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Write an expression for the volume charge density $\rho(\mathbf{r})$ of a point charge $q$ at $\mathbf{r}'$ in terms of the Dirac delta function.

My attempt:

Let $\mathbf{r}-\mathbf{r'}=\mathbf{R}$.

$$\mathbf{E}(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\frac{q}{R^2}\hat{\mathbf R}$$

Taking the divergence of both sides, \begin{align} \nabla\cdot\mathbf{E}(\mathbf{r}) & =\frac{q}{4\pi\epsilon_0}\nabla\cdot \left(\frac{\hat{\mathbf{R}}}{R^2}\right) \\ \implies\frac{\rho(\mathbf{r})}{\epsilon_0} & =\frac{q}{4\pi\epsilon_0} 4\pi\delta^3(\mathbf{R}) \\ \implies\rho(\mathbf{r}) & =q\delta^3(\mathbf{r}-\mathbf{r}') \tag{1} \end{align}

However, the dimensions of the left hand side are 'charge per volume' whereas the dimensions of the right hand side are just 'charge' as $\delta^3(\mathbf{r}-\mathbf{r}')$ is dimensionless. Where have I gone wrong?

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    $\begingroup$ You should not need to derive it in this complicated way. The expression for $\rho$ is obvious if you understand delta functions. It represents a point charge at $\mathbf{r}’$ because the delta function is zero elsewhere. And the charge of that point charge is $q$ because that is the value of the integral of the charge density over space. $\endgroup$ – G. Smith May 6 at 5:29
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Consider \begin{align} \int_{\mathrm{all \; space}} \delta^{3}(\mathbf{r}-\mathbf{r^{\prime}}) \, d^{3}\mathbf{r} \; = \; 1 \end{align}

The right hand side (i.e. 1) is dimensionless, and the differential element $d^{3}\mathbf{r}$ has the dimension of volume. Therefore $\delta^{3}$ has the dimension of "per volume".

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    $\begingroup$ Alternatively, the relationship used by OP, $\nabla\cdot \left(\frac{\hat{\mathbf{R}}}{R^2}\right) = 4\pi \delta^3(\mathbf{R})$, also implies that $\delta^3(\mathbf{R})$ has inverse-volume dimensions. $\endgroup$ – Emilio Pisanty May 6 at 14:07
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You got the right answer. The 3D delta function is not dimensionless; it has dimensions of inverse volume. Think about how its volume integral must give the dimensionless value 1.

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