# What is the charge density of a pure electric dipole?

I've only found two definitions of the charge density of a pure electric dipole: $$\rho(\vec{r})=\vec{p}\cdot\nabla\delta(\vec{r}-\vec{r}_0)$$ and $$\rho(\vec{r})=-\vec{p}\cdot\nabla\delta(\vec{r}-\vec{r}_0).$$

Neither of these came with reasons or derivations, so I'm not really sure which to use?

• This seems to be a dipole in the limit of zero distance between the charges and finite dipole moment. I am not sure what you mean by "pure" dipole. Is that the opposite of an "impure" dipole? Commented Feb 6, 2018 at 2:21
• @freecharly pure dipole is just a term for a "point" dipole in the same sense as a "point" charge.
– Karl
Commented Feb 6, 2018 at 16:45

In electrostatics, when you write the multipole expansion of the potential you find $$\Phi(\vec{x})=\frac{1}{4 \pi \epsilon_0} \left[ \frac{q}{r} + \frac{\vec{p} \cdot \vec{x}}{r^3} + ... \right] \, ,$$ where $r=|\vec{x}|$ and $...$ indicates the higher order multipole terms. We can find the effective charge distribution with $$\nabla^2 \Phi = - \frac{\rho}{\epsilon_0} \, .$$
Now we just need appropriate expressions for the laplacian of the terms we have. Remember that $$\frac{\vec{x}}{r^3} = - \nabla \frac{1}{r} \quad \mathrm{and} \quad \nabla^2 \frac{1}{r} = - 4 \pi \, \delta(\vec{x}) \, .$$ Now we will be able to reproduce the result if we notice that
Check each equality carefully. Hence we find the effective charge density $$\rho(\vec{x}) = q \, \delta(\vec{x}) - \vec{p} \cdot \nabla \delta(\vec{x}) \, + \, ... \, .$$