My textbook provides the following problem:

Consider a spherical conducting shell with inner radius $R_2$ and outer radius $R_3$, that has other spherical conductor inside it with radius $R_1$ (this one is solid). Initially the 2 spheres are connected by a wire. We put a positive charge $Q$ on the sphere and after some time we remove the wire that connected the 2 spheres.

What is the charge distribution of charge on the 2 spheres after we remove the potential? If we have charged the inner sphere instead, would the result be different?

The answer provided was that the charge would all flow to the outer surface of the spherical shell, and the same would happen if we had placed $Q$ on the inside of the conductor shell.

Now I'm trying to find an explanation to this.

At first I thought that because the electric field between $R_2$ and $R_3$ needs to be zero by Gauss law (we are inside a conductor), then the interior charge needs to be zero. Because of that all the charge will flow to the outside. But I don't think that makes sense because we could still have charge outside the inner sphere and on the inside surface of the shell if those charges would be on equilibrium... Can someone please clarify me about this? Also I don't understand what would happen if we didn't add charge $Q$ to our system, would the result be different if we didn't add it?


1 Answer 1


Consider this case: give the outer shell a charge $q_b$ and the inner shell $q_b$. also let $k=\frac{1}{4\pi \epsilon}$ and $R_3=q_b$ and $R_1=q_a$.potential on inner shell my be given by $V_a=$ $k[\frac{q_a}{r_a}-\frac{q_b}{r_b}]$ and for the outer one $V_b=$ $k[\frac{q_a}{r_b}-\frac{q_b}{r_b}]$( asuming the sphere to be a conductor). convince yourself this first. now if you calculate the potential difference you get $V_a -V_b=$$k q_a [\frac{1}{r_a}-\frac{1}{r_b}]$.note that this result is independent of the carge on the outer sphere($q_b$). if $q_a$ is +ve then $V_a -V_b$ is +ve (as $r_a>r_b$), i.e. $V_a>V_b$. so if the spheres are conected by a conducting wire the charges floe form the inner sphere to the outer as +ve charges flow form high potential to lower.. this flow will continue till $V_a=V_b$ is achieved, in which case $q_a$ needs to bee necessarily be zero, which implies that all the charges has flowed from the inner shell to the outer. you can also think likewise for -ve charges. the amusing this thing is you will get the same answer.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.