Calculating the surface charge of a sphere and a conducting shell

[Potatochip911]

Homework Statement

(Problem 2.38 From Griffth's Electrodynamics): A metal sphere of radius R, carrying charge q, is surrounded by a thick concentric metal shell (inner radius a, outer radius b). The shell carries no net charge.

Find the surface charge density $\sigma$ at R, a, and b.

Homework Equations

$\sigma = \frac{\mbox{Charge}}{\mbox{Surface Area}}$

The Attempt at a Solution

Since the metal sphere of radius R contains charge q, in order for the electric field to be 0 inside the conducting shell there must be charge -q at radius a which implies charge +q at radius b as the shell carries no net charge which gives $$\sigma_a=-\frac{q}{4\pi a^2}\\\sigma_b=\frac{q}{4\pi b^2}$$

Now what I'm confused about is that it just mentions that the metal sphere of radius R carries charge q and not whether it is a surface charge distribution or volume charge distribution. In the solutions manual they just give $\sigma_R=\frac{q}{4\pi R^2}$ as if all the charge is on the surface although I'm not sure this makes sense.

 

[conscience]

Homework Statement

(Problem 2.38 From Griffth's Electrodynamics): A metal sphere of radius R, carrying charge q, is surrounded by a thick concentric metal shell (inner radius a, outer radius b). The shell carries no net charge.

Find the surface charge density $\sigma$ at R, a, and b.

Homework Equations

$\sigma = \frac{\mbox{Charge}}{\mbox{Surface Area}}$

The Attempt at a Solution

Since the metal sphere of radius R contains charge q, in order for the electric field to be 0 inside the conducting shell there must be charge -q at radius a which implies charge +q at radius b as the shell carries no net charge which gives $$\sigma_a=-\frac{q}{4\pi a^2}\\\sigma_b=\frac{q}{4\pi b^2}$$

Now what I'm confused about is that it just mentions that the metal sphere of radius R carries charge q and not whether it is a surface charge distribution or volume charge distribution. In the solutions manual they just give $\sigma_R=\frac{q}{4\pi R^2}$ as if all the charge is on the surface although I'm not sure this makes sense.

Can charge reside within the volume of the metal sphere ?

You are missing a very important property of conductors under electrostatic conditions .

 

[Potatochip911]

Can charge reside within the volume of the metal sphere ?

You are missing a very important property of conductors under electrostatic conditions .

I thought that charge only entirely resided on the surface of conductors otherwise why would they mention this as a property of conductors and not just in general?

After looking around it seems like the charge will always distribute across the surface of anything in order to minimize the potential energy.

 

[conscience]

After looking around it seems like the charge will always distribute across the surface of anything in order to minimize the potential energy.

Is that the case if charge is given to an insulator ? Will charge reside on the surface of a non conductor as well ?

By the way , is there any confusion in the metal sphere being a conductor ?

 

[Potatochip911]

Is that the case if charge is given to an insulator ? Will charge reside on the surface of a non conductor as well ?

By the way , is there any confusion in the metal sphere being a conductor ?

So in an insulator the electrons can't flow freely therefore they won't be able to redistribute across the surface?

Yes, is it just a conductor because it's metal?

 

[conscience]

So in an insulator the electrons can't flow freely therefore they won't be able to redistribute across the surface?

Yes . Charges aren't mobile in an insulator unlike conductors . In conductors , whatever charge is given ends up on the surface .