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This relates to David Tong's notes on electromagnetism, specifically page 18.

We have as our charge density a delta function:

$$\rho(\vec x)=Q\delta^{(3)}(\vec x).$$

We want to solve:

$$\nabla^2\phi=-\frac{Q}{\epsilon_0}\delta^{(3)}(\vec x) \tag{2.14}.$$

To do this we postulate that:

$$\phi=\frac{\alpha}{r},$$

solves $\nabla^2\phi=0$, and after some vector calculus and some algebra we find that it does. However, next we use Gauss' law (with $S$ a sphere centred on the delta function) and find that:

$$\int_S\nabla\phi\cdot d\vec S=Q/\epsilon_0,$$

which, after working through the calculation and inserting our equation for $\phi$ leads to:

$$\phi=\frac{Q}{4\pi r\epsilon_0} \tag{2.16}.$$

I see how we do this, but, if I apply the divergence theorem to the above integral, I find:

$$\int_V \nabla^2\phi \text{ }dV=0,$$

since, as already established, $\nabla^2\phi=0$? What is going on here?

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Function

$$ \phi = \frac{\alpha}{r} $$

obeys the Laplace equation

$$ \Delta \phi = 0 $$

everywhere except at at $\vec{x} = \vec{0}$. At $\vec{0}$, $\phi$ has a singularity and so does $\Delta \phi$.

So when we integrate $\Delta \phi$ we cannot just put in $\Delta \phi = \vec{0}$. That would ignore the singularity.

The Gauss divergence theorem does not work if we restrict our considerations to cases where $\Delta \phi$ is a function.

But it works again if we define $\Delta\phi$ to be a distribution. The correct expression to put in is $$\Delta \phi = -\frac{Q}{\epsilon_0}\delta^{(3)}(\vec{x}) $$ as assumed in 2.14.

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  • $\begingroup$ Solved, thank you very much! $\endgroup$ – Charlie Jan 3 at 14:07

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