# Homework Help: Finding the bound charge in a dielectric ATTEMPT 2

1. Nov 17, 2013

### xophergrunge

1. The problem statement, all variables and given/known data
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. Relevant equations

$C=\frac{Q}{V}$

$C=\frac{Aε_{0}}{d}$

$D=ϵE$

$D=ε_{0}E+P$

$\int D\bullet da = Q_{f}$

$σ_{b}=P⋅\widehat{n}$

$ρ_{b}=−∇⋅P$

3. The attempt at a solution

First I found $ε_{r}$ as a function of x, $ε_{r}=\frac{x}{d}+1$.
Assuming that the area of each plate is A, I said that the bound charge must be $Q_b=A(\sigma_b+\int\rho_{b}dx)$.
Next, using $D=ε_{0}E+P$ and $D=εE$ I get that $P=D(\frac{x}{x+d})$.
Using $\int D\bullet da = Q_{f}$ I get that $D=\frac{Q_{f}}{A}$.
Nowing using $σ_{b}=P⋅\widehat{n}$ and $ρ_{b}=−∇⋅P$ I get that $\sigma_b=0$ at x=0, $\sigma_b=-\frac{Q_f}{2A}$ at x=d and $\rho_b=\frac{Q_f}{A}\frac{x}{(x+d)^2}$.
I integrate $\rho_b$ from 0 to d and I get $\frac{Q_f}{A}(\ln 2 -\frac{1}{2})$.
So, now plugging those values into $Q_b=A(\sigma_b+\int\rho_{b}dx)$ I get $Q_f(\ln 2 -1)$, 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 $C=\frac{Q}{V}$, $C=\frac{Aε_{0}}{d}$, and $D=ϵE$ I can get the capacitance $C=\frac{A\epsilon_0}{d\ln 2}$ and $Q_f=\frac{A\epsilon_{0}V}{d\ln 2}$ but I don't see how that will help me find the bound charge.

Last edited: Nov 17, 2013
2. Nov 17, 2013

### TSny

I think that's correct.

Should $\sigma_b$ have a negative sign at x = d? I guess it depends on your sign conventions and which plate is positively charged.

I don't agree with the numerator $x$ in this expression.

3. Nov 17, 2013

### xophergrunge

I get negative because $\hat{n}=-\hat{i}$ since x is increasing as you move from the first plate (x=0) to the second (x=d).

I just re-did my calculation for $\rho_b$ and I got d in the numerator now, and the whole thing has a negative sign. This will change my $\int \rho_b dx$ and hopefully give me an answer that makes sense. Is that what you got?

Thanks.

4. Nov 17, 2013

### TSny

Yes, I got $d$ in the numerator instead of $x$. I got positive for $\sigma_b$ at $x = d$ and I got $\rho_b$ to be negative. I assumed the lower plate (x = 0) is the positively charged plate so that D, E, and P all point in the positive x direction.

5. Nov 17, 2013

### xophergrunge

Great, I am getting that $Q_b=0$ now, with the bound volume charge equal to negative one half and the bound surface charge to be negative one half. Thank you so much.

6. Nov 17, 2013

### TSny

OK, but how do two negative quantities add to zero?

7. Nov 17, 2013

### xophergrunge

Sorry, that was a typo. Only the volume bound charge was meant to be negative.

8. Nov 17, 2013

### xophergrunge

And I meant one half $Q_f$ for both cases.

9. Nov 17, 2013

### xophergrunge

And I realized my mistake with the negative sign on $\sigma_b$, I was thinking of the surface of the plate rather than the surface of the dielectric.

10. Nov 17, 2013

### TSny

Great. Good work.

11. Nov 17, 2013

### xophergrunge

Thanks again.

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