# Plate capacitor

## Homework Statement

charge a plate capacitor, disconnect the battery, increase the distance between the plates: capacitance decreases (C=keA/d, as d increases, C must decrease),
V increases (q=CV, q is constant, C decreases, V must increase),
electric field stays constant (E=V/d, both V and d increase proportionally),
energy stored increases (energy= 1/2 C V*V, C decreases, but V increases and squares).

see above.

## The Attempt at a Solution

does it make sense that both the voltage and the energy increase as the distance between the plates increases? conceptually, I would think they would both decrease. can you help clarify? Thanks...

## The Attempt at a Solution

kuruman
Homework Helper
Gold Member
If the charge is fixed, the electric field (volts per meter) between the plates is constant and uniform regardless of the separation. So if you increase the distance, you would have fewer volts per meter unless the voltage increases.

The plates attract each other. To separate them some more, you need to do work on them by pulling them farther apart against the electrostatic attraction. That work appears as additional potential energy stored in the capacitor. It's like stretching an already stretched spring even more.

Thanks and I have some follow on question.

for the voltage increase: I see your point and it make sense mathematically and logistically. It just runs a little against the grain to think the voltage would increase when the plates are pulled apart.
for the energy stored increase: so you mean the energy increase is kinetic instead of electric?