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## Homework Statement

A parallel plate capacitor has 2 square plates of side l, separated by a distance d (l is a lot bigger than d). Between the plates, there is a linear and non-homogeneous dielectric with constant $$\epsilon_R=1+ay$$ with $$a$$a positive constant. Calculate the electric field between the plates knowing that the capacitor is connected to a battery of voltage V (positive armor at y=0).

Here's an image to help understanding the geometry of the problem:

## Homework Equations

Gauss Law:

$$\oint_S {E_n dA = \frac{1}{{\varepsilon _0 }}} Q_{inside}$$

Generalized Gauss law:

\oint_S {D_n dA = Q_{inside free}$$

$$D=\epsilon E$$

## The Attempt at a Solution

So first what I thought was that because the capacitor is connected to a battery we will have a an accumulated charge $$+Q$$ in the positive armor and $$-Q$$ in the negative armor-

Now in a capacitor with the empty space between the plates, the electric field is $$E=\frac{Q}{\epsilon*A}$$ adapting to our capacitor $$E=\frac{Q}{(1+ay)\epsilon*l^2}$$.

Now I thought I was done, but my textbook provided a different answer

$$E=\frac{Va}{(1+ay)\log{1+ad}}$$

Now I have no idea on how to relate my expression to the expression I got. I know there is probably a relation between Voltage V and the charge Q and that's probably how I will eliminate the $$\epsilon_0$$. I also know that 1+ad is the constant in the upper plate. But how do I include it in the field? Why do you have a logarithm?

Can someone help me please?

Thanks!