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## Homework Statement

Hi there, I got a little problem with this one:

plate capacitor is filled with 2 different dielectrics with permittivities [tex]\epsilon_1, \epsilon_2[/tex] (surrounding environment is [tex] \epsilon_0 [/tex]) and their respective thicknesses are [tex] d_1 [/tex] and [tex] d_2 [/tex]. 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 ?

## Homework Equations

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

## The Attempt at a Solution

I tried this:

electric displacement field is the same in whole capacitor [tex] D = \sigma [/tex] where [tex] \sigma [/tex] is surface density of charge on plates.

The voltage is then [tex] U = \frac{D}{\epsilon_1}d_1 + \frac{D}{\epsilon_2}d_2 [/tex].

From the last equation, using the 1st one, the charge density is:

[tex] \sigma = U \frac{\epsilon_1 \epsilon_2}{d_1 \epsilon_2 + d_2 \epsilon_1} [/tex].

And the force on plate at dielectric 1: [tex] 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 [/tex]

[tex]E_0[/tex] 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

[tex] F = \frac{\epsilon_0^2}{2 \epsilon_1} (\frac{U}{d_1 + d_2})^2 S [/tex].

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 ;)]

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