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Integral of Electric Field is Zero

  1. Jan 14, 2014 #1
    Hi everyone. This isn't a homework problem. Rather, I'm trying to understand how the δ term arises from the field of a dipole.

    1. The problem statement, all variables and given/known data

    Greiner supplies the following one-line derivation, which is easy to follow I guess, but doesn't make logical sense to me. Specifically, I don't understand how Gauss' law implies that ∫ E dV = 0. Gauss' law says that div(E) = 0, and if you actually do calculate that in spherical coordinates, you'll find that the dipole field indeed has 0 divergence.

    E = 2p cos θ /r3 er + p sin θ /r3 eθ

    ∇·E = 1/r2 ∂/∂r (2p cos θ /r) + 1/(r sin θ) ∂/∂θ (p sin2 θ/r3)
    = -2p cos θ /r4 + [1/(r sin θ)] 2p sin θ cos θ/r3
    = -2p cos θ /r4 + 2p cos θ /r4= 0

    Logically, I wouldn't expect ∫ E dV = 0, because right in the center of two opposite point charges, there should clearly be an electric field pointing in the -z direction. Therefore, the integral over dz should always be nonzero. What am I doing wrong?

    https://imagizer.imageshack.us/v2/534x313q90/46/gjlo.png [Broken]
    https://imagizer.imageshack.us/v2/534x85q90/607/0io5.png [Broken]

    Eq. (1.21):
    https://imagizer.imageshack.us/v2/534x152q90/855/w4zh.png [Broken]

    https://imagizer.imageshack.us/v2/418x295q90/829/lqdy.png [Broken]

    The Δ actually means ∇2
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
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