# Electrostatic repulsive force between parallel plates

Hi

I was thinking about two parallel plates with same charges (there's no electrical field between them). How can we calculate the repulsive force? Does anyone have any ideas besides numerical integration?

Thanks

## Answers and Replies

Simon Bridge
Science Advisor
Homework Helper
Welcome to PF,
For two infinite parallel planes of charge (note: not two real parallel pates) the net field between them is zero.

Consider a situation you are familiar with, two point charges: the field at each point charge is infinite - how do you normally compute the force between them (using fields)? It's the same for each plate.

• Arjang Sh
Thanks for your prompt response.

So what I understand from your suggestion is that I can have a first order estimation of repulsive force by calculating available charges on each plate (Q1 = Q2) and use Coulomb's law as k.Q1.Q2/(d^2), is that right?

But I'm a little confused here since I have reached a contradiction for attraction force between oppositely charged plates of a capacitor:

Approach 1 = Force is calculated by differentiating electrostatic energy as: Approach 2 = Coulomb's Law: Neglecting constant coefficients, the effect of distance "x" is 2 orders of magnitude different. What am I missing here? Would it be possible for you to clarify this for me?

Thanks

Welcome to PF,
For two infinite parallel planes of charge (note: not two real parallel pates) the net field between them is zero.

Consider a situation you are familiar with, two point charges: the field at each point charge is infinite - how do you normally compute the force between them (using fields)? It's the same for each plate.

Simon Bridge
Science Advisor
Homework Helper
So what I understand from your suggestion is that I can have a first order estimation of repulsive force by calculating available charges on each plate (Q1 = Q2) and use Coulomb's law as k.Q1.Q2/(d^2), is that right?
No.

I am saying that you should think more carefully about how you normally do other, simpler, electrostatic calculations.
In the case of 2 point charges, the electric field at each charge is infinite and yet you can still get a sensible answer for the force between them.
How do you do that?

This goes to the heart of your initial misunderstanding.
The net field is zero - between the plates, but you don't use the net field everywhere to calculate the force.
(Technically the net field at the position of the plates is undefined, anyway.)

In the case of 2 point charges, the electric field at each charge is infinite and yet you can still get a sensible answer for the force between them.
How do you do that?

Well in this case I know that electric field around each point charge is calculated as: kQ/(r2) (undefined at the position of point charge) and the force is defined as F=q.E which results in kQq/(r2).

I believe I can use the same approach for attracting parallel plates: Total net field between plates is Q/(ε*Area) and Force is equal to total charge on each plate multiplied by this electric field. So I can write F = Q.E = (Q2)/(ε*Area) and substituting for Q we end up having F=½*Area*ε*V2/d2. This is the exact same thing when I differentiate the electrostatic energy.

I don't know what I am missing here that I cannot relate this basic concept to repulsive parallel plate, as you mentioned this originates from my initial misunderstanding of the concept. I really appreciate if you could provide me with more hints so I can get back onto right track.
Thank you

Simon Bridge
Science Advisor
Homework Helper
Well in this case I know that electric field around each point charge is calculated as: kQ/(r2) (undefined at the position of point charge) and the force is defined as F=q.E which results in kQq/(r2).
... you skipped over a bit there ... please concentrate: what did you say the electric field at the position of the charge was?

Basically - the field at the charge is undefined, but that does not stop you calculating the force on the charge due to the other charge.
For the force on charge 1 due to charge two you write: ##\vec F = q_1\vec E_2##

Notice what E2 isn't? It is not the total field from both charges combined. It is the field from the second charge alone ... as if q1 were not there.
For this problem, the charge does not see its own electric field.

It's the same with the plates - the force on plate 1 due to plate 2 is what you want to calculate.
But in post #1 you observed that the field between the plates is zero! But the field due to plate 2 between the plates is not zero.
It is the total combined field that is zero.

If the field due to plate 1 is E1 and the field due to plate 2 is E2, then a charge q somewhere other than on one of the plates gets force ##\vec F = q(\vec E_1+\vec E_2)##, this will be zero between the plates... but the same charge at[*] the position of plate 1 gets force ##\vec F = q\vec E_2## ... see?

There are two basic ways to work out the force by brute, um, force... you can use coulombs law on each charge element, or you can work out how the energy stored in that system changes with the separation of the plates. I suspect this is something you should do ... compare the result with the electric fields for each plate separately.
Involves calculus I'm afraid - but after that you just remember the result.

The shortcut would be to realize that it is the same force as for oppositely charged plates, but with the opposite sign. In fact, we are assuming a uniform charge distribution - that only happens for an insulator.

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[*] ... when q is a part of the plate. All charges in the plate will see the forces due to all the other charges in the plate. But, since this is an electrostatic system, the internal forces all cancel out.

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