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## Homework Statement

I was attemping to solve problem 10.13 in Purcell's E&M Book: By considering how the introduction of a dielectric changes the energy stored in a capacitor, show that the correct expression for the energy density in a dielectric be [tex]\frac{\epsilon E^2}{8\pi}[/tex]

## Homework Equations

1 [tex] U = 1/2 C V^2[/tex]

2 [tex] Q_0 = Q + Q' = \epsilon Q_0 + Q'[/tex]

3 [tex] U = \frac{1}{8 \pi} \int_V E^2 d^3 x [/tex]

4 [tex] E_{tot} = E_{vac} - E_{pol} = \epsilon E - 4 \pi P[/tex]

## The Attempt at a Solution

It is not hard to derive Eq. 3 from Eq. 1. This is the energy stored in a vacuum capacitor.

Q_0 and E_0 are the charge and field in a vacuum capacitor before the dielectric is added

I attempted to use Eq 2 to derive the electric field inside a capacitor with a dielectric. The total electric field inside the capacitor (Eq 4)will be the superposition of the electric field of a vacuum capacitor of strength [tex]\epsilon E_0[/tex] and the field of a polarized dielectric, [tex] 4 \pi P[/tex]. Since P can be equated to the charge density per area, [tex]\sigma[/tex], you can solve equation 2 for Q', which is the surface charge density on the polarized dielectric's surface and plug that in for P in Eq. 4. You can also do this for the field of the vacuum capacitor:

[tex] E = \epsilon E_0 - 4\pi P = \frac{4 \pi \epsilon Q_0}{ A } - \frac{4 \pi Q_0(1-\epsilon)}{A} [/tex]

factoring out a 4 pi Q_0 /A yeilds E_0:

[tex]E = E_0 ( \epsilon - 1 + \epsilon ) = E_0(2 \epsilon - 1)[/tex]

plugging this into equation 2 after calculating the voltage from E yeilds

[tex]U = \frac{1}{8 \pi}E_0^2 (4 \epsilon^2 - 4\epsilon -1)[/tex]

this is not what was expected in the problem.

The only thing I can think of that would make this not work is equating P = Q' / A.

Please advise