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

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:


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?


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.

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

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