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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 = 1/r

= -2p cos θ /r

= -2p cos θ /r

Logically, I wouldn't expect ∫

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 ∇

## 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 /r^{3}**e**_{r}+ p sin θ /r^{3}**e**_{θ}∇·E = 1/r

^{2}∂/∂r (2p cos θ /r) + 1/(r sin θ) ∂/∂θ (p sin^{2}θ/r^{3})= -2p cos θ /r

^{4}+ [1/(r sin θ)] 2p sin θ cos θ/r^{3}= -2p cos θ /r

^{4}+ 2p cos θ /r^{4}= 0Logically, 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}
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