I first find the Green's function for the following PDE in $n=3$ dimensions, where $k:=|k|^2$. $$\nabla^2G(x,x')=\delta^3(x-x')$$ Upon Fourier transforming both sides, and inverting, I find that $$G(x,x')=-\int\frac{d^3k}{(2\pi)^3}\frac{e^{ik\cdot(x-x')}}{k^2}$$ Switching to spherical coordinates, we see that \begin{align} G(x,x')&=-(2\pi)^{-3}\int_0^{2\pi}d\phi\int_0^\infty dk \int_0^\pi d\theta \sin\theta\, e^{ik|x-x'|\cos\theta}\\ &=-\frac{2}{4\pi^2|x-x'|}\int_0^\infty dk \frac{\sin(k|x-x'|)}{k}\\ &=-\frac{1}{4\pi|x-x'|} \end{align} Poisson's equation for the electric potential is $$\nabla^2\phi=-\frac{\rho}{\epsilon_0}$$ With our Green's function, we can find $\phi$ for a case when it vanishes at $\infty$,
$$\phi(x)=-\frac{1}{4\pi\epsilon_0}\int d^3x' \frac{\rho(x')}{|x-x'|}$$ In the case of a point charge particle of charge $q$ at $x\in\mathbb{R}^3$, $\rho(x')=q\delta^3(x'-x)$. Thus, $$\phi(x)=-\frac{q}{4\pi\epsilon_0}\int d^3x' \frac{\delta^3(x'-x)}{|x-x'|}$$ However, this integral seems to not be well defined because the integrand is not finite when $x'=x$, or when the delta function is zero.

How do I recover the expected result that $$\phi(x)=-\frac{1}{4\pi\epsilon_0}\frac{q}{|x|}?$$

  • 3
    $\begingroup$ I think you have to put $\rho(x') = q\delta(x')$. If you put $\rho(x') = q\delta(x'-x)$, the point charge is a the same point as the potential it is asked for. Evidently, the result would be not defined. $\endgroup$ Jul 10, 2019 at 5:35

1 Answer 1


Your mistake is using $x$ for both the location of the point charge and the observation of the potential. Instead, introduce a third coordinate, say $x_0$, for the location of the point charge. Then you have

$\phi(x)=-\frac{q}{4\pi \epsilon_0}\int d^3x' \frac{\delta^3(x'-x_0)}{|x-x'|}$

Setting $x_0$ to the origin then produces the result you're looking for and coincides with the comment by Frederic Thomas.


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