Why Does Gauss' Law Imply Integral of Electric Field is Zero?

[Harrisonized]

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.

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 e_r + p sin θ /r³ e_θ

∇·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

 

[member 553083]

. Homework Equations Gauss' law (∇·E = 0) The Attempt at a SolutionI believe the answer to this question lies in the fact that Gauss' law is a vector equation, and thus it is being applied to the entire vector field, not just its components. This means that although the integral over dz may be nonzero, these values are cancelled out by the other components of the field. Therefore, the integral over the entire volume is indeed 0.