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xophergrunge
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MODERATOR'S NOTE: problem statement is mistaken, please see https://www.physicsforums.com/showthread.php?t=723342
The space between the plates of a parallel plate capacitor is filled with a dielectric material whose dielectric constant $\epsilon_r$ varies linearly from 1 at the bottom plate (x=0) to 2 at the top of the 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.
$C=\frac{Q}{V}$
$C=\frac{A\espilon_0}{d}$
$D=\epsilon E$
$\sigma_b=P\cdot\hat{n}$
$\rho_b=-\nabla \cdot P$
I am not sure if those are the relevant equations, I am really confused about what to do for this problem. What I have tried, with no success is to find the capacitance by assuming the plates are a distance L apart and have area A. First I find $\epsilon_r$ as a function of x, and it is $\epsilon_r = (\frac{x}{d}+1)$ and then saying $C=\int_0d (\frac{x}{d}+1)\frac{\epsilon_0A}{L}dx=\frac{3\epsilon_0 Ad}{2L}$ And since $C=\frac{Q}{V}$ then the free charge is $Q=\frac{3\epsilon_0 AdV}{2L}$. I don't really know where to go from here, or even if what I have done so far is how I want to approach it. I also thought about "slicing" the capacitor horizontally so that each little slice looked like a capacitor with a linear dielectric inside, so that I could use $P=\epsilon_0(\epsilon_r-1)E$. But, I again, don't know where I would go from there. Any help would be appreciated.
Homework Statement
The space between the plates of a parallel plate capacitor is filled with a dielectric material whose dielectric constant $\epsilon_r$ varies linearly from 1 at the bottom plate (x=0) to 2 at the top of the 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.
Homework Equations
$C=\frac{Q}{V}$
$C=\frac{A\espilon_0}{d}$
$D=\epsilon E$
$\sigma_b=P\cdot\hat{n}$
$\rho_b=-\nabla \cdot P$
The Attempt at a Solution
I am not sure if those are the relevant equations, I am really confused about what to do for this problem. What I have tried, with no success is to find the capacitance by assuming the plates are a distance L apart and have area A. First I find $\epsilon_r$ as a function of x, and it is $\epsilon_r = (\frac{x}{d}+1)$ and then saying $C=\int_0d (\frac{x}{d}+1)\frac{\epsilon_0A}{L}dx=\frac{3\epsilon_0 Ad}{2L}$ And since $C=\frac{Q}{V}$ then the free charge is $Q=\frac{3\epsilon_0 AdV}{2L}$. I don't really know where to go from here, or even if what I have done so far is how I want to approach it. I also thought about "slicing" the capacitor horizontally so that each little slice looked like a capacitor with a linear dielectric inside, so that I could use $P=\epsilon_0(\epsilon_r-1)E$. But, I again, don't know where I would go from there. Any help would be appreciated.
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