How to find distribution of charge in a conducting shell?

Question

A very long, non-conducting cylinder of radius R₁ has a uniform charge density of C₁ uC/m³. It is surrounded by a solid, conducting, cylindrical shell with an inner radius of R₁ and outer radius of R₂ and a linear charge density of L uC/m. The cylinder and the shell have the same geometric diameter.

Q) Numerically describe the distribution of charge in the conducting shell. Express values using units of uC/m.

Answer 1

To solve this problem you need to assume that the conductor has zero net charge,

meaning that it was not given any charge before the experiment.

Furthermore you need these two properties.

Property 1:

Electric field inside a conductor is 0.

The reason is that non-zero field would move any charges to the surface.

Property 2:

By Gauss's Law the electrical field flux across any closed surface

is proportional to the charge enclosed by the surface.

First let's describe qualitatively what happens.

Along the conductor's inner surface of radius R₁ there will accumulate some charges

of polarity opposite to the charge of the non-conducting cylinder.

And along the outer surface of radius R₂ there will accumulate an equal

amount of charge of same polarity as the non-conducting cylinder.

Let's form a Gaussian surface in the shape of a cylinder co-centric with the given cylinders.

Let its radius be some R between R₁ and R2 (R₁ < R < R₂).

And let its height be h.

Its top and bottom disc are parallel to the electrical field, and therefore the flux through them is 0.

The side of the Gaussian surface lies inside its conductor, where the electrical field is 0 (see Property 1).

Therefore the flux through the side is also 0.

Since the flux through the whole Gaussian surface is 0, the net charge inside must also be 0.

That means that the charge on the conductor's inner surface of radius R₁

must equal in magnitude the charge inside the nonconducting cylinder.

The amount of charge in the portion of the nonconducting cylinder enclosed

by the Gaussian surface is

Q = πR₁²hC₁

It is also the change along the length h of the conductor.

Therefore

L = Q/h = πR₁²C₁