# Homework Help: Calculating changes in stored energy of a capacitor

1. Feb 19, 2013

### Pinedas42

1. The problem statement, all variables and given/known data
Given a parallel plate capacitor with sides of size L connected to a constant potential (battery) separated by a distance D

A dielectric slab is inserted between the plates to a length x so that there are two areas
w/ dielectric constant K

L(L-x) for the space without a dielectric
L*x for the dielectric filled space
All epsilons are epsilon nought, permittivity of free space

I found the capacitance to be C=$\frac{ε * L}{D}$(L + x(K-1))

I found the change in energy

dU = $\frac{(K-1)ε V^{2} * L}{2D}$ dx

Question:
Suppose that before the slab is
moved by dx the plates are disconnected
from the battery, so that the
charges on the plates remain constant. Determine the magnitude of
the charge on each plate, and then show that when the slab is
moved dx farther into the space between the plates, the stored
energy changes by an amount that is the negative of the expression
for given dU (above)

2. Relevant equations

C=Q/V
U=1/2 * QV=1/2 CV^2=Q^2/2C

3. The attempt at a solution

I found the constant charge Q
Q = $\frac{ε * L * V}{D}$(L+x(K-1))

My problem is I am not sure at how to set this up.

$\Delta$U = U - U nought
$\Delta$U =$\frac{C*V^2}{2}$ - $\frac{Cnought *V^2}{2}$
Here I think to myself, the voltages wouldn't be equal would they?
SInce the charge Q is constant, for the capacitance to rise with a dielectric, the voltage has to drop right? Yet the solution manual starts with U=(1/2)CV^2

I have the solution manual and they start with:

U=$\frac{1}{2}$C*V^2 = $\frac{1}{2}$Cnought*V^2*$\frac{C}{Cnought}$

I can't think of the logic behind this expression at all.

BTW this is part (c) of problem 24.77 in University Physics 13th Ed. Young and Freedman

Last edited: Feb 19, 2013
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