Potential Energy and Force Between Capacitor Plates

In summary, the conversation revolves around the problem of determining the force between capacitor plates for a constant voltage, where pulling the plates apart would result in less stored energy in the field. The energy-based analysis suggests a repulsive force between the plates, but this conflicts with other sources and intuition that the force should be attractive due to opposite charges. It is ultimately concluded that the force is indeed attractive, but the lost energy when separating the plates at constant voltage is not taken into account. This energy is discussed further in Feynman's lectures.
  • #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?
 
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  • #2
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
MisterX said:
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
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
berkeman said:
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
diza 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
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...
 

1. What is potential energy in the context of capacitors?

Potential energy in the context of capacitors refers to the stored energy that exists between the two charged plates of a capacitor. This energy is a result of the electric field created by the opposite charges on the plates and is directly proportional to the amount of charge and the distance between the plates.

2. How is potential energy calculated for a capacitor?

The potential energy of a capacitor can be calculated using the formula U = 1/2 CV², where U is the potential energy, C is the capacitance of the capacitor, and V is the potential difference (voltage) between the plates.

3. What factors affect the potential energy of a capacitor?

The potential energy of a capacitor is affected by the amount of charge on the plates, the distance between the plates, and the dielectric material between the plates. As the charge or distance increases, the potential energy also increases. The type of dielectric material can also impact the potential energy.

4. How does the force between capacitor plates relate to potential energy?

The force between capacitor plates is directly related to the potential energy. The electric field created by the opposite charges on the plates exerts a force on the charges, causing them to move and creating potential energy. The greater the force between the plates, the higher the potential energy.

5. Can potential energy be converted into other forms of energy in a capacitor?

Yes, potential energy can be converted into other forms of energy in a capacitor. When a capacitor is connected to a circuit, the stored potential energy is released as electrical energy, which can then be used to power devices. However, some energy may also be lost as heat due to resistance in the circuit.

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