# Capacitors: Energy change

1. May 14, 2014

### teme92

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

(a)
The separation between the plates of a parallel plate capacitor is 2 mm and the potential diff erence between them is 45 V. Find the electric field between the plates. If the permittivity of the dielectric medium between the plates is 20 $\epsilon$0, fi nd the charge per unit area on each plate.

(b)

If the area of each plate in part (a) is 0.13 m2, fi nd the capacitance of the capacitor.

(c)
If the dielectric medium between the plates in parts (a) and (b) is removed, so that air now fi lls
the region between the plates, fi nd the resulting change in the energy stored in the capacitor if
the plates are electrically isolated.

(d)
If the medium between the plates in parts (a) and (b) is partially removed, so that half of the
area has air between the plates and the other half has the original dielectric medium between the
plates, find the capacitance of the capacitor.

2. Relevant equations

C=q/V=$\epsilon$A/d
E=$\sigma$/$\epsilon$
$\sigma$=q/A=E$\epsilon$
V=Ed
U=0.5CV2
3. The attempt at a solution

(a)
For Electric Field:
E=V/d=45/0.002=22500 Vm-1

For charge per unit area:
$\sigma$=E$\epsilon$=22500(20(8.85x10-12))=3.98x10-6 Cm-2

(b)
C=$\epsilon$A/d=20(8.85x10-12)(0.13)/0.002=1.15x10-8 F

(c)
So here's where I'm stuck a bit.

So the change in energy will be $\Delta$U=U1 - U2

I have worked out:

U1=1.16x10-5 J

I don't understand what is meant when the question says its electrically isolated. I assumed it was something to do with I=0 but I can't see where this effect my equation so I must be missing something. Any help would be greatly appreciated.

2. May 14, 2014

### Simon Bridge

"electrically isolated" just means the charge on the plates doesn't go anywhere.

3. May 15, 2014

### rude man

You didn't show how you got U1. Was it 1/2 C V^2? If so you'll need to think about how V changes when the dielectric is removed.