# Parallel plate capacitors

1. Feb 12, 2007

### map7s

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

A parallel-plate capacitor has plates with an area of 445 cm2 and an air-filled gap between the plates that is 1.51 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery. There is 4.311536113 x 10 ^-5 J of energy stored in the capacitor. The separation between the plates is now increased to 3.02 mm. How much energy is stored in the capacitor now?

2. Relevant equations

I know that since the capacitor is disconnected from the battery, V can change with C while Q, being trapped in the disconnected
capacitor, is constant. I calculated Q for the original
capacitor, which was about 1.5 x 10^-7 C. When the spacing, d, between the plates is increased, C changes, and V changes with it, so I know that I can't use the equation U=(1/2)CV^2 (with V=575 V) to find the energy, but I'm not sure exactly what I can use.

2. Feb 12, 2007

### antonantal

So let's see how they change.

3. Feb 12, 2007

### map7s

Well, I thought that maybe by treating the increased space as a dielectric field and using the dielectric constant I could figure out the new values of V and C and use the Q to calculate the new Energy, but apparently I wasn't on the right track...

4. Feb 12, 2007

### antonantal

How is the capacitance related to the distance between the plates?

5. Feb 12, 2007

### map7s

In calculating the capitance, the equation is C=(constant) x area/distance. So, b/c the battery is disconnected and the plates are pulled further away from each other, there needs to be a new equation for capitance...but I'm not sure what

6. Feb 12, 2007

### antonantal

Not a new equation. Only a new value for one parameter - the distance between the plates.

By how many times does the distance between the plates increase? How will this affect the capacitance? Will it become greater or smaller? By how many times?