# Capacitor filled with 2 different dielectrics

1. May 19, 2010

### _k_

1. The problem statement, all variables and given/known data
Hi there, I got a little problem with this one:
plate capacitor is filled with 2 different dielectrics with permittivities $$\epsilon_1, \epsilon_2$$ (surrounding environment is $$\epsilon_0$$) and their respective thicknesses are $$d_1$$ and $$d_2$$. The layers are parallel to the plates of the capacitor. The voltage between the electrodes is U and their surface is S. What are the forces acting upon plates ?

2. Relevant equations
gauss' law for electric displacement field; F = qE; U=d E (d = distance between plates)

3. The attempt at a solution
I tried this:
electric displacement field is the same in whole capacitor $$D = \sigma$$ where $$\sigma$$ is surface density of charge on plates.
The voltage is then $$U = \frac{D}{\epsilon_1}d_1 + \frac{D}{\epsilon_2}d_2$$.

From the last equation, using the 1st one, the charge density is:
$$\sigma = U \frac{\epsilon_1 \epsilon_2}{d_1 \epsilon_2 + d_2 \epsilon_1}$$.

And the force on plate at dielectric 1: $$F = S \sigma E_0 = \frac{S}{\epsilon_0} \sigma^2 = \frac{S}{\epsilon_0}U^2 (\frac{\epsilon_1 \epsilon_2}{d_1 \epsilon_2 + d_2 \epsilon_1})^2$$
$$E_0$$ is the field outside the dielectrics, since the plate is not in the dielectric.

Which, according to the book, is wrong. The correct answer should be
$$F = \frac{\epsilon_0^2}{2 \epsilon_1} (\frac{U}{d_1 + d_2})^2 S$$.

I'm completely lost. The book gives no detailed explanation, and after 2 hours of scratching my head I need help. Thanks in advance for any suggestions.

[edit: wrong names for electric field and electric displacement field... sorry for my english ;)]

Last edited: May 19, 2010