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Homework Help: Finding the bound charge in a dielectric

  1. Nov 16, 2013 #1
    MODERATOR'S NOTE: problem statement is mistaken, please see https://www.physicsforums.com/showthread.php?t=723342

    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 $\epsilon_r$ 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.

    2. Relevant equations

    $D=\epsilon E$
    $\rho_b=-\nabla \cdot P$

    3. 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.
    Last edited by a moderator: Nov 17, 2013
  2. jcsd
  3. Nov 16, 2013 #2
    How do I delete a post? I thought latex worked here and I think it would be easier just to start this over?
  4. Nov 16, 2013 #3


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    Hi there!

    We have a few relevant equations here indeed. So first off we want to focus on the auxiliary field ##\vec{D}## because we can use it to treat this just like any other parallel plate capacitor problem meaning that ##\vec{\nabla} \cdot \vec{D} = \rho_f\Rightarrow \oint \vec{D}\cdot d\vec{A} = Q_f## combined with the obvious symmetry of the system (assuming we can ignore edge effects) will straightforwardly give us ##\vec{D}##. Furthermore we know that ##\vec{D} = \epsilon\vec{E}## and ## \vec{D} = \epsilon_{0}\vec{E} + \vec{P}## as well as ##\rho_{b} = -\vec{\nabla}\cdot \vec{P}## so you just have to piece everything together.
  5. Nov 16, 2013 #4


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    I made your post readable. The slicing is a good idea, but the slices are capacitors connected in series, and they do not add. Use that the normal component of D is the same at every x, and the potential difference of the slice capacitors is dU = Edx=(D/ε) dx. Get the total voltage across the capacitor by integrating E.

    Last edited: Nov 17, 2013
  6. Nov 16, 2013 #5

    I was actually thinking of slicing perpendicularly to the plates, so that the dielectric in each slice had the same ε[itex]_{r}[/itex], and that way the slice capacitors would be connected in parallel.
  7. Nov 17, 2013 #6


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    The dielectric constant varies perpendicularly to the plates from 1 at x=0 (bottom plate) to 2 at x=d (top plate). At a given x, ε is constant on the plane parallel with the plates. You have to slice the capacitor parallel to its plates.

  8. Nov 17, 2013 #7
    No, you are misreading the question. The ε is constant in the plane perpendicular to the plates. The x-axis is running along the plate.
  9. Nov 17, 2013 #8
    I am sorry. I was misreading the question. I was thinking "bottom OF THE plate" and "top OF THE plate." Your interpretation of it is correct, it is the "bottom plate" and "top plate."

    Thanks. This may have been where all my confusion came from. I'll have to start fresh on it.
  10. Nov 17, 2013 #9
    Yeah, I realized right after I posted, sorry. I feel like an idiot, I wasted a lot of time attempting the wrong problem.
  11. Nov 17, 2013 #10


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    No worries. Start it again. It is simpler than the "vertical" slicing.

  12. Nov 17, 2013 #11

    I was looking at this problem again, and your suggestion. I don't understand how finding the voltage the way you describe helps me find bound charge. Could you please elaborate?
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