Finding the bound charge in a dielectric ATTEMPT 2

[xophergrunge]

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

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

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.

 

[TSny]

I get that P=D(\frac{x}{x+d}).
D=\frac{Q_{f}}{A}.

I think that's correct.

I get that \sigma_b=0 at x=0, \sigma_b=-\frac{Q_f}{2A} at x=d

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

and \rho_b=\frac{Q_f}{A}\frac{x}{(x+d)^2}.

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

 

[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.

 

[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.

 

[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.

 

[TSny]

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

 

[xophergrunge]

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

 

[xophergrunge]

And I meant one half Q_f for both cases.

 

[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.

 

[TSny]

Great. Good work.

 

[xophergrunge]

Thanks again.