thought experiment.  - Page 1
Forum section: electronics / general

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: December 03, 2012, 01:33:55 am
content: Take a variable capacitor.Charge it up.Then reduce the capacitance.Where does the energy go?Does the voltage increase to preserve E=CV2?

author: BravoV
author_rank: Super Contributor
timestamp: December 03, 2012, 01:50:28 am
content: Not an EE, but from math should be straight forward isn't it ? cmiiwE = C V2V = Sqrt (E / C) , if E is constant, then decreasing capacitance = more voltage.Assuming zero leakage of course.

author: Unknown
author_rank: Unknown
timestamp: December 03, 2012, 01:58:41 am
content: But should energy remain constant? Assume a parallel plate capacitor. Reducing the capacitance means increasing the distance between the plates (assuming constant surface and dielectric). This means work is performed that might increase or decrease the energy of the system. The work pulls the oppositely charged plates further apart (higher potential energy), and increases repulsion between charges on the plates (also higher energy), so I would say the energy in the capacitor increases: the electrons want out.I would rather approach it from the perspective of charge: Q=CV. The number of electrons on each of the plates is not going to change in our ideal capacitor, so as C goes down, V goes up. If C is reduced by a factor of two, than V is increased by the same factor two, and the energy is doubled.

author: BravoV
author_rank: Super Contributor
timestamp: December 03, 2012, 02:13:50 am
content: Quote from: alm on December 03, 2012, 01:58:41 amElectrical charge (Q) is not equal to energy, and you can not increase the "energy" just by decreasing the C, otherwise this will turned into perpetual free energy generating engine.

author: tom66
author_rank: Super Contributor
timestamp: December 03, 2012, 02:14:44 am
content: Well, a crappy circuit simulator I used once before I discovered LTSpice did this very same thing. It increased the voltage proportional to CV^2.

author: Unknown
author_rank: Unknown
timestamp: December 03, 2012, 02:25:24 am
content: Quote from: BravoV on December 03, 2012, 02:13:50 amI never stated that Q=E. Q is constant, so if C2 = C1 / 2, then V2 = V1 * 2 (Q = C1 * V1 = C2 * V2). E2 = C2 * V22= C1/2 * (V1*2)2= 2 * C1 * V12. This is not energy generation, since you performed work to pull the plates apart, or to turn the screw/knob of the variable capacitor. (Some of) this work is converted to potential energy in the capacitor.Lets assume your argument that E=C*V2should remain constant. C1 * V12= C2 * V22, so if C2 = C1 / 2, then V2 = V1 * sqrt(2). This means that Q2 = C2 * V2 = C1/2 * V1*sqrt(2) = C1 * V1 / sqrt(2) = Q1 / sqrt(2). Where did all these electrons go?

author: IanB
author_rank: Super Contributor
timestamp: December 03, 2012, 03:07:40 am
content: Quote from: mikeselectricstuff on December 03, 2012, 01:33:55 amYes, I would have to say the voltage increases. If the two plates are insulated and you move them apart then the charge differential between the plates doesn't change. If the capacitance reduces then the voltage difference between the plates must increase. At the limit you will get the voltage resulting from an isolated charged body in free space where you integrate the electric field to infinity to find the voltage.But as others have noted, a pair of oppositely charged capacitor plates have an attractive force between them due to the electric field, so it takes work to move them apart. This work will be stored as potential energy in the system (energy = force x distance), therefore the total energy of the system will increase as you move the plates further apart.

author: robrenz
author_rank: Super Contributor
timestamp: December 03, 2012, 03:34:42 am
content: What about when it is a rotary capacitor where the separation of the plates is fixed but the degrees of engagement vary? Isn't the charge going to have to actualy move to the smaller engaged surface area and therefore increase in density? Will that require a torque from electrical work (ignoring the mechanical frictions)

author: IanB
author_rank: Super Contributor
timestamp: December 03, 2012, 03:51:25 am
content: Yes, even if the capacitor has rotating interleaved plates, if you rotate the plates to decrease the overlap you will have to do mechanical work by rotating against an opposing torque. The is broadly the principle of electrostatic motors (but in reverse).

author: free_electron
author_rank: Super Contributor
timestamp: December 03, 2012, 05:32:46 am
content: Voltage will go up.You can actually do an electron density plot of the plate itself. If you have those old skool rotating caps where one plate rotates over the other ( plates are half moon ) then the charge density changes. Density is greatest where there is overlap. ADS can actually simulate such things. For very high frequency apps this is important as you get some fringe effect at the outlaying borders. Id have to talk to an ex collegue of mine that does satellite tuners (10ghz band). They did some simulations on variable caps where they used a field solver to pattern the charge dispersal in the plates.

author: TerraHertz
author_rank: Super Contributor
timestamp: December 03, 2012, 06:43:41 am
content: I think you are describing a VanDeGraff generator. Performing mechanical work, to carry charge further away (up the moving belt) from its initial location in a capacitor (where it was sprayed on the belt by a high local E field.) Just happens to be a continuous process as opposed to a once-off movement of a plate.

author: PA4TIM
author_rank: Super Contributor
timestamp: December 03, 2012, 08:32:41 am
content: Is the cap stil connected to the source ? In that case volage stays the same and the electric field decreases. Charge decreases because capacitance does. Electrons move back to generating source ( because in theory voltage goes up so the capacitor becomes a source itself for a moment)But without the source:But I do not think a conducter will hold it's charge very long ( if the dielectric is air)If you apply a voltage to it, the size, form distance and dielectric determine how strong the electric field will become. If you remove the voltage the field will collaps and the dieelectric ( an isolator) will probably hold some static charge. The collapsing field will induce current in a nearby conductor. If no conductor is in range energy is lost in the resistance of the space around it ( 370 Ohm if I remember well) you get one pulse.If you transmit on an antenna you have a similar situation. The antenna has capacitance between its legs and inductance in the length of wire. If the impedance is matched the collapsing fields cause EM waves that travell away from the antenna.If you move the legs closer to each other or further appart, the impedance changes and that causes part of the wave to reflect back to the source. The vswr. The impedance changes because the capacitance changes. Just like a variable capacitor.

author: G7PSK
author_rank: Super Contributor
timestamp: December 03, 2012, 08:47:18 am
content: Assuming that we have the perfect capacitor with zero leakage. Also the plates can be made to increase and decrease in size with no external influence. Then decreasing the capacity after charging will have to increase the voltage on them.

author: IanB
author_rank: Super Contributor
timestamp: December 03, 2012, 02:59:19 pm
content: Quote from: PA4TIM on December 03, 2012, 08:32:41 amRelated to that, this classic high school physics experiment is very interesting:http://www.lhup.edu/~dsimanek/scenario/e-stat.htmNote how you can take a capacitor apart and completely discharge the plates, then put it back together again. After you've put it back together, is it still charged?

author: PA4TIM
author_rank: Super Contributor
timestamp: December 03, 2012, 03:17:29 pm
content: This is what i mean. He tells it more clear. The insulator holds the charge ( if anything different is used as air)

author: G7PSK
author_rank: Super Contributor
timestamp: December 03, 2012, 04:05:55 pm
content: Ok what happens with an air gap capacitor or two plates in a vacuum. The dielectric cannot be holding the charge in this case but they still hold a charge. The leyden jar and VDG are using static so the plastic cup is taking a charge as well.It would be possible to charge the plastic cup and then add the metal plates after it would still discharge with a spark.

author: PA4TIM
author_rank: Super Contributor
timestamp: December 03, 2012, 04:46:04 pm
content: The disconnected plates will also hold some static charge ( the link also states it is not realy charging, you apply an electric field and removing the source without a dielectrium in between makes the field collaps) because the plates are isolated in that case. Like a car that is charged by airfriction.But In real this will be difficult because the plates are not floating. They are connected to something that holds them and that wil be an isolator ( like the tires of a car) . And that holds charge but even teflon has not an infinitive resistance so the charge will leak.It would be fun to try. Measuring the charge by use of an electrometer and make disconnectable plates. First place an EMF on the plates, then measure the field, remove the source and measure again, then remove the plates ( using a not charged isolator) and measure if the first isolators self are charged and also if the plates alone are charged. Then remove the plates from the temporary isolators and measure If they hold now a charge ( but it will be difficult i think because of leakage resistance)But the question was, what happens if you move plates closer so capacitance increases. Or further apart so C decreases. And I said that depends if the source is still connected or not. I assume it is disconnected. In that case capacitance changes so in that case Q=C x U. C changes, Q stays constant ( no source and ideal without leakage) and U changes by the amount to make the equation correct.

author: Colfaxmingo
author_rank: Contributor
timestamp: December 03, 2012, 04:47:57 pm
content: 1) assume a parallel plate capacitor with an inch thick Teflon (epsilon relative about 2.1) slab separating the plates2) deposit charge on the plates3) observe an electric field between the plates4) slide the slab out of the plates, (epsilon relative about 1)3) observe the electric field between the platesC = (epsilon_naught * epsilon_relative * A)/ dIn this case, epsilon_naught is a universal constant (cannot change), epsilon_relative is halved, Area does not change (same plates), and the distance is not change (1").Conclusion,The capacitance is halved when the Teflon slab is removed.Now, the definition of a capacitor:Q = C*Vwhere Q, the total charge in coulombs , and C is the capacitance inf Farads and V is voltage in volts.If the amount of charge Q is unchanged, and the capacitance doubles, the voltage must half.

author: G7PSK
author_rank: Super Contributor
timestamp: December 03, 2012, 05:00:51 pm
content: This sounds like an experiment for the space lab, the plates could then be free floating in a vacuum so no way the charge could leak.