Electrostatic force question - Page 1
Forum section: electronics / projects

author: fonograph
author_rank: Frequent Contributor
timestamp: April 21, 2018, 08:24:18 pm
content: We all know similiar charges repel each other due to electrostatic force.But how strong is this force per given voltage,surface area,and distance?My question is this : how many volts will two power supplies of same polarity need,to repel two electrodes with 1cm square surface area that are 1cm apart directly facing each other to produce repulsive force equal to 1 kg per square centimer,which is 1 atmosphere pressure?1. Ignore dielectric breakdown & field emission2. Dielectric constant is 1,assume there is vacuum or gas.

author: Marco
author_rank: Super Contributor
timestamp: April 21, 2018, 08:59:17 pm
content: http://physicstasks.eu/1535/force-acting-on-capacitor-plates

author: Cerebus
author_rank: Super Contributor
timestamp: April 21, 2018, 09:14:17 pm
content: Wrong initial question. It's not down to voltage, it's down to quantity of charge. The repulsive (attractive) force is given byCoulomb's law:Where F is the force, KeCoulomb's constant (8.9875×109N m2C−2), the q's are the two quantities of charge, and r is the distance between the two charges.The amount of charge you could get onto your plates is going to be determined by their capacitance, which in your case is 88.54 fF (femto Farads), and the voltage they are charged to. Air breaks down at around 3 MV/m, so your plates will spark over at 30 kV, at which point your capacitor will have a charge of q= CV, q = 88.54 fF x 30 kV = 2.65 nC (nano Coulombs).Calculating the force is left as an exercise for the student but as a clue I'll tell you that it is more than 1 billion times smaller than the force you are looking for.To achieve anything like the pressure you're looking for requiresmuchbigger plates and something with a higher dielectric constant and breakdown voltage than air between the plates.

author: chris_leyson
author_rank: Super Contributor
timestamp: April 21, 2018, 09:54:57 pm
content: http://www-eng.lbl.gov/~shuman/XENON/REFERENCES&OTHER_MISC/electric_forces.pdfI used equation 3 and got 2.6MV, assumed 1kg = 9.81N. Because the parallel plates are not infitnitely large there is a correction factor but I can't find a reference for it. Needless to say the forces are tiny but then again not that small because you can build practical electrostatic speakers and only need 100's of volts to drive them.

author: fonograph
author_rank: Frequent Contributor
timestamp: April 21, 2018, 10:05:02 pm
content: I disagree that it requires bigger plate.It doesnt matter if the plates are 1 kilometer long or 1 micrometer,for given voltage,the force per area aka pressure, is going to be always same.What you probably meant is total force,but I am not intersted in total,I am only intersted in force per area.I know about the air break down,still i appreciate that you warned me about it.For purposes of my question,completly ignore dielectric breakdown,field emission and other high voltage phenomena.As for dielectric constant,assume its in vacuum.

author: fonograph
author_rank: Frequent Contributor
timestamp: April 21, 2018, 10:14:55 pm
content: Quote from: chris_leyson on April 21, 2018, 09:54:57 pmThank you Chris,what do you mean by correction factor,I dont understand what it is,can you tell me what it is?

author: Cerebus
author_rank: Super Contributor
timestamp: April 21, 2018, 10:21:26 pm
content: Quote from: fonograph on April 21, 2018, 10:05:02 pmSorry I didn't realise you wanted to ignore the laws of physics in a physics question. Obviously, ignoring the laws of physics, the answer is elebenty jillion volts and a strawberry.Last time ya gets any help out of me...

author: fonograph
author_rank: Frequent Contributor
timestamp: April 21, 2018, 10:41:05 pm
content: Quote from: Cerebus on April 21, 2018, 10:21:26 pmWhy are you so triggered?I did nothing wrong,I liked your post as gesture of respect and gratefulness,I am happy you help me,no reason to be bitter.I wrote " I disagree",not "omg can you believe what this idiot wrote!? What a dumbass! I am so much smarter than him!".I dont see anything in my reply that would be considerd rude,disrespectful or ungrateful,if you dont want to help me in future,fine,but I dont act as if I did anything bad to you.If you cant handle that someone respectfully disagrees with you,then you are going to have hard time on internet.

author: Cerebus
author_rank: Super Contributor
timestamp: April 21, 2018, 10:59:11 pm
content: Quote from: fonograph on April 21, 2018, 10:41:05 pmI'm not triggered, I'm not some limp wristed millennial, I'm cross.You post a question that suggests you haven't a clue about the physics involved. I take the time to offer an answer that'll put you on the right path and between first post and then you've learned enough physics to tell me I'm wrong. If you know enough to know that I'm wrong you know enough to not need to ask the question in the first place. Then you add insult to injury by changing the rules and throwing away all of the rules of physics that affect a practical, accurate answer - no dielectric breakdown, no fringe effects, no higher order effects. If you know enough to know that those are issues then why are you asking a question that applying Coloumb's law and the formula for a basic plate capacitor will give you the answer for? To me that is just taking the piss, and if you can't figure out why that would annoy someone you've got much bigger problems than me being pissed off with you.

author: fonograph
author_rank: Frequent Contributor
timestamp: April 22, 2018, 12:12:16 am
content: Quote from: Cerebus on April 21, 2018, 10:21:26 pmA agree,I understand that from perspective of other person my question creates expectation that I "haven't a clue".The thing is,skill level in given field of knowledge isnt black and white,its alot more complex than that.For example,a teacher might know much more than student,but there might be one thing that student knows that the teacher doesnt know.Quote from: Cerebus on April 21, 2018, 10:21:26 pmI highly appreciate that you sacrificed your time to help,that is why I clicked SAY THANKS to give you thumb up for your effort.You claim I dismissed your post as "wrong",that is not completly true,I merely disagreed with one specific part.I agree with vast majority of your post.Furthermore,I in detail wrote why I disagree,your reply was emotional outburst devoided of any counter argumenting logic.If it is wrong,write how and why,if you present evidence to your argument,I will happily agree.I have no motive to be hard headed and stubbornly hold my position as unshakable truth.If I am proven to be wrong I will not feel defeated,I will feel enlightened and delighted that I got rid off wrong understanding that holded in the past,all I wish for is to learn the truth.Quote from: Cerebus on April 21, 2018, 10:21:26 pmThat is not true,my question was rather complex one,it requires multiple "pieces" of knowledge.The part I disagreed is about difference between force and pressure.Just becose I know what is difference between force and pressure does not mean I must know about anything else except definition of force and pressure.Just becose I disagreed on the force/pressure part does not mean I am master of electrostatic physics who is wasting time of people on internet by asking question which I already know.Quote from: Cerebus on April 21, 2018, 10:21:26 pmI appologize that I didnt clarified the dielectric constant and request to ignore electrical breakdown from the start.I forgot about it,as soon as I read your post I edited my original post and added this crucial information.That being said,I fail to see why this is such bad thing as you make it out to be.In my first reply to you I wrote "I know about the air break down,still i appreciate that you warned me about it." Ok,I forgot to clarify these things outright,then you tried to teach me thing I already knew,I wrote that I already know it,and... whats the issue there? You seem to have anger problems,very short temper.I am surprised you make big deal out of such little unimportant things,I would never expect anybody to get angry becose of this.

author: ejeffrey
author_rank: Super Contributor
timestamp: April 22, 2018, 02:51:34 am
content: Quote from: fonograph on April 21, 2018, 08:24:18 pmThis is not a meaningful question. The force (as you have described it), will always be attractive, never repulsive. You need a net positive (or net negative) charge on both plates. To get that, you need to take charges from somewhere else -- a third conductor (often called ground). The location and shape of that conductor, as well as the full shape of your plates -- especially the "outside" as that is where most of the charge will accumulate.

author: fonograph
author_rank: Frequent Contributor
timestamp: April 22, 2018, 03:17:04 am
content: Quote from: ejeffrey on April 22, 2018, 02:51:34 amWhy is it not meaningful question? Always attractive? Quote from my original post " how many volts will two power supplies of same polarity need,to repel two electrodes..." Two electrodes connected to two power supplies with same polarity are going to be "always attractive" how?You are describing the power supply side,which is irrelevant for the purpose of this question.

author: CopperCone
author_rank: Super Contributor
timestamp: April 22, 2018, 03:18:25 am
content: cant you bombard charges on with a electron beam?besides its a physics question, you can say there was a new universe formed which at time = 0 the plates with electron count of posters choice were in existence.

author: fonograph
author_rank: Frequent Contributor
timestamp: April 22, 2018, 03:24:43 am
content: Quote from: CopperCone on April 22, 2018, 03:18:25 amI think yes,the charge could for the purpose of this question as well landed from electron beam,or it simply was always there since begining of time.I am no expert,but I believe it makes no difference so its irrelevant how they got charged,the important thing is they are charged by specific amount and that is all that matters.

author: IanB
author_rank: Super Contributor
timestamp: April 22, 2018, 03:38:58 am
content: Quote from: fonograph on April 22, 2018, 03:17:04 amIf I take two power supplies each generating 1000 V DC and connect them back to back, negative to negative, positive to the electrodes, then the potential difference between the electrodes will be (+1000 V) - (+ 1000 V) = 0 V. With no potential difference between the electrodes, the force will be zero. If we change one of the power supplies to produce a different voltage, then there will be a potential difference between the electrodes, one more positive than the other, and they will attract each other.So as you can see, either there will be no force, or they will attract each other. They cannot repel each other.To achieve repulsion you need to consider a three body system, not a two body system as you have proposed.

author: fonograph
author_rank: Frequent Contributor
timestamp: April 22, 2018, 03:55:49 am
content: IanB Why there needs to be any potential difference between the plates? If both plates are charged by same amount and same polarity,they are going to repel.

author: IanB
author_rank: Super Contributor
timestamp: April 22, 2018, 04:08:14 am
content: Quote from: fonograph on April 22, 2018, 03:55:49 amYes, but you asked this question:Quote from: fonograph on April 21, 2018, 08:24:18 pmIf both plates are charged by the same amount and the same polarity, then each plate will have the same voltage, and therefore there will be no voltage difference between them, and therefore no force between them.Your question needs to be not about voltage, but about charge.

author: fonograph
author_rank: Frequent Contributor
timestamp: April 22, 2018, 04:33:05 am
content: Quote from: IanB on April 22, 2018, 04:08:14 am

author: ejeffrey
author_rank: Super Contributor
timestamp: April 22, 2018, 04:45:50 am
content: Quote from: fonograph on April 22, 2018, 03:17:04 amIt's not meaningful because the only parts you described aren't relevant, and you completely left out the things that are.The use of two power supplies is a red herring as well. You could just as easily do it with a single power supply.In particular, what matters is the shape of the *outside* of the conductors. There will be little to no free charge on the plates facing each other as there will be only a small electric field in the gap.Here is a problem that is easy to solve and similar to what you want:If you consider a pair of hemispheres separated by a thin gap, and each with charge Q/2, and assume that there are no nearby conductors (ground at infinity). For the moment, treat this as a single sphere with radius R and charge Q. The self capacitance of the sphere is C = 4*pi*eps_0*R, and the voltage (referenced to infinity) is V = q/C. The potential energy is U = 0.5 * C * V^2 = q^2/(2C). The pressure trying to "inflate" the sphere can be calculated from dU/dR / A where A is the surface area (4 * pi * R^2).If you work that out, you get: pressure = q^2 / (16 * pi^2 * eps_0 * R^4).So now, take that sphere and make a thin cut to return to the hemisphere. The force pushing the hemispheres apart is equal to the pressure times the cross sectional area (pi*R^2)F = q^2 / (16 * pi * eps_0 * R^2)q = 4*R * sqrt(pi * F * eps_0)V = q/(4 * pi * eps_0 * R)V = sqrt(F / pi * eps_0)This formula is only valid for hemispherical "plates", and only when the gap is much smaller than the radius of the sphere. Actually, it works just as well for a thin spherical shell -- what is inside the sphere doesn't matter at all. All that matters is that the outer shell is spherical and that the edge gap is small. You could do a similar calculation for other simple geometries, but this is the simplest.

author: fonograph
author_rank: Frequent Contributor
timestamp: April 22, 2018, 05:20:55 am
content: Quote from: ejeffrey on April 22, 2018, 04:45:50 amDescribe how are they not relevant,be specific.Show me the irrelevant part from my original post that you would omit without destroying the question.Quote from: ejeffrey on April 22, 2018, 04:45:50 amLike what for example? Show me single thing I left out that is essential for answering the question.Quote from: ejeffrey on April 22, 2018, 04:45:50 amI agree it was not necessary,not sure if I would describe it as red herring,it is not wrong though.Quote from: ejeffrey on April 22, 2018, 04:45:50 amBy conductor you mean the electrode plates? What do you mean free charge? If there are two plate electrodes facing each other,and both are charged to 2 000 000 volt,how is there going to be "small electric field in the gap"?Why did you start calculating repulsion of hemispheres? Why not just stick with the flat square like was included in my question?

author: ejeffrey
author_rank: Super Contributor
timestamp: April 22, 2018, 02:49:13 pm
content: Quote from: fonograph on April 22, 2018, 05:20:55 amThe outside shape of the electrodes, where most of the charge will be.QuoteElectric field corresponds to a change in potential. If you charge both plates to 2e6 volts, the potential difference is zero, and the field is small. Small not zero because there are still fringing fields due to the non-zero gap, but small. Another way to think of this: if you place a pair of metal objects nearby and place the same charge on both, the charges repel. Since metals are conductors, the free charge will move to the outside surface, away from the other object. No or small free charge on the surfaces facing each other means small electric field.QuoteBecause hemispheres is easier and if you understood my calculation you could figure out how to apply it to whatever shape you want. A sphere has the nice property that you can find the charge distribution by symmetry, so it is easy to calculate the self capacitance. Most other shapes can't be calculated analytically so you have to resort to numerical simulations.

author: ahbushnell
author_rank: Frequent Contributor
timestamp: April 22, 2018, 04:15:49 pm
content: Quote from: Cerebus on April 21, 2018, 10:21:26 pmIt's easy just read his question.I bet it was a homework problem.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: April 22, 2018, 07:32:56 pm
content: I'm guessing about 1.49E6 volts, but no matter what you do, the force will be attractive. Probably going to need some insulation.

author: westfw
author_rank: Super Contributor
timestamp: April 24, 2018, 04:17:28 am
content: QuoteThe force/charge equation posted by Cerebus is the thing to use, but it assumes that both "charges" are at "point" locations. In actuality, you're talking about individual electrons distributed across the plates, and (for example) the electron at one corner of one plate will repel the electron directly across the gap from it more than it will repel the electron in the diagonally opposite corner. Then you can add that the electrons on either plate are probably not distributed evenly across that plate, since they also repel each other AND there are probably spots that are QM-favorable places for them to be (dependent on the exact nature of the substance composing the plates.)I don't know offhand whether these factors ever result in "substantial" deviations from the simple model (ie big enough to affect a real-world application), and I'm pretty sure I can no longer do the math needed to figure it out. Or the QM.