- #1

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## Homework Statement:

- Am I tackling this problem in a correct way and if, is there any other way?

## Relevant Equations:

- ##\nabla \times \vec E = \nabla \times (-\nabla \phi) = 0##

Given $$\vec E = -\nabla \phi$$ there $$\vec d \rightarrow 0, \phi(\vec r) = \frac {\vec p \cdot \vec r} {r^3}$$ and ##\vec p## is the dipole moment defined as $$\vec p = q\vec d$$

It's quite trivial to show that ##\nabla \times \vec E = \nabla \times (-\nabla \phi) = 0##. However, I want to show that ##\nabla \cdot \vec E = - \nabla^2 \phi = 0##.

My assumption of this equal ##0##, is from Gauss Law $$ \iint_{\partial V} \vec E \cdot d\vec S = \iiint_V \nabla \cdot \vec E \, dV = \frac {Q_{encl}}

{\epsilon_{0}}$$

However, when ##\vec d \rightarrow 0## we can't enclose one of the charges without enclosing the other one with any surface, hence ##\Sigma Q = 0##,

thus ##Q_{encl} = 0 \leftrightarrow \nabla \cdot \vec E = 0##

It's quite trivial to show that ##\nabla \times \vec E = \nabla \times (-\nabla \phi) = 0##. However, I want to show that ##\nabla \cdot \vec E = - \nabla^2 \phi = 0##.

My assumption of this equal ##0##, is from Gauss Law $$ \iint_{\partial V} \vec E \cdot d\vec S = \iiint_V \nabla \cdot \vec E \, dV = \frac {Q_{encl}}

{\epsilon_{0}}$$

However, when ##\vec d \rightarrow 0## we can't enclose one of the charges without enclosing the other one with any surface, hence ##\Sigma Q = 0##,

thus ##Q_{encl} = 0 \leftrightarrow \nabla \cdot \vec E = 0##

**My questions:**Am I reasoning correct, and does it exist any other way to show it? I'd rather not do a bruteforce calculation of ##- \nabla^2 \phi ##. =D