Calculating the surface charge of a sphere and a conducting shell

In summary, the question asks to find the surface charge density at different radii of a metal sphere surrounded by a concentric metal shell. The shell carries no net charge. The solution reveals that the charge will distribute across the surface of the sphere and shell in order to minimize potential energy. This is a property of conductors. In contrast, an insulator does not allow for charge redistribution due to immobile electrons. The metal sphere is considered a conductor due to its ability to allow charge redistribution.
  • #1
Potatochip911
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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.
 
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  • #2
Potatochip911 said:

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 .
 
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  • #3
conscience said:
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.
 
  • #4
Potatochip911 said:
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 ?
 
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  • #5
conscience said:
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?
 
  • #6
Potatochip911 said:
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 .
 
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Related to Calculating the surface charge of a sphere and a conducting shell

1. How do you calculate the surface charge of a sphere?

To calculate the surface charge of a sphere, you will need to know the radius of the sphere and the charge density. You can then use the formula Q = 4πε0r2σ, where Q is the total charge, ε0 is the permittivity of free space, r is the radius, and σ is the charge density. This formula assumes a uniform charge distribution across the surface of the sphere.

2. Is the surface charge of a sphere affected by its size?

Yes, the surface charge of a sphere is affected by its size. As the radius of the sphere increases, the surface charge will also increase. This is because the surface area of a sphere increases with the square of its radius, while the charge remains constant, resulting in a larger surface charge.

3. How do you calculate the surface charge of a conducting shell?

The surface charge of a conducting shell can be calculated using the formula Q = σA, where Q is the total charge, σ is the charge density, and A is the surface area of the shell. This formula assumes a uniform charge distribution across the surface of the shell.

4. Can the surface charge of a conducting shell be negative?

Yes, the surface charge of a conducting shell can be negative. This can occur if the charge density is negative, or if the charge on the shell is distributed in a non-uniform manner. It is also possible for different regions of the shell to have different charges, resulting in a net negative surface charge.

5. What is the difference between the surface charge of a sphere and a conducting shell?

The main difference between the surface charge of a sphere and a conducting shell is their charge distribution. A sphere has a uniform charge distribution across its surface, while a conducting shell may have a non-uniform charge distribution or even a net charge due to the presence of an external electric field. Additionally, the surface charge of a sphere is affected by its size, while the surface charge of a conducting shell is only affected by its surface area and charge density.

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