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My attempt:

$$\rho(\overrightarrow r) = Q\delta^3 (\overrightarrow r)$$

$\overrightarrow r =$ separation vector from the origin to $\overrightarrow a$

$Q$ = the total charge of the electric dipole

Here's what I don't understand though. The total charge of the dipole is $-q + q=0$, isn't it? Therefore, my volume charge density should be zero, no?

Also, what volume are we talking about here? The question implies that we are either talking about some arbitrary volume defined when attempting to solve the problem or the volumes of the individual point charges, which are zero.

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Note that whatever your answer is, when you integrate over a region of volume $V$ containing the charge $-q$, but excluding the positive charge, your total charge should be $-q$, since there is exactly that much charge in that volume. If there is a total charge $-q$ in a volume $V$, the charge density in that region is $-q/V$. In other words, $$ \int\limits_{\text{region including $-q$, excluding $+q$}}d^3r\,\rho(\vec r) = -q $$

Similarly, when you integrate over a region of volume $V$ containing the charge $+q$ but excluding the negative charge, your total charge should be $+q$. Last but not least, when you integrate over a region that contains both charges, your net charge should be zero.

Can you formulate a density function $\rho(\vec r)$ that satisfies these three properties? Remember the sifting property of the three-dimensional delta function: $$ \int_\mathbf{R^3}d^3r\,\delta^3(\vec r - \vec a) = 1. $$

Answer: (try to figure it out before revealing the solution)

$\rho(\vec r) = q[\delta^3(\vec r - \vec a) - \delta^3(\vec r)]$

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