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I am trying to show that the electric field inside a homogeneous distribution of superficial charge is of the order of magnitude of $\delta$, with: $$V(\textbf{r})=\int d^3\textbf{r'}\frac{\rho(\textbf{r}')}{|\textbf{r-r'}|^{1+\delta}}$$

The electric field can be found as $E=-\nabla V$, but the notation throws me off. This is what I tried: $$E=-\frac{d}{dr}\left(\int d^3\textbf{r'}\frac{\rho(\textbf{r}')}{|\textbf{r-r'}|^{1+\delta}}\right)=-\frac{d}{dr}\left(\frac{Q_{total}}{|\textbf{r-r'}|^{1+\delta}}\right)=(1+\delta)(\frac{Q_{total}}{r^{2+\delta}})$$

And this result just feels wrong, because it blows up at $r=0$, and it does not reduce to the "normal" case of $\delta =0$, $E=0$ inside. I know that I probably messed up the calculation badly, but as I don't usually have this kind of notation, I am quite lost as to how to approach it.

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  • $\begingroup$ A homogeneous distribution in what shape? $\endgroup$ – G. Smith Aug 22 at 0:38
  • $\begingroup$ it is not specified, only that it is a "homogeneous distribution of superficial charge" $\endgroup$ – Nick Heumann Aug 22 at 0:39
  • $\begingroup$ The second equality in your second equation is wrong because you integrated the numerator ignoring the $\mathbf{r}’$ in the denominator. $\endgroup$ – G. Smith Aug 22 at 0:40
  • $\begingroup$ I don’t know what superficial charge is. Does your text make that clear somewhere? $\endgroup$ – G. Smith Aug 22 at 0:41
  • $\begingroup$ I was thinking about that integration step, but I'm not sure how to do it, as I don't really know what exactly my "r"s are, so I kind of ignored it because it made sense from the dimensions (As the potential should be 1/r). As to the superficial charge thing, it is not specified more than what I wrote, all I'm told is that it is on the surface of the object. $\endgroup$ – Nick Heumann Aug 22 at 0:43

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