# Parallel plate capacitor's plate distance is doubled, does the energy double?

1. Mar 1, 2012

### DerekZ10

1. The problem statement, all variables and given/known data

A parallel-plate capacitor is charged by being connected to a battery and is then disconnected from the battery. The separation between the plates is then doubled. How does the electric field change? The potential difference? The total energy? Explain your reasoning.

2. Relevant equations

σ=Q/A (Surface Charge Density)

E=σ/ε0 (Electric Field Between Plates)

V=Ed (Potential Difference)

U=(1/2)QV (Potential Energy)

3. The attempt at a solution

Q1=Q2

A1=A2

d1=x
d2=2x

Question 1: Electric Field Change?

E=Q/(Aε0)

No, d is not related.

Question 2: Potential Difference Change?

V=Ed

Yes, the voltage will double.

Question 3: Total Energy change?

U=(1/2)QV

Yes, it will double.

4. Hold up! That doesn't make sense!

So I have the answers, but I need someone to explain this to me because this looks to be breaking thermodynamics. The answers given is inline with what cramster and numerous other sources have said. I think the new capacitor will allow it to store twice the charge, but the problem makes it seem like the energy is just created. Or am I suppose to be assuming that by moving the plates farther apart, the capacitor is converting the physical motion energy into electrical energy?

My logic is telling me the final total energy should be the same as the initial. And that the electric field should have been halved. But I don't have an equation available for electric field intensity.

2. Mar 1, 2012

### Staff: Mentor

Welcome to the PF.

It looks like you are missing two key equations. The first relates the capacitance to the palte area A and separation distance d. Are you familiar with that equation?

The second equation relates the charge Q to the capacitance C and the voltage between the plates V. Are you familiar with that equation?

Does using those two equations make this easier for you?