Capacitor Energy with a Dielectric

1. May 31, 2014

JMaria

1. The problem statement, all variables and given/known data
A dielectric-filled parallel-plate capacitor has plate area A = 15.0cm2 , plate separation d = 5.00mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 5.00V . Throughout the problem, use ϵ0 = 8.85×10−12C2/N⋅m2 .

a)Find the energy U1 of the dielectric-filled capacitor.
b)The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.
c)The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.
d)In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric?

2. Relevant equations

a) U=1/2CV2
U=1/2(V2(kε0A)/2d)

b)((K+1)ε0AV2)/4d

c)[((K+1)2)ε0AV2]/8d

d)[((K2)-1)ε0AV2]/8d

3. The attempt at a solution

For part A, I got 9.96*10^-10 just by plugging those numbers given into the equation. I think I'm having an issue though with the scientific notation. I don't think I'm getting the right numbers. I used the A value and the d value in meters. I don't know if I was supposed to.

b)6.64*10-10

c) 1.32*10-9

d)6.64*10-10

2. May 31, 2014

tms

$A$ is an area; it should be in square meters.

3. May 31, 2014

JMaria

So should I square the value first then convert it to meters? Am I plugging this in right?
For part A:

[(1/2)(8.85*10^-12)(3)(.225)]/(2*5*10^-3)

4. May 31, 2014

ehild

No, the area is 15 cm2. What is it in m2?

ehild

5. May 31, 2014

JMaria

Is it .0015m^2?

6. May 31, 2014

Yes.