now address this concern. For n layers of dielectric, we use n+1 electrodes. 
         [0000]      ( n+ 1)T L   =T   D   eq. (15)
 
         [0000]      (200+1)(0.00093 in)=0.18693 in.= T   D  (total device thickness) 
         [0000]    eq. (16) 
         [0078]    We choose V S =200 volts, we obtain the electrostatic induction effects of 35 KV. Using V S =±200 volts in a push pull configuration we obtain the electrostatic induction effects of 70 KV. We do not expect electric discharge to be a problem, 70 KV over 0.030 in is equivalent to 2.333 KV per mil. As stated earlier, the C1011 dielectric material has a dielectric strength of 3.3 KV per mil. In the event discharge does become a problem, source voltage can be lowered. 
       E. Applications 
       [0079]    An electrostatic induction system that can produce large electric fields over air or vacuum insulation gaps on the order of 0.030 in, has applications for motors, generators and power transfer units. These applications typically require magnetic induction across a 0.030 in air gap (because these applications involve two objects, moving with respect to each other and involving rolling bearings and the safe clearance allowed in this circumstance is on the order of 0.030 in). An electrostatic induction system that can produce large electric fields over air or vacuum gaps can also be applied where electrets had been previously used, such as electrostatic microphones and oscillating power generators or motors. 
       1. Motor Application Using Motion Transverse to E-Field 
       [0080]    A motor application will now be described whereby a moveable, charge neutral conductor moves transverse to the E-Field projected into the air/vacuum gap C 1 , according to  FIGS. 10   a.  and  10   b.  This applies to rotary electrostatic motors where rotor moves transverse to an electric field and to linear actuators supported by low friction bearings which prevent the slide from sticking to the walls. Projecting an electric field in an air/vacuum gap C 1  and inducing charge in a remote structure beyond 1, stores electrical in the C 1  gap, according to  FIG. 10   a,  and a moveable member (labeled  6 ) moves transverse to the electric field and removes electrical energy from C 1 , by providing an easier path across C 1 . The rate of change of stored energy with respect to transverse motion of moveable member  6  determines the force on the moveable member. When the projected E-Field is collapsed, as in  FIG. 10   b,  the force is removed. This type of motor can use poles or it can act as solenoid with limited movement and is analogous to electromagnetic pole motor and solenoid devices. 
         [0081]    Energy stored in field reduces as capacitance of moving member increases. We want the amount of energy in an air gap. The force is the rate of change of energy in the air gap. The energy is 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       ( 
                       
                         
                           C 
                           11 
                         
                         + 
                         
                           C 
                           12 
                         
                         + 
                         
                           C 
                           13 
                         
                       
                       ) 
                     
                      
                     V 
                   
                   = 
                   
                     
                       Q 
                       1 
                     
                     = 
                     
                       
                         
                           V 
                           S 
                         
                          
                         
                           ( 
                           
                             
                               C 
                               1 
                             
                             + 
                             
                               C 
                               2 
                             
                           
                           ) 
                         
                       
                       = 
                       const 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     17 
                     ) 
                   
                 
               
             
             
               
                 
                   V 
                   = 
                   
                     
                       Q 
                       1 
                     
                     / 
                     
                       ( 
                       
                         
                           C 
                           11 
                         
                         + 
                         
                           C 
                           12 
                         
                         + 
                         
                           C 
                           13 
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     18 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     
                       1 
                       2 
                     
                      
                     
                       ( 
                       
                         
                           C 
                           11 
                         
                         + 
                         
                           C 
                           12 
                         
                       
                       ) 
                     
                      
                     
                       V 
                       2 
                     
                   
                   = 
                   
                     
                       E 
                       G 
                     
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         stored 
                          
                         
                             
                         
                          
                         energy 
                          
                         
                             
                         
                          
                         in 
                          
                         
                             
                         
                          
                         air 
                          
                         
                             
                         
                          
                         gap 
                       
                       ) 
                     
                   
                 
               
               
                 
                   
                     [ 
                     6 
                     ] 
                   
                    
                   
                     q 
                     . 
                     
                         
                     
                      
                     
                       ( 
                       19 
                       ) 
                     
                   
                 
               
             
           
         
       
     
       1a). Linear Motor 
       [0082]        dE   G   /dX={right arrow over (F)}   X   =V ( dV/dX )( C   11   +C   12 )+(½) V   2 ( dC   11   /dX+dC   12   /dX )  [7] eq. (20)
 
       Where: 
       [0083]        C   11 =ε 0   A   11   /d   11   , C   12 =ε 0   A   12   /d   12   , A   11   =WX, A   12   =W ( X   0   −X )
 
         [0000]        dC   11   /dX =ε 0   W/d   11   , dC   12   /dX =ε 0   W (−1)/ d   12 
 
       And: 
       [0084]    
       
         
           
             
               
                  
                 V 
               
               
                  
                 X 
               
             
             = 
             
               
                 
                   
                     ( 
                     
                       
                         C 
                         11 
                       
                       + 
                       
                         C 
                         12 
                       
                     
                     ) 
                   
                   
                     - 
                     1 
                   
                 
                  
                 
                   
                      
                     
                       Q 
                       1 
                     
                   
                   
                      
                     X 
                   
                 
               
               + 
               
                 
                   
                     Q 
                     1 
                   
                    
                   
                     ( 
                     
                       - 
                       1 
                     
                     ) 
                   
                 
                  
                 
                   
                     ( 
                     
                       
                         C 
                         11 
                       
                       + 
                       
                         C 
                         12 
                       
                     
                     ) 
                   
                   
                     - 
                     2 
                   
                 
                  
                 
                   ( 
                   
                     
                       
                          
                         
                           C 
                           11 
                         
                       
                       
                          
                         X 
                       
                     
                     + 
                     
                       
                          
                         
                           C 
                           12 
                         
                       
                       
                          
                         X 
                       
                     
                   
                   ) 
                 
               
             
           
         
       
       
         
           
             
               
                  
                 
                   Q 
                   1 
                 
               
               
                  
                 X 
               
             
             = 
             
               0 
                
               
                   
               
                
               
                 ( 
                 
                   because 
                    
                   
                       
                   
                    
                   trapped 
                    
                   
                       
                   
                    
                   charge 
                    
                   
                       
                   
                    
                   
                     Q 
                     1 
                   
                    
                   
                       
                   
                    
                   is 
                    
                   
                       
                   
                    
                   constant 
                 
                 ) 
               
             
           
         
       
     
       So: 
       [0085]        dV/dX=−Q   1 ( C   11   +C   12 ) −2 (ε 0   W )(1/ d   11 −1/ d   12 )
 
       Where: 
       [0086]    
       
         
           
             
               
                 C 
                 11 
               
               = 
               
                 
                   
                     ɛ 
                     0 
                   
                    
                   WX 
                 
                 
                   d 
                   11 
                 
               
             
             , 
             
               
 
             
              
             
               
                 C 
                 12 
               
               = 
               
                 
                   
                     ɛ 
                     0 
                   
                    
                   
                     W 
                      
                     
                       ( 
                       
                         L 
                         - 
                         X 
                       
                       ) 
                     
                   
                 
                 
                   d 
                   12 
                 
               
             
             , 
             
               
 
             
              
             
               L 
               = 
               Constant 
             
           
         
       
     
       So: 
       [0087]        dV/dX=−V   S ( C   1   +C   2 )(ε 0   W ) −2 (1/ d   11 −1/ d   12 ) −2 (ε 0   W )(1/ d   11 −1/ d   12 )
 
       Where: 
       [0088]        Q   1   =V   S ( C   1   +C   2 ) 
         [0000]    This simplifies to: 
         [0000]        dV/dX=−V   S ( C   1   +C   2 )(ε 0   W ) −1 (1/ d   11 −1/ d   12 ) −1 
 
         [0000]    We plug this into eq. 20 resulting in: 
         [0000]        {right arrow over (F)}   X   =−VV   S ( C   1   +C   2 )(ε 0   W ) −1 (1/ d   11 −1/ d   12 ) −1 +0.5  V   2 (ε 0   W )(1/ d   11 −1/ d   12 )  eq. (21)
 
       1b) Rotary Motor 
       [0089]    We will now examine the rotary motor case. 
         [0000]    
       
         
           
             
               
                 
                   E 
                   = 
                   
                     
                       1 
                       / 
                       2 
                     
                      
                     
                         
                     
                      
                     
                       CV 
                       2 
                     
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         energy 
                          
                         
                             
                         
                          
                         stored 
                          
                         
                             
                         
                          
                         in 
                          
                         
                             
                         
                          
                         air 
                          
                         
                             
                         
                          
                         gap 
                       
                       ) 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     22 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     
                       
                          
                         E 
                       
                       
                          
                         θ 
                       
                     
                     = 
                     
                       
                         F 
                          
                         
                           
                             a 
                             -&gt; 
                           
                           θ 
                         
                       
                       = 
                       
                         
                           
                             
                               V 
                               2 
                             
                             2 
                           
                            
                           
                             
                                
                               C 
                             
                             
                                
                               θ 
                             
                           
                         
                         + 
                         
                           CV 
                            
                           
                             
                                
                               V 
                             
                             
                                
                               θ 
                             
                           
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     Where 
                      
                     
                       : 
                     
                   
                 
               
               
                 
                   
                     [ 
                     8 
                     ] 
                   
                    
                   
                       
                   
                    
                   
                     eq 
                     . 
                     
                         
                     
                      
                     
                       ( 
                       23 
                       ) 
                     
                   
                 
               
             
             
               
                 
                   C 
                   = 
                   
                     
                       C 
                       11 
                     
                     + 
                     
                       C 
                       12 
                     
                     + 
                     
                       
                         C 
                         2 
                       
                       n 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     24 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     
                       
                         ( 
                         
                           
                             C 
                             11 
                           
                           + 
                           
                             C 
                             12 
                           
                           + 
                           
                             
                               C 
                               2 
                             
                             n 
                           
                         
                         ) 
                       
                        
                       V 
                     
                     = 
                     
                       
                         V 
                         S 
                       
                        
                       
                         C 
                         2 
                       
                        
                       
                           
                       
                        
                       
                         ( 
                         constant 
                         ) 
                       
                        
                       
                           
                       
                        
                       and 
                     
                   
                    
                   
                       
                   
                    
                   
                     
 
                   
                    
                   
                     
                       
                         C 
                         2 
                       
                       n 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     constant 
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     25 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     
                       
                         
                            
                           
                             ( 
                             
                               
                                 C 
                                 11 
                               
                               + 
                               
                                 C 
                                 12 
                               
                               + 
                               
                                 
                                   C 
                                   2 
                                 
                                 n 
                               
                             
                             ) 
                           
                         
                         
                            
                           θ 
                         
                       
                        
                       V 
                     
                     + 
                     
                       
                         
                            
                           V 
                         
                         
                            
                           θ 
                         
                       
                        
                       
                         ( 
                         
                           
                             C 
                             11 
                           
                           + 
                           
                             C 
                             12 
                           
                           + 
                           
                             
                               C 
                               2 
                             
                             n 
                           
                         
                         ) 
                       
                     
                   
                   = 
                   0 
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     26 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     
                       
                          
                         C 
                       
                       
                          
                         θ 
                       
                     
                      
                     V 
                   
                   = 
                   
                     
                       - 
                       
                         
                            
                           V 
                         
                         
                            
                           θ 
                         
                       
                     
                      
                     C 
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     27 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     F 
                      
                     
                       
                         a 
                         -&gt; 
                       
                       θ 
                     
                   
                   = 
                   
                     
                       
                         
                           
                             V 
                             2 
                           
                           2 
                         
                          
                         
                           
                              
                             C 
                           
                           
                              
                             θ 
                           
                         
                       
                       - 
                       
                         
                           V 
                           2 
                         
                          
                         
                           
                              
                             C 
                           
                           
                              
                             θ 
                           
                         
                       
                     
                     = 
                     
                       
                         - 
                         
                           
                             V 
                             2 
                           
                           2 
                         
                       
                        
                       
                         
                            
                           C 
                         
                         
                            
                           θ 
                         
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     28 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     
                        
                       T 
                     
                      
                     
                       
                         a 
                         -&gt; 
                       
                       θ 
                     
                   
                   = 
                   
                     R 
                      
                     
                        
                       F 
                     
                      
                     
                       
                         a 
                         -&gt; 
                       
                       θ 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     29 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     
                       
                          
                         F 
                       
                       
                          
                         R 
                       
                     
                      
                     
                        
                       R 
                     
                   
                   = 
                   
                     
                       
                         - 
                         
                           
                             V 
                             2 
                           
                           2 
                         
                       
                        
                       
                         
                           
                              
                             2 
                           
                            
                           C 
                         
                         
                           
                              
                             θ 
                           
                            
                           
                              
                             R 
                           
                         
                       
                        
                       
                          
                         R 
                       
                     
                     = 
                     
                        
                       
                         F 
                          
                         
                           
 
                         
                         ( 
                         
                           V 
                            
                           
                               
                           
                            
                           is 
                            
                           
                               
                           
                            
                           considered 
                            
                           
                               
                           
                            
                           invariant 
                            
                           
                               
                           
                            
                           over 
                            
                           
                               
                           
                            
                           R 
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     30 
                     ) 
                   
                 
               
             
             
               
                 
                   
                      
                     T 
                   
                   = 
                   
                     R 
                      
                     
                        
                       F 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     31 
                     ) 
                   
                 
               
             
           
         
       
     
         [0000]    We now perform steps to determine dF. We begin by determining C. 
         [0000]    
       
         
           
             
               
                 
                   C 
                   = 
                   
                     
                       
                         
                           ε 
                           0 
                         
                         
                           d 
                           11 
                         
                       
                        
                       
                         
                           ∫ 
                           
                             
                               R 
                                
                               
                                   
                               
                                
                               1 
                             
                             , 
                           
                           
                             R2 
                             , 
                           
                         
                          
                         
                           
                             ∫ 
                             0 
                             θ 
                           
                            
                           
                             R 
                              
                             
                                 
                             
                              
                             
                                
                               θ 
                             
                              
                             
                                 
                             
                              
                             
                                
                               R 
                             
                           
                         
                       
                     
                     + 
                     
                       
                         
                           ɛ 
                           0 
                         
                         
                           d 
                           12 
                         
                       
                        
                       
                         
                           ∫ 
                           
                             
                               R 
                                
                               
                                   
                               
                                
                               1 
                             
                             , 
                           
                           
                             
                               R 
                                
                               
                                   
                               
                                
                               2 
                             
                             , 
                           
                         
                          
                         
                           
                             ∫ 
                             θ 
                             
                               θ 
                               0 
                             
                           
                            
                           
                             R 
                              
                             
                                 
                             
                              
                             
                                
                               θ 
                             
                              
                             
                                 
                             
                              
                             
                                
                               R 
                             
                           
                         
                       
                     
                     + 
                     
                       
                         C 
                         2 
                       
                       n 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     32 
                     ) 
                   
                 
               
             
             
               
                 
                   OR 
                    
                   
                     : 
                   
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   C 
                   = 
                   
                     
                       
                         
                           ε 
                           0 
                         
                         
                           d 
                           11 
                         
                       
                        
                       θ 
                        
                       
                         
                           ∫ 
                           
                             R 
                              
                             
                                 
                             
                              
                             1 
                           
                           
                             R 
                              
                             
                                 
                             
                              
                             2 
                           
                         
                          
                         
                           R 
                            
                           
                               
                           
                            
                           
                              
                             R 
                           
                         
                       
                     
                     + 
                     
                       
                         
                           ε 
                           0 
                         
                         
                           d 
                           11 
                         
                       
                        
                       
                         ( 
                         
                           
                             θ 
                             0 
                           
                           - 
                           θ 
                         
                         ) 
                       
                        
                       
                         
                           ∫ 
                           
                             R 
                              
                             
                                 
                             
                              
                             1 
                           
                           
                             R 
                              
                             
                                 
                             
                              
                             2 
                           
                         
                          
                         
                           R 
                            
                           
                               
                           
                            
                           
                              
                             R 
                           
                         
                       
                     
                     + 
                     
                       
                         C 
                         2 
                       
                       n 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     33 
                     ) 
                   
                 
               
             
           
         
       
     
       Resulting in: 
       [0090]    
       
         
           
             
               
                 
                   
                     
                       
                         
                            
                           2 
                         
                          
                         C 
                       
                       
                         
                            
                           θ 
                         
                          
                         
                            
                           R 
                         
                       
                     
                     = 
                     
                       
                         
                           
                             ε 
                             0 
                           
                           
                             d 
                             11 
                           
                         
                          
                         R 
                       
                       - 
                       
                         
                           
                             ɛ 
                             0 
                           
                           
                             d 
                             12 
                           
                         
                          
                         R 
                       
                     
                   
                   , 
                   
                     
                       where 
                        
                       
                           
                       
                        
                       
                         
                            
                           
                             
                               C 
                               2 
                             
                             n 
                           
                         
                         
                            
                           θ 
                         
                       
                     
                     = 
                     0 
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     34 
                     ) 
                   
                 
               
             
           
         
       
     
         [0000]    Substituting the results of eq. (34) into eq. (20) results in: 
         [0000]    
       
         
           
             
               
                 
                   
                      
                     F 
                   
                   = 
                   
                     
                       - 
                       
                         
                           V 
                           2 
                         
                         2 
                       
                     
                      
                     
                       ( 
                       
                         
                           
                             
                               ε 
                               0 
                             
                             
                               d 
                               11 
                             
                           
                            
                           R 
                         
                         - 
                         
                           
                             
                               ɛ 
                               0 
                             
                             
                               d 
                               12 
                             
                           
                            
                           R 
                         
                       
                       ) 
                     
                      
                     
                        
                       R 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     35 
                     ) 
                   
                 
               
             
           
         
       
     
         [0000]    Substituting the results of eq. (35) into eq. (31) results in: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                        
                       T 
                     
                     = 
                     
                       
                         R 
                          
                         
                            
                           F 
                         
                       
                       = 
                       
                         
                           - 
                           
                             
                               V 
                               2 
                             
                             2 
                           
                         
                          
                         
                           
                             ɛ 
                             0 
                           
                            
                           
                             ( 
                             
                               
                                 1 
                                 
                                   d 
                                   11 
                                 
                               
                               - 
                               
                                 1 
                                 
                                   d 
                                   12 
                                 
                               
                             
                             ) 
                           
                         
                          
                         
                           R 
                           2 
                         
                          
                         
                            
                           R 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     So 
                      
                     
                       : 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     36 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     T 
                     → 
                   
                   = 
                   
                     
                       
                         ∫ 
                         
                           R 
                            
                           
                               
                           
                            
                           1 
                         
                         
                           R 
                            
                           
                               
                           
                            
                           2 
                         
                       
                        
                       
                           
                       
                        
                       
                          
                         T 
                       
                     
                     = 
                     
                       
                         - 
                         
                           
                             V 
                             2 
                           
                           2 
                         
                       
                        
                       
                         
                           ɛ 
                           0 
                         
                          
                         
                           ( 
                           
                             
                               1 
                               
                                 d 
                                 11 
                               
                             
                             - 
                             
                               1 
                               
                                 d 
                                 12 
                               
                             
                           
                           ) 
                         
                       
                        
                       
                         
                           R 
                           3 
                         
                         3 
                       
                        
                       
                         
                           d 
                           → 
                         
                         z 
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     37 
                     ) 
                   
                 
               
             
           
         
       
     
         [0091]    We know that V is a function of θ so we calculate V for the angle we are considering, using known design parameters and eq. 39. We then substitute the value for V back into eq 38 to calculate torque. 
         [0000]    From eq. (33) we have: 
         [0000]    
       
         
           
             
               
                 
                   C 
                   = 
                   
                     
                       
                         
                           ε 
                           0 
                         
                         
                           d 
                           11 
                         
                       
                        
                       θ 
                        
                       
                         
                           ∫ 
                           
                             R 
                              
                             
                                 
                             
                              
                             1 
                           
                           
                             R 
                              
                             
                                 
                             
                              
                             2 
                           
                         
                          
                         
                           R 
                            
                           
                               
                           
                            
                           
                              
                             R 
                           
                         
                       
                     
                     + 
                     
                       
                         
                           ε 
                           0 
                         
                         
                           d 
                           11 
                         
                       
                        
                       
                         ( 
                         
                           
                             θ 
                             0 
                           
                           - 
                           θ 
                         
                         ) 
                       
                        
                       
                         
                           ∫ 
                           
                             R 
                              
                             
                                 
                             
                              
                             1 
                           
                           
                             R 
                              
                             
                                 
                             
                              
                             2 
                           
                         
                          
                         
                           R 
                            
                           
                               
                           
                            
                           
                              
                             R 
                           
                         
                       
                     
                     + 
                     
                       
                         C 
                         2 
                       
                       n 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     33 
                     ) 
                   
                 
               
             
           
         
       
     
         [0000]    This computes to: 
         [0000]    
       
         
           
             
               
                 
                   C 
                   = 
                   
                     
                       
                         ( 
                         
                           
                             
                               R 
                               2 
                               2 
                             
                             - 
                             
                               R 
                               1 
                               2 
                             
                           
                           2 
                         
                         ) 
                       
                        
                       
                         ( 
                         
                           
                             
                               ɛ 
                               0 
                             
                              
                             
                               θ 
                               
                                 d 
                                 11 
                               
                             
                           
                           + 
                           
                             
                               ɛ 
                               0 
                             
                              
                             
                               
                                 
                                   θ 
                                   0 
                                 
                                 - 
                                 θ 
                               
                               
                                 d 
                                 12 
                               
                             
                           
                         
                         ) 
                       
                     
                     + 
                     
                       
                         C 
                         2 
                       
                       n 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     38 
                     ) 
                   
                 
               
             
           
         
       
     
         [0000]    We are working with a fixed amount of trapped Charge V S C 2  which will distribute itself between C 11  and C 12  as per: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       V 
                       ( 
                       
                         
                           
                             R 
                             2 
                             2 
                           
                           - 
                           
                             R 
                             1 
                             2 
                           
                         
                         2 
                       
                       ) 
                     
                      
                     
                       ( 
                       
                         
                           
                             ɛ 
                             0 
                           
                            
                           
                             θ 
                             
                               d 
                               11 
                             
                           
                         
                         + 
                         
                           
                             ɛ 
                             0 
                           
                            
                           
                             
                               
                                 θ 
                                 0 
                               
                               - 
                               θ 
                             
                             
                               d 
                               12 
                             
                           
                         
                         + 
                         
                           
                             C 
                             2 
                           
                           n 
                         
                       
                       ) 
                     
                   
                   ≅ 
                   
                     
                       V 
                       S 
                     
                      
                     
                       C 
                       2 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     39 
                     ) 
                   
                 
               
             
           
         
       
     
         [0000]    Thus we can calculate V for any θ using eq. (39) and can substitute that V into eq. (37) to determine {right arrow over (T)}. 
         2 . Motor Application Using Motion Parallel to the E-Field 
       [0092]    A motor application will now be described whereby a moveable, grounded electrical conductor  1  moves to increase or decrease the air/vacuum gap according to  FIGS. 5   a  and  5   b.  This applies to oscillation type motors, to electrets microphone type devices and to energy conversion devices (mechanical to electrical or electrical to mechanical). When a moveable grounded electrical conductor moves to reduce the size of an air/vacuum gap as per  FIG. 5   a,  the stored electrical energy in the gap is reduced and force is applied to the conductor  1  proportional to the rate of change of energy stored in gap C 1  divided by rate of change of gap size. When the E-Field is collapsed according to  FIG. 5   b,  the force on  1  is removed and the moveable conductor  1  is free to return to its starting position, possibly by spring return. 
         [0000]    
       
         
           
             
               
                 
                   
                       
                   
                    
                   
                     
                       
                         V 
                         X 
                       
                        
                       
                         ( 
                         
                           
                             C 
                             X 
                           
                           + 
                           
                             
                               C 
                               2 
                             
                             n 
                           
                         
                         ) 
                       
                     
                     = 
                     
                       
                         
                           V 
                           S 
                         
                          
                         
                           ( 
                           
                             
                               C 
                               0 
                             
                             + 
                             
                               C 
                               2 
                             
                           
                           ) 
                         
                       
                       = 
                       const 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     40 
                     ) 
                   
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     
                       C 
                       X 
                     
                     = 
                     
                       
                         ɛ 
                         0 
                       
                        
                       
                         A 
                         X 
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     41 
                     ) 
                   
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     
                       
                         V 
                         X 
                       
                        
                       
                         ( 
                         
                           
                             
                               
                                 ɛ 
                                 0 
                               
                                
                               A 
                             
                             X 
                           
                           + 
                           
                             
                               C 
                               2 
                             
                             n 
                           
                         
                         ) 
                       
                     
                     = 
                     
                       
                         V 
                         S 
                       
                        
                       
                         ( 
                         
                           
                             C 
                             0 
                           
                           + 
                           
                             C 
                             2 
                           
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     42 
                     ) 
                   
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     
                       V 
                       X 
                     
                     = 
                     
                       
                         
                           V 
                           S 
                         
                          
                         
                           ( 
                           
                             
                               C 
                               0 
                             
                             + 
                             
                               C 
                               2 
                             
                           
                           ) 
                         
                       
                       
                         ( 
                         
                           
                             
                               
                                 ɛ 
                                 0 
                               
                                
                               A 
                             
                             X 
                           
                           + 
                           
                             
                               C 
                               2 
                             
                             n 
                           
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     43 
                     ) 
                   
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     E 
                     = 
                     
                       
                         1 
                         2 
                       
                        
                       
                         ( 
                         
                           
                             C 
                             X 
                           
                           + 
                           
                             
                               C 
                               2 
                             
                             n 
                           
                         
                         ) 
                       
                        
                       
                         V 
                         X 
                         2 
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     44 
                     ) 
                   
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     
                       
                         F 
                         → 
                       
                       X 
                     
                     = 
                     
                       
                         
                            
                           E 
                         
                         
                            
                           X 
                         
                       
                       = 
                       
                         
                           
                             
                               V 
                               X 
                               2 
                             
                             2 
                           
                            
                           
                             
                                
                               
                                 ( 
                                 
                                   
                                     C 
                                     X 
                                   
                                   + 
                                   
                                     
                                       C 
                                       2 
                                     
                                     n 
                                   
                                 
                                 ) 
                               
                             
                             
                                
                               X 
                             
                           
                         
                         + 
                         
                           
                             ( 
                             
                               
                                 C 
                                 X 
                               
                               + 
                               
                                 
                                   C 
                                   2 
                                 
                                 n 
                               
                             
                             ) 
                           
                            
                           
                             V 
                             X 
                           
                            
                           
                             
                                
                               
                                 V 
                                 X 
                               
                             
                             
                                
                               X 
                             
                           
                         
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     45 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     
                        
                       
                         
                           V 
                           X 
                         
                          
                         
                           ( 
                           
                             
                               C 
                               X 
                             
                             + 
                             
                               
                                 C 
                                 2 
                               
                               n 
                             
                           
                           ) 
                         
                       
                     
                     
                        
                       X 
                     
                   
                   = 
                   
                     
                       
                         
                           V 
                           X 
                         
                          
                         
                           
                              
                             
                               ( 
                               
                                 
                                   C 
                                   X 
                                 
                                 + 
                                 
                                   
                                     C 
                                     2 
                                   
                                   n 
                                 
                               
                               ) 
                             
                           
                           
                              
                             X 
                           
                         
                       
                       + 
                       
                         
                           ( 
                           
                             
                               C 
                               X 
                             
                             + 
                             
                               
                                 C 
                                 2 
                               
                               n 
                             
                           
                           ) 
                         
                          
                         
                           
                              
                             
                               V 
                               X 
                             
                           
                           
                              
                             X 
                           
                         
                       
                     
                     = 
                     0 
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     46 
                     ) 
                   
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     
                       
                         V 
                         X 
                       
                        
                       
                         
                            
                           
                             ( 
                             
                               
                                 C 
                                 X 
                               
                               + 
                               
                                 
                                   C 
                                   2 
                                 
                                 n 
                               
                             
                             ) 
                           
                         
                         
                            
                           X 
                         
                       
                     
                     = 
                     
                       
                         - 
                         
                           ( 
                           
                             
                               C 
                               X 
                             
                             + 
                             
                               
                                 C 
                                 2 
                               
                               n 
                             
                           
                           ) 
                         
                       
                        
                       
                         
                            
                           
                             V 
                             X 
                           
                         
                         
                            
                           X 
                         
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     47 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     
                       F 
                       → 
                     
                     X 
                   
                   = 
                   
                     
                       
                         
                           
                             V 
                             X 
                             2 
                           
                           2 
                         
                          
                         
                           
                              
                             
                               ( 
                               
                                 
                                   C 
                                   X 
                                 
                                 + 
                                 
                                   
                                     C 
                                     2 
                                   
                                   n 
                                 
                               
                               ) 
                             
                           
                           
                              
                             X 
                           
                         
                       
                       - 
                       
                         
                           V 
                           X 
                           2 
                         
                          
                         
                           
                              
                             
                               ( 
                               
                                 
                                   C 
                                   X 
                                 
                                 + 
                                 
                                   
                                     C 
                                     2 
                                   
                                   n 
                                 
                               
                               ) 
                             
                           
                           
                              
                             X 
                           
                         
                       
                     
                     = 
                     
                       
                         - 
                         
                           
                             V 
                             X 
                             2 
                           
                           2 
                         
                       
                        
                       
                         
                            
                           
                             ( 
                             
                               
                                 C 
                                 X 
                               
                               + 
                               
                                 
                                   C 
                                   2 
                                 
                                 n 
                               
                             
                             ) 
                           
                         
                         
                            
                           X 
                         
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     48 
                     ) 
                   
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     
                       F 
                       → 
                     
                     = 
                     
                       
                         - 
                         
                           
                             
                               1 
                               2 
                             
                             [ 
                             
                               
                                 
                                   V 
                                   S 
                                 
                                  
                                 
                                   ( 
                                   
                                     
                                       C 
                                       0 
                                     
                                     + 
                                     
                                       C 
                                       2 
                                     
                                   
                                   ) 
                                 
                               
                               
                                 ( 
                                 
                                   
                                     
                                       
                                         ɛ 
                                         0 
                                       
                                        
                                       A 
                                     
                                     X 
                                   
                                   + 
                                   
                                     
                                       C 
                                       2 
                                     
                                     n 
                                   
                                 
                                 ) 
                               
                             
                             ] 
                           
                           2 
                         
                       
                        
                       
                         ɛ 
                         0 
                       
                        
                       
                         A 
                         X 
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     49 
                     ) 
                   
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     E 
                     = 
                     
                       
                         1 
                         2 
                       
                        
                       
                         
                           
                             ( 
                             
                               
                                 
                                   ɛ 
                                   0 
                                 
                                  
                                 
                                   A 
                                   X 
                                 
                               
                               + 
                               
                                 
                                   C 
                                   2 
                                 
                                 n 
                               
                             
                             ) 
                           
                           [ 
                           
                             
                               
                                 V 
                                 S 
                               
                                
                               
                                 ( 
                                 
                                   
                                     C 
                                     0 
                                   
                                   + 
                                   
                                     C 
                                     2 
                                   
                                 
                                 ) 
                               
                             
                             
                               ( 
                               
                                 
                                   
                                     
                                       ɛ 
                                       0 
                                     
                                      
                                     A 
                                   
                                   X 
                                 
                                 + 
                                 
                                   
                                     C 
                                     2 
                                   
                                   n 
                                 
                               
                               ) 
                             
                           
                           ] 
                         
                         2 
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     50 
                     ) 
                   
                 
               
             
           
         
       
     
       3. Power Transfer Application 
       [0093]    A Power Transfer application, according to  FIGS. 12   a  and  12   b  will now be described. When an E-Field is projected into an air/vacuum gap C 3 , opposite charge is induced in conductor  3  according to  FIG. 12   a.  This charge comes from the grounded structure to electrode  3  through switch SD 1 , while switch SD 2  is left open to isolate activities on electrode  3  from electrodes  1 L and  2 L. Also, switches S 2 Ig, S 22   g,  S 2 I and S 1 I are left open to leave electrodes  1 L and  2 L isolated from load (Z L ) and from ground. At the same time, switches SO 1 P, SO 1 N, SO 1   g  and SO 2   g  are left open to facilitate trapping charge on electrode  2 O 1  to power the charge induction on electrode  3 . When the E-Field is removed, according to  FIG. 12   b,  the induced charge on electrode  3  experiences forces of dispersion. Switches SD 2  and SI 1   g  are, then, closed and SD 1  is opened to allow the electrode  3  charge to disperse to electrode  2 L and to attract equal and opposite charge on  1 L. The capacitance between  1 L and  2 L is large to maximize the charge on  1 L and  2 L and to minimize the charge remaining on electrode  3 . This process can be continued for several cycles in a charge pumping action until charge on electrodes  1 L and  2 L are equal to the peak charge on electrode  3 . When sufficient charge is accumulated on electrodes  1 L and  2 L, this stored charge can be used to power load Z L . As shown in  FIG. 12 , the load is powered by positive charge when S 1 Ig is opened, S 1 I is closed and  2 Ig is closed. 
         [0094]    The load is powered by negative charge when S 2 Ig is opened, S 2 I is closed to load and S 1 I is opened and S 1 Ig is closed. When discharge through Z L  is completed, the system can be reconfigured to begin charging again. 
         [0095]    With this introduction we will introduce the equations for predicting performance and providing design guidance for specific device applications. We will, first estimate the amount of charge that can be induced across an air/vacuum gap followed by the power that can be transferred across the gap. The power transfer function equations are similar to those used for a motor application where movement is in the direction of the electric field except that there is no motion and the capacitance of the air/vacuum gap is constant and electrostatic force across the air/vacuum gap is not a factor. 
         [0000]    
       
         
           
             
               
                 
                   
                       
                   
                    
                   
                     
                       
                         
                           V 
                           X 
                         
                          
                         
                           ( 
                           
                             
                               C 
                               1 
                             
                             + 
                             
                               
                                 C 
                                 2 
                               
                               n 
                             
                           
                           ) 
                         
                       
                       = 
                       
                         
                           
                             V 
                             S 
                           
                            
                           
                             ( 
                             
                               
                                 C 
                                 1 
                               
                               + 
                               
                                 C 
                                 2 
                               
                             
                             ) 
                           
                         
                         = 
                         const 
                       
                     
                      
                     
                       
 
                     
                      
                     
                         
                     
                      
                     
                       
                         
                           V 
                           S 
                         
                          
                         
                           ( 
                           
                             
                               C 
                               1 
                             
                             + 
                             
                               C 
                               2 
                             
                           
                           ) 
                         
                       
                       = 
                       
                         charge 
                          
                         
                             
                         
                          
                         trapped 
                          
                         
                             
                         
                          
                         on 
                          
                         
                             
                         
                          
                         201 
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     51 
                     ) 
                   
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     
                       C 
                       1 
                     
                     = 
                     
                       
                         ɛ 
                         0 
                       
                        
                       
                         A 
                         X 
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     52 
                     ) 
                   
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     
                       
                         V 
                         X 
                       
                        
                       
                         ( 
                         
                           
                             
                               
                                 ɛ 
                                 0 
                               
                                
                               A 
                             
                             X 
                           
                           + 
                           
                             
                               C 
                               2 
                             
                             n 
                           
                         
                         ) 
                       
                     
                     = 
                     
                       
                         V 
                         S 
                       
                        
                       
                         ( 
                         
                           
                             C 
                             1 
                           
                           + 
                           
                             C 
                             2 
                           
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     53 
                     ) 
                   
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     
                       V 
                       X 
                     
                     = 
                     
                       
                         
                           V 
                           S 
                         
                          
                         
                           ( 
                           
                             
                               C 
                               1 
                             
                             + 
                             
                               C 
                               2 
                             
                           
                           ) 
                         
                       
                       
                         ( 
                         
                           
                             
                               
                                 ɛ 
                                 0 
                               
                                
                               A 
                             
                             X 
                           
                           + 
                           
                             
                               C 
                               2 
                             
                             n 
                           
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     54 
                     ) 
                   
                 
               
             
             
               
                 
                   E 
                   = 
                   
                     
                       1 
                       2 
                     
                      
                     
                       ( 
                       
                         
                           C 
                           1 
                         
                         + 
                         
                           
                             C 
                             2 
                           
                           n 
                         
                       
                       ) 
                     
                      
                     
                       V 
                       X 
                       2 
                     
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         total 
                          
                         
                             
                         
                          
                         electric 
                          
                         
                             
                         
                          
                         energy 
                          
                         
                             
                         
                          
                         stored 
                          
                         
                             
                         
                          
                         in 
                          
                         
                             
                         
                          
                         system 
                       
                       ) 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     55 
                     ) 
                   
                 
               
             
             
               
                 
                   E 
                   = 
                   
                     
                       1 
                       2 
                     
                      
                     
                       ( 
                       
                         C 
                         1 
                       
                       ) 
                     
                      
                     
                       V 
                       X 
                       2 
                     
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         electric 
                          
                         
                             
                         
                          
                         energy 
                          
                         
                             
                         
                          
                         stored 
                          
                         
                             
                         
                          
                         in 
                          
                         
                             
                         
                          
                         
                           air 
                           / 
                           vacuum 
                         
                          
                         
                             
                         
                          
                         gap 
                       
                       ) 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     56 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     Q 
                     1 
                   
                   = 
                   
                     
                       V 
                       X 
                     
                      
                     
                       C 
                       X 
                     
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         electric 
                          
                         
                             
                         
                          
                         charge 
                          
                         
                             
                         
                          
                         induced 
                          
                         
                             
                         
                          
                         across 
                          
                         
                             
                         
                          
                         
                           air 
                           / 
                           vacuum 
                         
                          
                         
                             
                         
                          
                         gap 
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     57 
                     ) 
                   
                 
               
             
           
         
       
     
       3a). Dispersion 
       [0096]    With V X  set to zero, Q 1  disperses and seeks the nearest ground. It has two choices, C 1  and C 2  and it will prefer C 2  by a C 2 /C 1  ratio. 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         V 
                         X 
                       
                        
                       
                         C 
                         1 
                       
                     
                     + 
                     
                       
                         V 
                         X 
                       
                        
                       
                         C 
                         2 
                       
                     
                   
                   = 
                   
                     
                       Q 
                       1 
                     
                     = 
                     
                       
                         
                           V 
                           X 
                         
                          
                         
                           ( 
                           
                             
                               C 
                               1 
                             
                             + 
                             
                               C 
                               2 
                             
                           
                           ) 
                         
                       
                       = 
                       
                         
                           V 
                           X 
                         
                          
                         
                           
                             C 
                             1 
                           
                            
                           
                             ( 
                             
                               1 
                               + 
                               
                                 
                                   C 
                                   2 
                                 
                                 
                                   C 
                                   1 
                                 
                               
                             
                             ) 
                           
                         
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     58 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     
                       Q 
                       
                         2 
                          
                         
                             
                         
                          
                         L 
                       
                     
                     = 
                     
                       
                         V 
                         X 
                       
                        
                       
                         C 
                         2 
                       
                     
                   
                   , 
                   
                     
                       Q 
                       11 
                     
                     = 
                     
                       
                         
                           V 
                           X 
                         
                          
                         
                           C 
                           1 
                         
                       
                       = 
                       
                         
                           Q 
                           1 
                         
                         - 
                         
                           
                             Q 
                             
                               2 
                                
                               
                                   
                               
                                
                               L 
                             
                           
                            
                           
                             ( 
                             
                               
                                 C 
                                 2 
                               
                                
                               
                                 C 
                                 1 
                               
                             
                             ) 
                           
                         
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     59 
                     ) 
                   
                 
               
             
           
         
       
     
         [0000]    Equation 59 shows Q 2L &gt;&gt;Q 11  so most of the charge on electrode  1  moves to electrode  2 L during dispersion, with multiple pumping steps not needed. This means nearly all the charge is transferred to  1 L and  2 L during each step and a relatively large current can be supplied to power Z L  on a continuous basis. 
       3b). Power 
       [0097]    We start with the energy stored in a capacitor with electrodes  1 L and  2 L and examine how much power can be supplied to a load with this stored energy. We also examine how fast we can resupply the stored energy and how much sustained power can be delivered. 
         [0000]    
       
         
           
             
               
                 
                   E 
                   = 
                   
                     
                       
                         Q 
                         2 
                       
                       
                         2 
                          
                         
                             
                         
                          
                         
                           C 
                           2 
                         
                       
                     
                     = 
                     
                       
                         
                           ( 
                           
                             
                               V 
                               X 
                             
                              
                             
                               C 
                               1 
                             
                           
                           ) 
                         
                         2 
                       
                       
                         2 
                          
                         
                             
                         
                          
                         
                           C 
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     60 
                     ) 
                   
                 
               
             
           
         
       
     
         [0098]    This capacitor must cycle between being charged and supplying current to Z L  and the speed of this cycle determines the rate of energy, or power, transferred. 
       Cycle Sequence 
       [0000]    
       
         1. The Stack is charged and charge is induced in electrode  1 . 
         2. The Stack is discharged and the electric field from the Stack is collapsed. Simultaneously, charge on electrode  1  moves to electrode  1 L, with equal and opposite charge simultaneous induced in  2 L by capacitance. 
         3. Electric power is supplied to the load from energy stored in the capacitor using electrodes  1 L and  2 L and simultaneously and independently electrode  1  is charged as per step  1 . 
         4. Repeat step 2. 
         5. Repeat step 3. 
       
     
         [0104]    The cycle sequence uses two steps, but one of the steps requires charging the Stack, which requires three steps. Thus, in effect, we have a four step for each burst of energy that is supplied to the load. We choose a step frequency so we can achieve an ac load frequency of 25% of the step frequency. This suggests we speed up the drive frequency by a factor of four to obtain suitable power transfer frequency. Power transfer frequencies on the order of a ghz seem possible. 
       4. Generator Applications 
       [0105]    Generation of electricity can be achieved in a reverse application of the electrostatic motors described in  1   a  and  1   b  above. In the generator application, an electric field is maintained in the air or vacuum gap by a charged stack of capacitors, the field is periodically changed by using mechanical power, induced ac electrical power is induced in the process and that induced electrical power is stored to be used as needed. The charged stack of capacitors maintains the electric field in the air or vacuum gap without external electrical power because it has charge trapped in place so we have a method and device for converting mechanical power to electrical energy or an electrostatic generator. 
         [0106]    The charged Stack of Capacitors stores electrical energy in the air or vacuum gap as per eq. 56. 
         [0000]    
       
         
           
             
               
                 
                   E 
                   = 
                   
                     
                       1 
                       2 
                     
                      
                     
                       ( 
                       
                         C 
                         1 
                       
                       ) 
                     
                      
                     
                       V 
                       X 
                       2 
                     
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         electric 
                          
                         
                             
                         
                          
                         energy 
                          
                         
                             
                         
                          
                         stored 
                          
                         
                             
                         
                          
                         in 
                          
                         
                             
                         
                          
                         
                           air 
                           / 
                           vacuum 
                         
                          
                         
                             
                         
                          
                         gap 
                       
                       ) 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     56 
                     ) 
                   
                 
               
             
           
         
       
     
         [0000]    4a). Electrostatic Generators in which the Moving Member moves in a direction transverse to the electric field which stores energy in the air or vacuum gap.
 
4a1). Rotary Case
 
         [0107]    For a rotary electrostatic generator, mechanical power can be used to rotate a moving member as through an air or vacuum gap as shown in  FIG. 10   a.  As shown in  FIG. 10   a,  the electrical energy stored in the air or vacuum gap is disturbed as the Moving Member passes through it. This physically changes C 1  as per eq. 55 and, with it, E and V X . When the Moving Member leaves the gap, E, C 1  V X  return to their original values and when the Moving Member re-enters the air or vacuum gap, the values of E, C 1  and V X  return to load values. These conditions can be physically changed in a periodic manner to produce ac induced current and this current, in turn, can be stored as electrical energy to be applied as needed. The frequency of the induced ac current is limited only by the physical rotation speed of the Moving Member. The output frequency of the stored electrical energy can be faster than Moving Member rotation rate. For example the stored electrical energy can be used to power an oscillator at a higher frequency and this oscillator can serve as the source of the output ac current. 
         [0000]    4a2). Linear Case 
         [0108]    A linear electrostatic generator can use mechanical power to oscillate back and forth through a region where the air or vacuum gap contains stored electric energy. This motion serves to generate induced ac current and stored electrical energy similar to that done by a rotary electrostatic generator. The circumstances and effects of  FIG. 10   a  apply as do the effects as per eq. 56. 
         [0000]    4b. Electrostatic Generators in which the Moving Member moves in a direction parallel to the electric field which stores energy in the air or vacuum gap. 
         [0109]    In this application, a grounded Moving Member is moved by mechanical force in an air or vacuum gap with electric energy stored therein, with the movement parallel to the direction of the electric field supplying the electric energy in the gap. The movement in one direction causes the air or vacuum gap to decrease and the movement in the opposite direction causes the gap to increase as per  FIGS. 11   a  and  11   b.  When the gap is decreased, electrical energy is taken from the gap and when the gap is increased, electric energy is added to the gap. The electric energy that is taken from the gap is stored external to the Electrostatic Generator hardware in a capacitor with the resultant electric field polarity opposite to the polarity of the electric field in the air or vacuum gap. The electric energy that is added to the air or vacuum gap induces a stored energy in a capacitor with the resultant field polarity in the direction of the polarity of the electric field stored in the air or vacuum gap. The circumstances and effects of  FIGS. 11   a  and  11   b  apply as do the effects as per eq. 56 except that mechanical force and work is used to generate electrical energy rather than using electrical energy to cause mechanical force and motion. Equations 40 through 50 also apply except that force and displacement are to be interpreted as inputs and changes in electrical energy are to be interpreted as outputs. 
       4c). Charge-Driven Electrostatic Induction Generators in Energy Scavenger Applications 
       [0110]    The Energy Scavenger application, according to  FIGS. 15   a,    15   b,    16  and  9 , induces charge on the Charged Stack ground electrode  202  when the Target Conductor  1  moves closer to Charged Stack electrode  201  and removes charge from electrode  202  when Target Conductor  1  moves away from electrode  201 . Thus, charge on electrode  202  increases and decreases with the in and out motion of Target Conductor  1 . The Energy Scavenger application uses a one-way diode between ground and  202  such that charge, originating from ground, can pass through the one-way diode to  202  during charge increase but, cannot return during periods of charge decrease. A second one-way diode allows charge to leave  202  by dispersion en route to a capacitor C 5  ( FIG. 13   a ) where it attracts equal and opposite charge from ground and stores energy in C 5 . This energy is available for use when switch S 1  is closed ( FIG. 13   a ) and C 5  can recharged when S 1  is opened. The limiting factor is when the charge stored in C 5  has voltage equal to the dispersion voltage on  202 . At this point, the excess charge on  202  cannot leave and the energy harvesting ceases. Using the stored energy keeps the energy harvesting process going.  FIG. 13   a  shows a situation where negative charge is stored on  202  but, positive charge can be used by reversing the polarity of the diodes ( FIG. 13   a ) in combination with using negative trapped charge on  201  ( FIG. 9 ). Diodes are passive components and the trapped charge stored in the capacitive stack (labeled  4  in  FIG. 1 ) so the entire energy scavenger system need not require external power and can remain in a sleep mode until sufficient energy is stored in C 5 . At this point the stored energy can be used to activate and operate electronic and electromechanical systems. 
       4d). Deformable Charge-Driven Electrostatic Induction Generators in Energy Scavenger Applications 
       [0111]    The Energy Scavenger application ( FIGS. 15   a,    15   b,    16  and  9 ) can take advantage of flexible structures to generate electrical energy. For example, articles of clothing can use human activity to generate electrical energy and power electrical and electromechanical devices. Such articles of clothing can be constructed in layers which are flexible in bending, with a thin insulation gap between the outer layers and inner layers such that the insulation gap is changes in response to wearer activity. The outer layer must be electrically conductive so it will act as a moveable target conductor, but the conductor can be very thin and can be embedded in the clothing material such that the clothing material can perform full function in its clothing role. The insulation gap must be capable of storing electrical energy, must be capable of allowing motion of the target conductor relative to the inner layers and must be capable of elastically restoring the target conductor to its original position when the external forces are removed. The inner layers would comprise a composite of stacked ultra-thin capacitors and clothing material such that inner layer composite can elastically bend, with full range of motion, and can perform both its clothing and generator functions, without hampering user motion. The stack of ultra-thin capacitors must be thin enough that it remains flexible in bending, but it must provide sufficient separation between the charge trapped on the outer layer nearest the insulation gap and the charge on the outer layer nearest the wearers&#39; body. This can be compensated by reducing the thickness of the insulation layer, so enough additional trapped charge is being attracted to the moveable target conductor to compensate the loss of charge separation. Wearable Charge-Driven Electrostatic Generators require the dielectric film between the electrodes in the stack be able to withstand bending without cracking. This can be accomplished in a number of ways: 1. A stiff dielectric material can be used in regions where wearer movement requires limited bending with significant compression, 2. A composite of stiff and flexible binder can be used, 3. A flexible dielectric material can be used with reduced dielectric capabilities. 
       5. Sensor Applications 
       [0112]    A Charge-Driven Electrostatic generator can function as a sensor. As shown in 4. GENERATOR APPLICATIONS, a Charge-Driven Electrostatic generator can provide stored electric energy in an insulation gap between an outer electrode on the generator and a grounded conductive moving member, with power off. When the generator moving member moves in response to an external time variable mechanical force, the electric energy stored in the gap is disturbed and the charge on the grounded outer electrode of the generator changes in a time variable manner. When a load is inserted between ground and the outer electrode, time varying electric power is driven through the load or storage device. The time varying electrical power going through the load produces a time varying voltage across the load and current through the load that can be sensed in frequency, phase and amplitude and information about the nature of the mechanical force driving the moving member is provided. The result is a sensor which measures the behavior of the mechanical force acting on the moving member. 
       a). Microphone Application 
       [0113]    When the moving member is a diaphragm being driven by sound, we have a microphone and this microphone is analogous to an electret microphone with some notable advantages. An electret is permanently polarized with a fixed surface charge. This charge attracts foreign elements with opposite charge which tend to degrade performance over time. A Charge-Driven Electrostatic Induction sensor can simply change the charge every so often and clean the offending foreign elements by repulsion. Also, current electret technology provides a limited ε R  so, the surface charge available is limited, which limits microphone performance as well. Embedded capacitor technology has a much better energy storage capability with an ε R =20 and dielectric thicknesses as thin as 0.00043 in. [5]. This means that large amounts of electrical energy can be stored in a stack of capacitors using modest voltage sources. This also means that surface charge trapped on the outer electrode nearest the diaphragm can be large and microphone performance improved as a result. Another advantage for Charge-Driven Electrostatic Induction microphones are in their ability to measure signal at the grounded outer electrode. The charge on the grounded outer electrode changes with diaphragm movement so measurements can be taken by inserting a measuring system between the grounded outer electrode and ground. The electronics in such a measuring system would be relatively stationary and would be in a protected position. Diaphragm requirements for Charge-Driven Electrostatic Induction microphones and electret microphones are similar and are well within the state of the art. A passive sleep mode (as described in section 4c) above) can be used to reduce or eliminate battery requirements for a Charge-Driven Electrostatic Induction microphone. The method as applied to a microphone would be similar to that described in section 4c, but with application specific adjustments. 
         [0000]    5a1). Sensing Method 
         [0114]    An application is illustrated, according to  FIGS. 13   a,    13   b  and  14  where a diaphragm (or moveable target conductor) moves and its mechanical oscillation is converted to electrical energy and where this electrical energy is manifested by charge addition or subtraction on either electrode  2 O 2  or the moveable target conductor  1 . In  FIGS. 13   a,    13   b  and  14  we illustrate the case where the charge addition and charge subtraction on electrode  2 O 2  is directed by switches SO 2 G and SO 3 G to terminate on a driven ground [9], with this termination resulting in amplified voltage output according to  FIG. 14 . This is a charge breathing cycle in which charge is inhaled from ground through SO 3 G to terminate on the driven ground. Attaching the switches SO 2 G and SO 3 G and driven ground termination to 1 (moveable target conductor) provides the same result as attaching same to electrode  2 O 2 . 
         [0000]    5a2). Sensing Method Using Passive Electric Components 
         [0115]    An application is illustrated, according to  FIGS. 15   a,    15   b  and  16  where electrical energy can be stored in a capacitor  701  while the microphone is in a sleep or passive mode. When sufficient energy is stored in  701 , the switching system, SO 2 G and SO 3 G can be activated along with the driven ground termination and voltage output and the system can be operated like 5a1). This provides a method and apparatus to eliminate or reduce battery requirements. For continuous microphone operation without a battery, the input mechanical power supplied as electrical power to capacitor  701  must be equal to or greater than the electrical signal output power plus the electrical power needed to operate the electronics (including switches and op-amps) for continuous operation. Otherwise, operation must be intermittent. 
         [0000]    5a3). Sensing Method Options 
         [0116]    There is also the option of using 5a1) and 5a2) in combination whereby the device is operated in sleep mode until sufficient energy is accumulated to operate in active mode. 
       5b). Sensor Options 
       [0117]    Any force or energy source which can disturb the electrostatic energy stored in the insulation gap can be sensed and measured, particularly if it is time variable. Some of these energy sources mechanically act on the moveable member. Some of these act on the moveable member indirectly through an intermediate member. Some of these operate on the electric energy stored in the insulation gap without moving the moveable member. 
         [0000]    5b1) Sensors Using Mechanical Energy to Move the Moveable Member
       (a) Direct mechanical force can move the moving member and disturb the electric energy stored in the insulation gap. Strain gauges, pressure gauges, weight measuring devices, accelerometers, etc.   (b) Indirect mechanical energy can be used to move the moveable member through an intermediate medium. In this way, temperature changes can be measured by introducing an expandable gas container between the moveable member and the heat source. The heat source expands the gas and the container expands with it. The container movement moves the moveable member and this movement is measured. In the process, temperature is measured.
 
5b2) Sensors Using Sensing Methods that do not Move the Moveable Member
   Energy sources can operate on the electric energy stored in the insulation gap without moving the moveable member. When electric energy is stored in the insulation gap, charge is trapped on the nearest Charge-Driven Electrostatic Induction outer electrode and opposite charge is induced on the grounded moveable member. When the moveable member is disconnected from ground the charge induced by the outer electrode is trapped on the moveable member. When a time varying electric field is introduced on the moveable member, a back time varying voltage is induced which affects the electric energy and can be measured by the Charge-Driven Electrostatic Induction sensor.       
 
       F. Nuances and Error Approximations 
       [0121]    In this section we examine the error that attends the assumption of net zero charge on the internal electrodes. We find that there is a slight net charge on each of the internal electrodes and that this net charge progressively increases as we go down the stack of electrodes. On the other hand, we also find that the error is not great and the net charge helps performance so we are left with confidence that our assumption is good to &lt;5% for the applications described. 
         [0122]    The argument in this detailed description assumes that when an interior electrode is charged positive, the negative charge on that interior electrode is held in place by the trapped positive charge on the previously charged electrode. This is an approximation and more accurately we can expect that most, but not all of the negative charge will be held in place. (The argument applies for negative charge voltage as well.) With the first set of 3 electrodes (1 outer and 2 inner), we see the first signs of error in our approximation. In charging the 2 inner electrodes, we raise the effective voltage of the outer electrode nearest the target conductor. This results in more trapped positive charge on the outer electrode coupling with the target conductor and less coupling with the nearest inner electrode. At the same time, the other surface of the effected inner electrode receives a full positive charge, so we have a slight net positive charge on the first inner electrode rather than the net zero charge used in our approximation. When we charge the next pair of inner electrodes, we raise the voltage of the outer electrode again and divert more trapped positive charge to couple with the target conductor. Again, less negative charge is held in place and again the net charge on this new inner electrode is slightly positive, this time slightly more so. This continues as the entire stack of electrodes is charged. With each charge sequence, we gather more and more net positive charge. 
         [0123]    We begin by examining the loss of negative charge in the first set of 3 electrodes, outer electrode  2 O 1  and inner electrodes  2 I 1  and  2 I 2 . We examine the loss of negative charge on  2 I 1  when  2 I 1  and  2 I 2  are charged. 
         [0124]    With our initial charge sequence, we apply V S  to electrode  2 O 2  and ground electrode  2 I 1 . With the target conductor also grounded we have positive charge induced on both surface of  2 O 1 . Most of the charge is between  2 O 1  and  2 I 1  because C 2 &gt;&gt;C 1 . 
         [0000]        V   S ( C 1+ C 2)= Q   2O1   , V   2O1   =V   S  Charge trapped on 2O1  (eq. 60)
 
         [0000]      V S C1=Q 10  (original charge across air gap)  (eq. 61)
 
       And 
       [0125]      V S C2=Q 11  (original charge induced between 2O1 and 2I1)  (eq. 62)
 
         [0000]    We trap the positive charge on  2 O 1  and apply V S  to  2 I 1  with  2 I 2  grounded. This causes the voltage potential of  2 O 1  to be raised above V S . 
         [0000]      ( V   2O1   −V   S ) C 2 +V   2O1   C 1= Q   2O1   =V   S ( C 1+ C 2)  (eq. 63)
 
       With 
       [0126]      ( V   2O1   −V   S ) C 2=negative charge held in place on 2I1  (eq. 64)
 
       And 
       [0127]      V 201 C1=charge now induced across air gap to target conductor (eq. 65) 
         [0000]    We rearrange eq. 63 to solve for V 2O1 . 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         V 
                         201 
                       
                        
                       
                         ( 
                         
                           
                             C 
                              
                             
                                 
                             
                              
                             1 
                           
                           + 
                           
                             C 
                              
                             
                                 
                             
                              
                             2 
                           
                         
                         ) 
                       
                     
                     = 
                     
                       
                         
                           V 
                           S 
                         
                          
                         
                           ( 
                           
                             
                               2 
                                
                               
                                   
                               
                                
                               C 
                                
                               
                                   
                               
                                
                               2 
                             
                             + 
                             
                               C 
                                
                               
                                   
                               
                                
                               1 
                             
                           
                           ) 
                         
                       
                       = 
                       
                         
                           V 
                           S 
                         
                          
                         
                           ( 
                           
                             
                               C 
                                
                               
                                   
                               
                                
                               1 
                             
                             + 
                             
                               C 
                                
                               
                                   
                               
                                
                               2 
                             
                             + 
                             
                               C 
                                
                               
                                   
                               
                                
                               2 
                             
                           
                           ) 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   Or 
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     66 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     V 
                     201 
                   
                   = 
                   
                     
                       V 
                       S 
                     
                      
                     
                       ( 
                       
                         1 
                         + 
                         
                           
                             C 
                              
                             
                                 
                             
                              
                             2 
                           
                           
                             
                               C 
                                
                               
                                   
                               
                                
                               1 
                             
                             + 
                             
                               C 
                                
                               
                                   
                               
                                
                               2 
                             
                           
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     67 
                     ) 
                   
                 
               
             
           
         
       
     
       Resulting in: 
       [0128]    
       
         
           
             
               
                 
                   
                     
                       
                         V 
                         201 
                       
                        
                       C 
                        
                       
                           
                       
                        
                       1 
                     
                     = 
                     
                       
                         
                           V 
                           S 
                         
                          
                         
                           ( 
                           
                             1 
                             + 
                             
                               
                                 C 
                                  
                                 
                                     
                                 
                                  
                                 2 
                               
                               
                                 
                                   C 
                                    
                                   
                                       
                                   
                                    
                                   1 
                                 
                                 + 
                                 
                                   C 
                                    
                                   
                                       
                                   
                                    
                                   2 
                                 
                               
                             
                           
                           ) 
                         
                       
                        
                       C 
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       
                         ( 
                         
                           new 
                            
                           
                               
                           
                            
                           induced 
                            
                           
                               
                           
                            
                           charge 
                            
                           
                               
                           
                            
                           across 
                            
                           
                               
                           
                            
                           air 
                            
                           
                               
                           
                            
                           gap 
                         
                         ) 
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                       
                   
                    
                   
                     And 
                      
                     
                       : 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     68 
                     ) 
                   
                 
               
             
             
               
                 
                   
                     
                       ( 
                       
                         
                           V 
                           201 
                         
                         - 
                         
                           V 
                           S 
                         
                       
                       ) 
                     
                      
                     C 
                      
                     
                         
                     
                      
                     2 
                   
                   = 
                   
                     
                       
                         V 
                         S 
                       
                        
                       
                         ( 
                         
                           
                             C 
                              
                             
                                 
                             
                              
                             2 
                           
                           
                             
                               C 
                                
                               
                                   
                               
                                
                               1 
                             
                             + 
                             
                               C 
                                
                               
                                   
                               
                                
                               2 
                             
                           
                         
                         ) 
                       
                     
                      
                     C 
                      
                     
                         
                     
                      
                     2 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         new 
                          
                         
                             
                         
                          
                         negative 
                          
                         
                             
                         
                          
                         charge 
                          
                         
                             
                         
                          
                         held 
                          
                         
                             
                         
                          
                         in 
                          
                         
                             
                         
                          
                         place 
                          
                         
                             
                         
                          
                         on 
                          
                         
                             
                         
                          
                         
                           2 
                           / 
                           1 
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     69 
                     ) 
                   
                 
               
             
           
         
       
     
       While: 
       [0129]      V S C2=(positive charge held in place on 2I1)  (eq. 70)
 
         [0000]    Subtracting (eq. 69) from (eq. 70) yields the net charge on  2 I 1  when  2 I 1  is disconnected from ground and the net charge is trapped. 
         [0000]    
       
         
           
             
               
                 
                   Net 
                    
                   
                       
                   
                    
                   charge 
                    
                   
                       
                   
                    
                   on 
                    
                   
                       
                   
                    
                   
                     2 
                     / 
                     1 
                   
                    
                   
                       
                   
                    
                   is 
                    
                   
                       
                   
                    
                   
                     V 
                     S 
                   
                    
                   C 
                    
                   
                       
                   
                    
                   2 
                    
                   
                       
                   
                    
                   
                     ( 
                     
                       1 
                       - 
                       
                         
                           C 
                            
                           
                               
                           
                            
                           2 
                         
                         
                           
                             C 
                              
                             
                                 
                             
                              
                             1 
                           
                           + 
                           
                             C 
                              
                             
                                 
                             
                              
                             2 
                           
                         
                       
                     
                     ) 
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     71 
                     ) 
                   
                 
               
             
           
         
       
     
         [0000]    We will apply some expected values to see what these equations tell us. We expect C 1  to be part of an air or vacuum gap typically 0.020 in. We expect C 2  to be part of a high capacitance material, such as 3M with an insulation gap of 0.00043 in and ε R =20. 
       So: 
       [0130]    
       
         
           
             
               
                 
                   
                     0.00043 
                     
                       0.03 
                       · 
                       20 
                     
                   
                   = 
                   
                     
                       
                         C 
                          
                         
                             
                         
                          
                         1 
                       
                       
                         C 
                          
                         
                             
                         
                          
                         2 
                       
                     
                      
                     
                         
                     
                      
                     0.0007166666667 
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     72 
                     ) 
                   
                 
               
             
           
         
       
     
         [0000]    And results using (eq. 71) might typically be on the order of: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       V 
                       S 
                     
                      
                     C 
                      
                     
                         
                     
                      
                     2 
                      
                     
                       ( 
                       
                         1 
                         - 
                         
                           1 
                           
                             1 
                             + 
                             0.0007166666667 
                           
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       V 
                       S 
                     
                      
                     C 
                      
                     
                         
                     
                      
                     2 
                      
                     
                       ( 
                       0.0007161534234 
                     
                   
                 
               
               
                 
                   eq 
                   . 
                   
                       
                   
                    
                   
                     ( 
                     73 
                     ) 
                   
                 
               
             
           
         
       
     
         [0131]    Net charge is positive, but almost zero (0.072% net+) If this trend continues on a linear basis, we have n 0.0007161534234 (net positive charge accumulated in the stack of electrodes. For n=200, we accumulate 0.14323068468% positive charge in the stack which we will neglect. We do not expect the relationship to be linear, preliminary indications are it is a mathematical series. A more proper solution would probably involve an electrostatic simulation. We conclude the approximation in which we assume charge trapped on electrode  2 O 1  arranges itself according to the relationship between C 1  and nC 2  is conservative. This assumes no net positive charge on the interior electrodes. The additional positive charge can only help as per Gauss&#39; Law of Charge. Our approximation looks good to &lt;5% for an air gap of 0.030 in and n=200. 
         [0132]    Although the invention has been described with reference to certain preferred embodiments, it will be appreciated that many other variations and modifications thereof may be devised in accordance with the principles disclosed herein. The invention, including the described embodiments and all variations and modifications thereof within the scope and spirit of the invention, is defined in the following claims.