- #1
rakso
- 18
- 0
- 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##
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
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