This is what I have drawn (see attachment). Is the basic idea here to integrate the electric field of the outer Gaussian surface from b to a? If so, what about the inner Gaussian surface?
My professor said to in order to solve this, integrate the electric field to find the electric potential...
\Delta V = -\int\stackrel{\rightarrow}{E}dl
My question is, using Gauss's Law, (\oint E_{n}dA = \frac{Q}{\epsilon}), how do I go about finding Q?
Isn't Q just the charge of the shell?
So I suppose V(x) = kq(\frac{2}{\sqrt{x^{2}+a^{2}}} + \frac{1}{a-x}) if x < a and V(x) = kq(\frac{2}{\sqrt{x^{2}+a^{2}}} + \frac{1}{x-a}) if x > a.
I think I can compute E_{x}(x) from here.. Thanks for all your help!
Hi all. I'm having a very hard time understanding this portion of Physics so please bear with me.
The furthest I got with this problem is deciding to use the sum of the potentials at each point to calculate the potential of the system. Something like...
\frac{kq_{1}}{r_{1}} +...