Widening plates in a capacitor

[Seung Tai Kang]

Homework Statement

What happens if the capacitor plates are pulled farther apart?

Homework Equations

C=Q/V=EoA/d

The Attempt at a Solution

The answer is that the charge stays the same, and the potential difference increases, which results in increased PE.
I saw multiple explanations elsewhere. But their explanations are based on equations. Can some explain using electrons and forces?
Sometimes textbook uses battery potential difference to refer to the V in the equation Q/V and sometimes use the potential difference within a capacitor like when explaining dielectric. Which potential difference does V in the equation Q/V refers to?

 

[scottdave]

The textbook may be referring to different situations. Can you give examples?
Mechanically you have put energy in the form of work (using a force to move the plates a distance). The energy is now in the form of increased electrical potential energy.

 

[CWatters]

Consider a capacitor charged to a voltage V. It doesn't matter how it's charged, but afterwards the charger is removed. Then the following holds...

Q = VC

Where V is the voltage on the capacitor, C is the capacitance and Q is the charge.

If you then pull the plates apart C will reduce. Q is constant so V must increase.

The energy stored in the capacitor is

E=0.5*CV^2

The change in V is inversely proportional to the change in C so because of the squared term the energy stored increases. In other words it takes energy to pull the plates apart.