Calculating changes in stored energy of a capacitor

[Pinedas42]

Homework Statement

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 energydU = \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)

Homework Equations

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

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.

\DeltaU = U - U nought
\DeltaU =\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

 

[member 553083]

, so if anyone has that book, perhaps they can help me out.A:You have the formula for the capacitance:$$C=\frac{\epsilon V}{D}\left[L+x\left(K-1\right)\right]$$In words: The capacitance is proportional to the voltage applied to it and inversely proportional to the distance between the plates. The constant of proportionality depends on the length of the plates and the dielectric constant of the material in between them.The energy stored in a capacitor is $U=\frac{1}{2}CV^2$. So when you increase the capacitance, the stored energy increases.When the plates are disconnected from the battery, the voltage across the plates is zero and thus the energy stored in the plates is zero.Now you insert a dielectric slab between the plates and the capacitance increases. Thus the energy stored in the plates also increases.The energy stored in a capacitor with capacitance $C$ and voltage $V$ is given by $$U=\frac{1}{2}CV^2$$The change in energy is then given by$$\Delta U=\frac{1}{2}\Delta CV^2$$where $\Delta C$ is the change in capacitance.Using the expression we found for the capacitance above, we get$$\Delta C=\frac{\epsilon V}{D}\left[x\left(K-1\right)\right]$$thus$$\Delta U=\frac{1}{2}\frac{\epsilon V^2}{D}\left[x\left(K-1\right)\right]$$or$$\Delta U=\frac{\epsilon V^2 L}{2D}\left[\frac{x}{L}\left(K-1\right)\right]$$which is the same as the expression you found for the change in energy.