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Harrisonized
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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.
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
https://imagizer.imageshack.us/v2/534x85q90/607/0io5.png
Eq. (1.21):
https://imagizer.imageshack.us/v2/534x152q90/855/w4zh.png
https://imagizer.imageshack.us/v2/418x295q90/829/lqdy.png
The Δ actually means ∇2
Homework Statement
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
https://imagizer.imageshack.us/v2/534x85q90/607/0io5.png
Eq. (1.21):
https://imagizer.imageshack.us/v2/534x152q90/855/w4zh.png
https://imagizer.imageshack.us/v2/418x295q90/829/lqdy.png
The Δ actually means ∇2
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