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The original problem:

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I read this one, but there's something that is still unclear, even after going through all their answers and comments.

Why doesn't the expression use the 3-dimensional Dirac Delta function? That is, shouldn't the expression be: $$\rho (\overrightarrow r ) = q \delta^3 (\overrightarrow r - \overrightarrow r')$$

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That's just a notational issue. $\delta(\vec{r} - \vec{r}')$ does refer to the 3D delta function, as you can tell because it has vectors inside of it. It's probably more clear to write it as $\delta^3(\vec{r} - \vec{r}')$ but it means the same thing.

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  • $\begingroup$ Wait, so are there only 1-D and 3-D Dirac Delta function definitions? Because I was going to say: What if those vectors are only in the xy-plane? Therefore, the expression would be ambiguous. $\endgroup$ – whatwhatwhat Sep 5 '16 at 21:35
  • $\begingroup$ @whatwhatwhat You can have delta functions in any number of dimensions - the dimension of the delta function is the same as the dimension of the vector inside of it. $\endgroup$ – tparker Sep 5 '16 at 21:38

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