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Field strength giving rise to a polarization charge of -s.cos(theta) on a cavity

  1. Dec 9, 2008 #1

    xn_

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    Hello,
    I'm learning EM from Bleaney & Bleaney and got stuck on Ex2.1 (can do Ex2.2-2.7, though...) - If the polarization charge on the surface of a spherical cavity is -s.cos(theta), prove that the field strength at the centre is s/3e0. If I expand V(r) within the cavity as A.r.cos(theta) + B.r^-2.cos(theta), then I agree B=0 or else V(0) would be singular, but that leaves the A term and surface charge = e0Er = -e0(dV/dr) = -e0A.cos(theta) and hence A = s/e0 not s/3e0 - what am I doing wrong?
    Thanks,
    Julia
     
  2. jcsd
  3. Dec 10, 2008 #2

    gabbagabbahey

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    Check your boundary condition. Is that statement really true when the potential outside the boundary is not constant in 'r'?:wink:
     
  4. Dec 11, 2008 #3

    xn_

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    I've been thinking that the potential outside the cavity (radius R) could be written Vout(r) = B_1.r.cos(theta)+B_2.r^-2.cos(theta), then, and the boundary conditions are that Vout(R)=Vin(R) which implies:
    A = B_1 + B_2 R^-2
    and that e0E_in,radial(R) = eE_out,radial(R) (where e is the permittivity of the medium surrounding the cavity) which implies:
    -e0.A = -e(B_1 - 2B_2.R^-3)
    But your reply (thanks!) suggests that I'm doing something fundamentally wrong here - am I wrong to take Vin = A.r.cos(theta) because the Vout depends on r? I see that something of the form B.r would help, but I thought each power of r had do go with a certain power of cos(theta) to make a valid spherical harmonic?
    Thanks, Julia
     
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