Recent content by babyEigenshep

That was indeed what I was trying to argue. I feel quite silly now. Thankyou for your help

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...

Actually, looking at your work again, you already found r\leq \sqrt{1/2}. Since this is the only restriction on r, the minimum will be 0.

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) =...