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xophergrunge

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## Homework Statement

The space between the plates of a parallel plate capacitor is filled with a dielectric material whose dielectric constant ϵr varies linearly from 1 at the bottom plate (x=0) to 2 at the top plate (x=d). The capacitor is connected to a battery of voltage V. Find all the bound charge, and check that the total is zero.

**2. Homework Equations**

[itex]C=\frac{Q}{V}[/itex]

[itex]C=\frac{Aε_{0}}{d}[/itex]

[itex]D=ϵE[/itex]

[itex]D=ε_{0}E+P[/itex]

[itex]\int D\bullet da = Q_{f}[/itex]

[itex]σ_{b}=P⋅\widehat{n}[/itex]

[itex]ρ_{b}=−∇⋅P[/itex]

## The Attempt at a Solution

First I found [itex]ε_{r}[/itex] as a function of x, [itex]ε_{r}=\frac{x}{d}+1[/itex].

Assuming that the area of each plate is A, I said that the bound charge must be [itex]Q_b=A(\sigma_b+\int\rho_{b}dx)[/itex].

Next, using [itex]D=ε_{0}E+P[/itex] and [itex]D=εE[/itex] I get that [itex]P=D(\frac{x}{x+d})[/itex].

Using [itex]\int D\bullet da = Q_{f}[/itex] I get that [itex]D=\frac{Q_{f}}{A}[/itex].

Nowing using [itex]σ_{b}=P⋅\widehat{n}[/itex] and [itex]ρ_{b}=−∇⋅P[/itex] I get that [itex]\sigma_b=0[/itex] at x=0, [itex]\sigma_b=-\frac{Q_f}{2A}[/itex] at x=d and [itex]\rho_b=\frac{Q_f}{A}\frac{x}{(x+d)^2}[/itex].

I integrate [itex]\rho_b[/itex] from 0 to d and I get [itex] \frac{Q_f}{A}(\ln 2 -\frac{1}{2})[/itex].

So, now plugging those values into [itex]Q_b=A(\sigma_b+\int\rho_{b}dx)[/itex] I get [itex]Q_f(\ln 2 -1)[/itex], which only equals zero when there is no free charge on the plates of the capacitor, so I know I am doing something wrong. I am pretty sure I am going about this completely wrong, but I don't see any other way to do it. Any help would be really appreciated. Thank you.

Also, using [itex]C=\frac{Q}{V}[/itex], [itex]C=\frac{Aε_{0}}{d}[/itex], and [itex]D=ϵE[/itex] I can get the capacitance [itex]C=\frac{A\epsilon_0}{d\ln 2}[/itex] and [itex]Q_f=\frac{A\epsilon_{0}V}{d\ln 2}[/itex] but I don't see how that will help me find the bound charge.

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