# Electric field inside a homogenous distribution for slightly different Coulomb's law

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.

• A homogeneous distribution in what shape? – G. Smith Aug 22 at 0:38
• it is not specified, only that it is a "homogeneous distribution of superficial charge" – Nick Heumann Aug 22 at 0:39
• The second equality in your second equation is wrong because you integrated the numerator ignoring the $\mathbf{r}’$ in the denominator. – G. Smith Aug 22 at 0:40
• I don’t know what superficial charge is. Does your text make that clear somewhere? – G. Smith Aug 22 at 0:41
• 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. – Nick Heumann Aug 22 at 0:43