Potential Energy and Force Between Capacitor Plates

  • #1
diza
Hello guys, came across some Analytical Physics lectures (http://www.physics.rutgers.edu/ugrad/227_f11/classes/lect08.pdf) that got me confused.

It is described in the lecture the problem of determining the force between capacitor plates for a constant voltage (ie. connected to a battery). For constant voltage, pulling the plates apart would result in less stored energy in the field, since:

U = CV²/2 = eps*AV²/2d (fixed voltage, parallel plates)

Since the force between the plates is in the direction to reduce the potential energy, this would seem to indicate a repulsive force between the plates (in the direction of increasing d).

This energy-based analysis of the problem seems correct. However, this conflicts with some other sources saying the force is attractive and my intuition that the opposite charges on the two electrodes tend to attract each other.

Can someone guide me in the right direction?
 

Answers and Replies

  • #2
764
71
Separating the plates at constant voltage is different than separating them with constant charge. When the voltage is constant that means charges can enter or leave the two sides the capacitor.
 
  • #3
diza
Separating the plates at constant voltage is different than separating them with constant charge. When the voltage is constant that means charges can enter or leave the two sides the capacitor.

I understand that. When you pull plates apart you are actually charging the battery because some of the charge goes back into it. I don't see how this affects the energy-based argument, though.

The force should be in the direction of less potential energy either way, therefore it should be repulsive. However, most textbooks describe this force as attractive (even for constant voltage, variable charge etc).

Also what I don't understand is: what is the origin of this physical force? Coulomb's law says that this force should be attractive unless the plates are both charged at the same polarity.
 
  • #4
berkeman
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Assume you charge the plates to V and remove the battery.

There is an attractive force between the plates. Now pull the plates apart, which takes energy put into the system to overcome the attractive force.

Q=CV, so what happens to V as C is decreased by pulling the plates apart?

And the energy on the capacitor is E=1/2CV^2, so what happens to the energy of the capacitor when you pull the plates apart with the constant charge on the plates? :smile:
 
  • #5
diza
Assume you charge the plates to V and remove the battery.

There is an attractive force between the plates. Now pull the plates apart, which takes energy put into the system to overcome the attractive force.

Q=CV, so what happens to V as C is decreased by pulling the plates apart?

And the energy on the capacitor is E=1/2CV^2, so what happens to the energy of the capacitor when you pull the plates apart with the constant charge on the plates? :smile:

Suppose we are moving the capacitor plates from d->2d.

For fixed charge
U' = Q²2d/εA = 2U
ΔU = 2U - U = U

That seems right. Capacitance goes down. Voltage goes up. And it takes energy put into the system to overcome the attractive force, like you said.

However, if we keep the battery connected, the voltage is fixed and therefore as capacitance goes down, charge goes down as well, and:
U' = εAV²/4d = U/2
ΔU = U/2 - U = -U/2

That would mean the force is repulsive since you took energy from the system to move the plates apart.

However, I still don't understand the origin of this repulsive force and this goes against our intuition that opposite charges attract.

Am I missing something?
 
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  • #6
berkeman
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However, if we keep the battery connected, the voltage is fixed and therefore as capacitance goes down, charge goes down as well, and:
U' = εAV²/4d = U/2
deltaU = U/2 - U = -U/2

That would mean the force is repulsive since you took energy from the system to move the plates apart.
No, the force is attractive because opposite charges attract. You are not accounting for where the lost energy goes...
 

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