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## Homework Statement

Show that with [itex]d[/itex] spatial dimensions the potential [itex]\phi[/itex] due to a point charge [itex]q[/itex] is given by

[tex]\phi (r) = \frac{\Gamma(\frac{d}{2}-1)}{4\pi^{d/2}}\frac{q}{r^{d-2}}[/tex]

## Homework Equations

## The Attempt at a Solution

The electric field strength is known to be:

[tex]E(r) = \frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}}[/tex]

[tex]E(r) = \sqrt{(\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}})^2}[/tex]

[tex]E(r) = \sqrt{\frac{1}{d}\sum_{i=1}^d(\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}})^2}[/tex]

[tex]E(r) = \sqrt{\sum_{i=1}^d(\frac{1}{\sqrt{d}}\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}})^2}[/tex]

Since the magnetic field will be the same in all directions, we know that

[tex]\vec {E}(r) = (\frac{1}{\sqrt{d}}\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}},\frac{1}{\sqrt{d}}\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}},\cdots)[/tex]

[tex]\phi(r) = -\int\frac{1}{\sqrt{d}}\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}}dr[/tex]

[tex]\phi(r) = \frac{1}{\sqrt{d}}\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{1}{(d-2)}\frac{q}{r^{d-2}}[/tex]

[tex]\phi(r) = \frac{1}{\sqrt{d}}\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{1}{2(\frac{1}{2}d-1)}\frac{q}{r^{d-2}}[/tex]

[tex]\phi(r) = \frac{1}{\sqrt{d}}\frac{\Gamma(\frac{d}{2}-1)}{4 \pi^{d/2}}\frac{q}{r^{d-2}}[/tex]

Why am I getting an extra [itex]\frac{1}{\sqrt{d}}?[/itex]