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Surface-charge of a uniformly polarized sphere

  1. Jun 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    Please take a look at #2 in http://www.physics.utoronto.ca/~hori/Courses/P352F/hw6.pdf

    (this is not homework, I'm just training).

    Ok, I can find the electric field inside the cavity y superposition and it is E = (-1/ 3*e_o)*P. I know that the bound surface charge is P*cos(theta), where theta is the usual spherical coordinate. My question is: Is the total charge on the inner surface P*cos(theta)*4*Pi*(R_0)^2?
  2. jcsd
  3. Jun 30, 2008 #2
    Where is [tex]\theta[/tex] in your last sentence? In order to find the total charge, you must integrate - not multiply - since the surface charge varies with respect to [tex]\theta[/tex]. The proper way to find the surface charge would be

    [tex]Q = \int_S \sigma(\theta) \, da = \int_0^{2\pi} d\phi \int_0^{\pi} \cos \theta \sin \theta d\theta = 0[/tex]

    It's zero, because on the northern hemisphere the bound charge is positive, which is compensated by the symmetrically negative bound charge on the southern hemisphere.

    In conclusion, the answer is no.
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