# How to verify generally that the formula for electric potential is a solution to Poisson's equation?

$$V(\vec{a})=\frac{1}{4 \pi \epsilon_0}\int_\tau \frac{\rho(\vec{r})}{l}d\tau$$

This is the formula for the potential at a general point $$\vec{a}$$. Note that $$l$$ in this formula is the magnitude of the vector $$\vec{l}=\vec{a}-\vec{r}$$.

According to Poisson's equation, the equation above should satisfy:

$$\nabla^2V=-\frac{\rho(\vec{r})}{\epsilon_0}$$

Is it possible to verify generally that the potential formula is in fact a solution to Poisson's differential equation, by applying the Laplacian to both sides? If so, how would this be done?

## 1 Answer

Yes. This relies on the fact that $$\nabla^2_\vec{a} \left(\frac{1}{l} \right) = - 4\pi \delta^{(3)}(\vec{a} - \vec{r}),$$ where $$\delta^{(3)}(\vec{a} - \vec{r})$$ is the three-dimensional Dirac delta function, and $$\nabla^2_\vec{a}$$ is the Laplacian with respect to $$a_x$$, $$a_y$$, and $$a_z$$. If you apply this identity, the "picking property" of the delta functions causes the desired relation $$\nabla^2 V = - \rho/\epsilon_0$$ to fall out immediately.

There are a couple of ways to prove this; the easiest way is to consider the related quantity $$\nabla^2 \left( \frac{1}{r} \right) = \vec{\nabla} \cdot \left[ \vec{\nabla} \left( \frac{1}{r} \right) \right],$$ integrate it over a spherical region of radius $$R$$, and show that the result is always $$-4\pi$$ regardless of $$R$$. Since this is exactly the property that the delta-function has, you can conclude that the equation above is valid in a distributional sense.

• By $\nabla ^2_\vec{a}$ do you mean $\frac{\partial}{\partial a_x}\vec{i}+\frac{\partial}{\partial a_y}\vec{j}+\frac{\partial}{\partial a_z}\vec{k}$? – Pancake_Senpai Apr 13 at 19:19
• @Pancake_Senpai: No, in terms of coordinates it would be $\frac{\partial^2}{\partial a_x^2} + \frac{\partial^2}{\partial a_y^2} + \frac{\partial^2}{\partial a_z^2}$. – Michael Seifert Apr 13 at 22:09
• Oh sorry, that's my bad. That's what I meant to write. Thank you for your response. – Pancake_Senpai Apr 14 at 12:59