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Homework Help: Polarization Charge on the surface of a spherical cavity

  1. Jul 23, 2011 #1
    1. The problem statement, all variables and given/known data
    The polarizatiob charge on the surface of a spherical capacitor is [tex] -\sigma_e \cos(\theta), [/tex] at a point whose radius vector from the centre makes an angle [itex] \theta [/itex] witha given axis Oz. Prove that the field strength at the centre is [tex] \frac{\sigma_e}{3 \epsilon_0}, [/tex]

    2. Relevant equations

    3. The attempt at a solution

    Well I not entirly sure how to aproach this problem. I tried exapanding the potentials inside and outside the sphere as:
    [tex] V_{in} = A_1 r \cos(\theta) + \frac{A_2}{r^2}\cos(\theta)
    V_{out} = B_1 r \cos(\theta) + \frac{B_2}{r^2}\cos(\theta) [/tex]

    Then since [itex] V_{in} \neq \infty, [/itex] [itex] A_2 = 0 ,[/itex]

    then saying that at r=R: [itex] D^{perpendicular}_{in} - D^{perpendicular}_{out} = \sigma_f [/itex] and assuming that both inside and outside have the same permitivitty then [tex] E^{radial}_{in} - E^{radial}_{out} = \frac{\sigma_f}{\epsilon_0} [/tex].

    [tex] A_1 + B_1 - \frac{B_1}{R^2} = \frac{\sigma_e}{\epsilon_0} [/tex]

    I am not entirely sure if I am aproaching this problem in the right way. Any help or advice on how to g o about solving this problem or problems like this would very much apreciated.
    Last edited: Jul 23, 2011
  2. jcsd
  3. Jul 23, 2011 #2
    While I am sure there is some clever way to apply spherical harmonic expansion here, the problem is more straight forwardly done by just use integration. Just do a surface integral over the electric field contribution of each area element. There is clear azimuthal rotation symmetry so that you only needs to compute the [itex]z[/itex] component. [tex]E_z = \frac{1}{4\pi \epsilon_0} \int \frac{\sigma(\theta) \cos\theta \, da}{R^2} \,,[/tex] where [itex]R[/itex] is radius of the sphere.
  4. Jul 23, 2011 #3
    Fistly thank yo for your help. Secondly I am having trouble seeing how yuo get to this integral and then how you solve it. Am I correct in saying this is a result of Gauss' Law?

    I have tried to compute the integral and i am unfortunately at a loss as to how to do this to get the desired answer:

    [tex] E_z = \frac{1}{4\pi \epsilon_0} \int \frac{\sigma(\theta) \cos\theta \, da}{R^2} [/tex]

    [tex] = \frac{1}{4\pi \epsilon_0} \int\int \frac{\sigma(\theta) \cos\theta \, rdrd\theta}{R^2}[/tex]

    [tex] = \frac{1}{4\pi \epsilon_0} \int r dr \int \frac{\sigma(\theta) \cos\theta \, d\theta}{R^2}[/tex]

    here is where i am stuck. Do I take limits, and if so how do I determine them, I have tried taking r=R to r=0 and [itex] \theta = 0, 2\pi. [/itex]? Thank you for you help.
  5. Jul 23, 2011 #4
    You have the wrong area element of a sphere. It should be: [itex]da = R^2 \sin \theta d\theta d\varphi[/itex].

    There is no variation in radial distance, so no [itex]dr[/itex].
    Last edited: Jul 23, 2011
  6. Jul 24, 2011 #5
    Ah ok that makes a lot more sense, thank you for your help.
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