Capacitor and Force

Trying to remember some fundamentals...

Isn't Force = dE/dx

change of energy with spacing?

...Jim Thompson

What's your value of K? I can't recall exactly what it would be, but it would include 'epsilon-0', an electrostatic constant, which is very small, somewhere around 1.0E-07 IIRC. Electrostatic forces are weak. You can get a feel for the situation by doing one of the classic electrostatic experiments like picking up tiny pieces of lint with a charged plastic or glass rod. The voltages involved are kilovolts.

I don't think that the force between two point charges is of much relevance to the force between the two parallel plates of a capacitor.

An easier way of getting there is to note that the energy stored in a capacitor is 0,5*C*V^2, or 8 microjoules for a 40nF capacitor.

Energy is force times distance, and doubling the gap to 0.2 micron would halve the capacitance and halve the stored energy, which gives an attractive force of the order of ten newtons - about one kilogram weight - but only over about 0.1 micron. Obviously the force is going to decrease very rapidly as the gap gets larger ...

Hope this helps.

-- Bill Sloman, Nijmegen

The (CV)^2 term alone has a value of 16*10^(-14), so the force can't be anywhere near 3.6N

I listen to my capacitors

Martin

Look at Jim Thompson's comment about calculating force.

For a reality check, consider that when you do electrostatics experiments you're happy to lift little bitty styrofoam balls with multi-kV of static electricity -- what in heck are you going to manage to do with a measly

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

lol. No. F = q*E. E = grad(V) which is probably what you were thinking.

k = 9*10^9 and is actually very large(it does have e in it but in the denominator.. I think its 1/(4pie_0) or something).

k is what makes the electrostatic force so strong compared with the gravitational force and the gravitational constant G.

Electrostatic forces in your example are no weak... unless you are comparing it to the nuclear forces or something like that. You have to realize that in your example the amount of charge is extremely small.

"the gravitational attraction between the two protons is roughly a trillion trillion trillion times weaker than the electrostatic repulsion."

um.... Did you pass highschool algebra?

seriously?

What makes you think that that is the only term that governs the order? there are two other factors involved.

Its relevant for an order approximation. Each plate has a charge Q on it. Its a fact of like that columbs law works so since we have two charges, equal and opposite in this case, we have -kQ^2/r^2 for the force. That force must exist. No way around it unless somehow it is getting neuralized but I don't see how.

If I do the same calculations I will arrive at the same answer...

C*V^2/

Well, I was using the dielectric in the ceramic cap. It doesn't quite matter what the capacitance is cause I measured it and its quite strong. So even if I cannot measure r I can still measure the capacitance. Sure my F might be off but I'm pretty sure its within < 1mm.

Jesus christ... I thought you guys were like MIT grads or something?

Do you realize that the charges involved in "electrostatics" are on the order of nC's if your lucky?

"Two balloons are charged with an identical quantity and type of charge: -6.25 nC"

"The force of repulsion of two +1.00 Coulomb charges held 1.00 meter apart is 9 billion Newtons."

Seeing that the parallel plate capacitors have a charge of ~1000 times implies the force is 1M times larger(since it depends on Q^2).

Well, I should get the same answer either way but I don't... I'll have to see whats going on later.

Its not air gapped as its a ceramic cap. The dielectric has a width of about

Of course cause its 1/r^2.

Even with simple electrostatics with charges much smaller one can see large forces. The voltages are much higher than this but voltage has nothing to do with it since we know the charge(of course the voltage is related to the force)

I don't know. Maybe my logic is flawed somewhere. (obviously something is flawed cause I'm not getting the proper results)

If you are putting external electrodes around a piece of hi K dielectric material, it seems to me the charge is stored inside the material, not in the airgaps. So when you move the electrodes, you are not changing ( much ) the charge stored in the dielectric, so no force. You would have to expand the dielectric itself to get a measurable force.

That might be true. My calculations are assuming that the dielectric is constant for all distances and thats not true.

But, charge is charge and that its what matters in calculating the force. There is no question about that since F = k*q1*q2/r^2.

If I were to charge the plates up when the distance is minimal(i.e., when the capacitance is actually high) then I will get the proper charge that I'm using. Now when I move the plates apart the capacitance rapidly decreases decreasing the charge and hence decreasing the force.

BUT, If the charge is fixed on the plates(so the power source is removed) then there is no way it can change for any distance and the force must therefor be constant too.

So potentially what is happening when I disconnect the power supply is that all the charge is leaking away to fast for me to measure any force. This was my initial idea but I don't know if its correct or not.

e.g., take a very large plate capacitor with capacitance C. That is, you fix some distance r and dielectric and measure C. Charge the plates and disconnect the power source. F = k*(C*V)^2/r and that is a fact. Now if Q is drastically decreasing then F is too and that is the only real problem because if Q is fixed then F depends on r which I have control of.

Chances are its cause the charge is leaking off though but again, this is somewhat of a guess. I guess I would have to make much larger plates and somehow reduce the leakage.

Jon

and

application.

by

is

The formula I have is, Newtons attract = [8.856e-12 x relative permeability x Area in mtr^2 x Voltage^2] / [2 x gap in mtrs] Would seem my 10n 400V multilayer poly cap has 1/4 ounce squeezing it together at max voltage. Maybe this is the source of dielectric absorbsion (sp?). E.g. the more rigid dielectrics will 'spring' together under a voltage strain and relax at leisure. Suggests a dielectric with similar cold flow properties to putty would be ideal. I.e once the plates have been squeezed together a few times then no more compaction can take place.

huh? e*A*V^2/2/d?

e*A/d is the capacitance and so you have C*V^2/2 which is the energy stored in the electric field. This is not the same as the force but W = F*d so one should theoretically be able to calculate the force due to the energy but I'm not sure if this is the same as coulumbs force as it is the force required to move one charge against the electric field.

Coulumbs law is quite simple and is analogous to the gravitational field. F = k*q1*q2/r^2. The charge q1 and q2 are the same in a parallel plate capacitor and can be found if the capacitance and applied voltage is known since Q = C*V.

Its actually quite simple

Q = CV F = k*Q^2/r^2

C and V are measured and hence F can be found.

By that logic _you_ are an MIT grad -- so what was it like? I graduated from WPI where we actually did things.

Since you're obviously quoting, why aren't you putting in your sources?

By your own math, with C = 20pF and V = 20V, you get q = 400pC. That's less than a nano-Coulomb.

Assuming your paper was 5 mils thick you get (9e9)(400e-12)^2/(127e-6)^3 =

huh? Are you serious?

Find them yourself... You don't seem to believe the logic of coulumbs law so want can I say?

"So for a ceramic capacitor of 20nF"

Jesus christ... your off by a factor of 1000.

Remember, I did two experiements. One with pennies which as about 20pF and one with a ceramic cap which was 40nF. Theres a huge difference and this is what I said the first time but you obviously didn't read it.

To be clear just so you don't keep going down the wrong path

C = 40nF!!!!!!!!!!!!!!!!!!!!!!! V = 20V!!!!!!!!!!!!!!!!!!!!!!!

NOW COMPUTE Q AND TELL ME WHAT YOU GET!!!

NOW COMPUTE F AND TELL ME WHAT YOU GET!!!

I bet you will be supprised.

Yes, you are correct... for C = 20pF. This is similar to the calculation of got with two pennies and a piece of paper. It is insiginificant. That is why I used half of a ceramic capacitor which has C = 40nF which is over 1000 times more. Now do the the calculation and see what you get. I bet you will be astonished by the force... its not weak is it? Your first post is total BS about how weak the force is.

Remember, that force is for V = 20 and not those voltages normally associated with static electricity. Even if you take the distance much larger, since I have a dielectric in the way, say its 2mm, then the force is still > 1

Seriously. Compute the force for C = 20pF and C = 40nF for r = 1mm and note the HUGE difference.(or 10pF and 10nF) Since F depends on Q^2 the difference in C is magnified by the square.

So if we have C1 = 20pF and C2 = 20nF = 1000*C1, that factor of 1000 becomes

I suppose maybe you should go back to WPI and ask for a refund?

It amazes me how much you guys bitch about my grammar and spelling but don't understand basic physics ;/

I mean, the first comment you made about how weak the electric force is just pure idiocity. I'm sure 8th grade physics mentions just how strong it is. Hell, most people only know 2 forces and so either you have gravitational force or you have electric force and and the electric force wins hands down. Of course maybe you had some other force in mind? Maybe the strong nuclear force?

In any case the force for the 40nF cap is large enough to observe so there is another problem. I imagine that the problem is actually keeping the charge on the caps. Either its leaking(but it must be doing it at the speed of light) or the charge is flowing back into the power supply when I disconnect it from the cap. (this is probably understandable since even diodes probably won't stop the flow of this small amount of charge)

I suppose I could try to electrostatically charge the caps and maybe that will work but I don't know since the plates are conductors.

I'm sorry, you clearly live so much closer to God than I do than nothing I could say would be acceptable.

Plonk.