The first equation is
E(r) = \sqrt{\frac{1}{d}\sum_{i=1}^d\left(\frac{\Gamma(\frac{d}{2})}{2 \pi^{d/2}}\frac{q}{r^{d-1}}\right)^2}
taking \frac{1}{d} through the sum yields
E(r) = \sqrt{\sum_{i=1}^d\frac{1}{d} \left(\frac{\Gamma(\frac{d}{2})}{2...
19:
You have very nearly solved this one. You have found z \geq r and z \leq \sqrt{1-r^2} , so all you need to do is find the minimum and maximum values for r and you will have the triple integral
\int_{\min(r)}^{\max{r}}\int_r^{\sqrt{1-r^2}}\int_0^{2\pi}rd\theta dz dr
13:
You forgot...
Homework Statement
Show that with d spatial dimensions the potential \phi due to a point charge q is given by
\phi (r) = \frac{\Gamma(\frac{d}{2}-1)}{4\pi^{d/2}}\frac{q}{r^{d-2}}
Homework Equations
The Attempt at a Solution
The electric field strength is known to be:
E(r) =...