Question

A solid sphere of radius R is made of a metallic conductor. A hollow spherical shell of the same radius R is made of the same conducting material. An excess charge Q is deposited on each.
1)Which object has the greatest surface charge density?


2) Briefly explain your answer to the previous question.


3)A solid sphere of radius R is made of a metallic conductor. Another solid sphere of radius R is made of an insulating material. An excess charge Q is deposited on each. Which object has the greatest surface charge density?


4) explain your answer to the previous question.

Answer 1

1-2) They have same surface charge density

3-4) The metallic conductor has greatest surface charge density

Explanation:

1-2)

In a conductor, the charge carriers (mainly electrons) are free to move. Therefore, as a result, they tend to move at the largest possible distance from each other, because of the repulsive force that they exert on each other.

The configuration that maximize the distance between the charge carriers for a solid sphere of metallic conductor is the one in which all the electrons are on the surface, and they are equally spaced between each other. This means that for the solid sphere of radius R, the excess charge Q will be entirely spread over the surface of the sphere.

Similarly, the excess charge Q on the hollow spherical shell (which is also made of the same conducting material) will also be spread over the surface with the charge carriers at the maximum distance from each other. Therefore, the surface charge density for both objects will be

$\sigma = \frac{Q}{4\pi R^2}$

where R is the radius of the two spheres.

3-4)

In this case, the surface charge density on the two objects is different.

In fact, on the metallic sphere (conducting) the surface charge density is (as explained in part 1):

$\sigma = \frac{Q}{4\pi R^2}$

Hoever, the second sphere is made of an insulating material. In an insulator, the charge carriers are not free to move. If the initial charge Q is spread across the all sphere (which is not hollow), this means that some of the charge will actually also be inside the sphere. So the charge deposited on the surface, Q', will be less than the total charge Q. Therefore, the surface charge density will be

$\sigma' = \frac{Q'}{4\pi R^2}$

And since Q' < Q, this means that $\sigma'$, so the conducting sphere has a greatest surface charge  density.