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At class we have been told that in an area with no charged particles, the Laplace equation holds ($\phi$ is a scalar potential function) : $$\Delta\phi=0$$

I tried to take the example of a single charged particle at the origin, its potential function in relation to infinity is known as $$\phi(r)=\frac{kq}{r}$$

At point, let's say $r=2$, there are no charged particles, and therefore the Laplace equation should hold.

But I get that $\Delta\phi =\frac{kq}{r^3} \neq 0$. What is my misunderstanding?

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    $\begingroup$ The LHS of your equation for a point charge is a delta function. It's described here. $\endgroup$
    – StephenG - Help Ukraine
    Commented Feb 4, 2017 at 12:14
  • $\begingroup$ And since a single charged particle acts like a delta function, the Laplacian I was calculating is "incorrect" and that's why I receive this odd result? @StephenG $\endgroup$
    – Taru
    Commented Feb 4, 2017 at 12:25
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    $\begingroup$ @StephenG That's completely irrelevant here, the problem is that OP did not calculate the Laplacian in spherical coordinates correctly. $\endgroup$
    – ACuriousMind ♦
    Commented Feb 4, 2017 at 13:01

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You haven't computed the Laplacian correctly; presumably because you did not consider that the Laplacian in spherical coordinates is not simply taking the derivative w.r.t. $r$ twice.

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  • $\begingroup$ Actually I did use this formula and still received a non zero expression (also WolframAlpha received the same result). Could you verify? $\endgroup$
    – Taru
    Commented Feb 4, 2017 at 14:14
  • $\begingroup$ @Taru The expression becomes 0 rather straightforwardly using the correct formula since $r^2 \partial \phi / \partial r$ is constant. $\endgroup$
    – ACuriousMind ♦
    Commented Feb 4, 2017 at 14:16
  • $\begingroup$ You're right, I used cylindrical instead. English translation mistake. Thank you. $\endgroup$
    – Taru
    Commented Feb 4, 2017 at 14:18

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