- #1

pondzo

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## Homework Statement

I am not sure whether to put this in the introductory level or advanced. It seems to be relatively introductory in an electromagnetism course.

A spherical conductor of radius ##a## carries a charge ##q##. It is situated inside a concentric spherical conducting shell of inner radius ##b## and outer radius ##c##, ##a<b<c##, which carries a charge ##Q##. These two are in turn surrounded by a spherical conductor radius ##R##, which is earthed. Find the potentials at all points in the regions between these conductors.

## The Attempt at a Solution

I thought there might be three different ways I could do this, but it seems for one I get significantly different answers and I would like to know why (also I am not sure if any of my answers are correct). I will use the region ##c<r<R## to show each method.

**Method One:**Compute the electric field in the region using Gauss' law and then find the potential difference between two points using the path integral ##\phi_{ab}=\int_a^b\vec{E}\cdot\vec{dr}##.

In the region ##c<r<R## the electric field is: ##\vec{E}=\frac{Q+q}{4\pi\varepsilon_0r^2}\hat{r}##. Now the potential difference between a point r (c<r<R) and R is: $$\phi_{rR}=\int_r^R\frac{(Q+q)}{4\pi\varepsilon_0 r'^2}dr'=\frac{(Q+q)}{4\pi\varepsilon_0}\big(\frac{1}{r}-\frac{1}{R}\big)$$.

By some luck this works in that at r=R the potential is 0. But this is only because it is giving the potential difference between the same point. For this reason, and because this gives the potential relative to R, I don't think this is the answer the question seeks. (Consider if the potential at R was set to, say, 5. There are no boundary conditions I have the opportunity to use). Do you agree?

**Compute the electric field in the region using Gauss' law and then compute the potential using ##\vec{E}=-\vec{\nabla}\phi##.**

Method Two:

Method Two:

Assuming zenith and azimuthal symmetry then ##-\vec{\nabla}\phi## in spherical coordinates reduces to ##-\frac{\partial \phi}{\partial r}\hat{r}##. Which implies ##E(r)=-\frac{\partial \phi}{\partial r}##. So we have that in the region ##c<r\leq R## where the electric field is the same is mentioned before : ##\phi(r)=\frac{(Q+q)}{4\pi\varepsilon_0 r} +m## for some constant m. Using the boundary condition ##\phi(R)=0##, I get that ##m =\frac{-(Q+q)}{4\pi\varepsilon_0}## so that ##\phi(r)=\frac{(Q+q)}{4\pi\varepsilon_0}\big(\frac{1}{r} -\frac{1}{R}\big)##.

This agrees with method one, but not by luck. For this reason I think that this method is probably the most correct. What do you think?

**Method Three:**Solve laplaces equation for the potential in the region ##c<r<R## with the boundary conditions ##\phi(R)=0## and ##\phi(c)=V_c##.

Assuming spherical symmetry and no angular dependence, ##\nabla^2\phi=0\implies \frac{1}{r^2}\frac{\partial}{\partial r}\big(r^2\frac{\partial \phi}{\partial r}\big)=0\implies \phi=\frac{-m}{r}+n## for some constants m and n. Using the boundary conditions I conclude that ##\phi=\frac{cV_c}{c-R}\big(1-\frac{R}{r}\big)##.

This potential works in that when r=R the potential is zero, and at r=c the potential equals Vc=Vb, but only because I constructed it so. The problem I have with this method is that 1. The answer appears to be significantly different to that of method 2 (which I believe to be correct) and 2. There seems to be some ambiguity in what Vc, Vb and Va is, and it doesn't seem that I can calculate them using this same method, but maybe if I used method two. Or maybe this method is completely wrong. What do you think? Why does method 2 and 3 seemingly give different answers?