Capacitor Energy with a Dielectric

[JMaria]

Homework Statement

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?

Homework 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

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

 

[tms]

I used the A value and the d value in meters. I don't know if I was supposed to.

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

 

[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)

 

[ehild]

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


ehild

 

[JMaria]

Is it .0015m^2?

 

[ehild]

Yes.