Potential Difference between concentric shells

Homework Statement

The figure shows two concentric shells of radii R and 2R. A charge q is initially imparted to the inner shell. After the keys k1 and k2 are alternately closed n times, find the potential difference between the shells.

The Attempt at a Solution

Initially, the charge q is induced on the outer shell so that it has -q charge on its inner surface.
When the key k1 is pressed, the outer shell gets earthed.
So the potential which V1 between them is Kq/2R
Now when k2 is closed (after opening k1) the charge q on inner shell disappears so that it gains 0 potential. The charge -q which was induced on the outer surface remains as it is.
Now the potential is given by V2 = -Kq/2R

When k1 is closed again, outer shell gains 0 potential, But the inner shell has an induced charge of +Q. So potential is Kq/2R
In this way, the potential alternates with + and - sign. At the end of 'n', the potential is -Kq/2R

Any hint appreciated.

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rl.bhat
Homework Helper
Potential of the concentric spherical shell with radii a and b when outer shell is earthed is

V = q(b-a)/4πεοab.

When the inner shell is earthed

V = q(b-a)/4πεοb^2.

How did you get this expression-

"When the inner shell is earthed

V = q(b-a)/4πεοb^2. "

What is wrong with my approach? I would like to tackle the problem conceptually rather than using just formulae.

rl.bhat
Homework Helper
When you open k1, net charge on A is -q and on B is Q.
when you ground B, net potential of B is zero. It is due to potential of B, potential due to inner surface of A and potential due to outer surface of A.
Let -q' be the charge in the inner surface of A, x be the charge on the outer surface of A and +q' on B. So the potential on B is ( assuming the thickness of the shell is small compared to 2R.)
Kq'/R -Kq'/2R + Kx/2R = 0
From the charge conservation of the outer shell, -q = x -q' or x = q' - q.
So Kq'/R -Kq'/2R + K(q'-q)/2R = 0 or q' = q/2.

So V2 = Kq/2*2R

Last edited:
Thank you very much Sir!...you cleared my misconception. I thought that if potential on a surface is 0, q must be 0...which is not so.
Thanks once again!