1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook