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Cylindrical capacitor with varying dielectric

  1. Jul 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider a long cylindrical coaxial capacitor with an inner conductor of radius a, and outer conductor of radius b, and a dielectric with a relative electric permittivity or dielectric ε(r), varying with the cylindrical radius. The capacitor is charged to the voltage V. Choose the radial dependence of ε(r) such that the energy density in the capacitor is constant.
    Calculate the electric field inside the capacitor.

    2. Relevant equations
    Energy density equation

    3. The attempt at a solution
    from the energy density eq. which is defined by [itex]u = \frac{1}{2} \textbf{E} \circ \textbf{D}[/itex] we get that
    1/ε(r) [itex]\propto \left| \textbf{E} \right|^{2}[/itex]

    [itex]\Phi(b) - \Phi(a) = V[/itex]

    Honestly I'm not quite sure which ansatz one can use. I was trying to solve it with the Gauss's law but didn't find a satisfying solution.
     
  2. jcsd
  3. Jul 14, 2012 #2

    TSny

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    You might try using the basic equation [itex]\cdot[/itex]D = [itex]\cdot[/itex](εE) = [itex]\rho[/itex] along with the energy condition ε(r)E2 = constant to get a differential equation for ε(r).

    ρ represents the free charge density which would be zero between the plates.
     
  4. Jul 14, 2012 #3

    gabbagabbahey

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    I think Gauss' Law for dielectrics is probably the easiest way to go here. You say you did not get a satisfying answer....can you post your attempt so we can see where you may be going wrong?
     
  5. Jul 14, 2012 #4

    TSny

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    brainslush,

    Take gabbagabbahey's advice! My suggestion will get you there, but not as easily.:redface:
     
  6. Jul 16, 2012 #5
    [itex]\nabla \circ \textbf{D} = \epsilon (\nabla \circ \textbf{E}) + (\nabla \circ \epsilon) \textbf{E} = 0[/itex]

    Now this gives the diferential eq. (we only care about the radial direction)

    [itex]\frac{\frac{\partial}{\partial \rho} \epsilon (r)}{\epsilon (r)} = -\frac{\frac{\partial}{\partial r} E(r)}{E(r)}[/itex]

    which then resolves into

    [itex]\epsilon = \frac{c}{\left|E\right|}[/itex]

    Which somewhat makes no sense to me, since this would imply that E is constant and therefore epsilon is constant. I probably messed something up.
     
  7. Jul 16, 2012 #6

    ehild

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    Your equation is wrong. Check out divergence in cylindrical coordinates.
    http://en.wikipedia.org/wiki/Divergence

    Anyway, it would be easier to use Gauss' Theorem for the displacement vector D. Edit: in integral form...

    ehild
     
    Last edited: Jul 16, 2012
  8. Jul 16, 2012 #7

    TSny

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    ε
    Sorry for leading you off onto this path! If you really want to do it this way, then as echild pointed out you need to express the divergence in cylindrical coordinates. [(By the way, in the last term of your first equation you should have a gradiant of ε rather than a divergence (which wouldn't make sense).] Anyway, it is easier to first solve the differential equation ∇[itex]\cdot[/itex]D = 0 for D in the region between the plates and then relate E to D via the dielectric constant. You can then discover the functional form of ε by the requirement of constant energy density. This avoids having to find a differential equation for ε(r), although it's not particularly difficult to do so.

    Of course ∇[itex]\cdot[/itex]D = ρ is just the differential form of Gauss' law for D. The integral form of Gauss' law for D will allow you to easily find the form of D between the plates, as others have suggested.
     
  9. Jul 16, 2012 #8

    gabbagabbahey

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    Just to clarify, my suggestion was to use Gauss' Law in integral form. Using it in differential form is also very easy provided you are comfortable with taking the divergence of a function in cylindrical coordinates - if you just apply the divergence directly to the product [itex]\mathbf{D}=\epsilon(r)E(r)\hat{r}[/itex] without expanding it using the product rule, the differential equation you get will be trivial to solve.
     
  10. Jul 16, 2012 #9
    Of course...

    [itex]\nabla \circ (\epsilon \textbf{E}) = (\nabla \epsilon) \circ \textbf{E} + \epsilon (\nabla \circ \textbf{E}) \Rightarrow [/itex]

    [itex]\frac{\frac{\partial}{\partial \rho} \epsilon (\rho)}{\epsilon (\rho)} = -\frac{\frac{\partial}{\partial \rho} (\rho E(\rho))}{\rho E(\rho)} = -\frac{1}{\rho}-\frac{\frac{\partial}{\partial \rho} E(\rho)}{E(\rho)} \Rightarrow[/itex]

    [itex]\epsilon = \frac{c_{0}}{\rho \left| E(\rho) \right|} [/itex] by using the condition [itex]\epsilon \propto \frac{1}{\left| \textbf{E} \right|^{2}}[/itex] we get

    [itex]\epsilon = \left(\frac{c_{1}}{\rho}\right)^{2}[/itex]

    So we can say that [itex] E \propto \rho[/itex]

    then we can use [itex]\int_{a}^{b} E d\rho = V = \int_{a}^{b} c \rho d\rho[/itex]

    to get [itex]c = \frac{2V}{b^{2}-a^{2}}[/itex]

    [itex]E = \frac{2V \rho}{a^{2}-b^{2}}[/itex]
     
  11. Jul 16, 2012 #10

    TSny

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    Looks good. Note that you could also write ∇[itex]\cdot[/itex](εE) = 0 as [itex]\frac{1}{ρ}\frac{d(ρεE)}{dρ}[/itex] = 0 which immediately gives ρεE = co
     
  12. Jul 17, 2012 #11
    Thanks for the help
     
  13. Jul 17, 2012 #12

    gabbagabbahey

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    Careful with your signs :wink:
     
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