Question

Through cunning physics trickery I have trapped a point charge Q at the center of a thin-walled hollow conductive sphere of radius R, which itself carries a net charge of -3Q.

a.    How much charge will collect on the inside surface of the hollow sphere?

b.   How much charge will collect on the outside surface of the hollow sphere?

c.    Draw a picture of this arrangement, including a depiction of any electric field lines.

d.   Graph the resulting electric field E as a function of distance from the center, r.

Answer 1

The trick with understanding conductors on these Gauss's law problems, is to realize that once electrostatic equilibrium is reached, the electric field throughout MOST of the body of the conductive material, must be zero.

This means that a Gaussian surface through the center of the material that makes up the outer sphere, must enclose a net charge of zero, so that it has a net electric flux of zero. The charge of this outer sphere will distribute in such a way that some of it is on the inside surface, and some of it is on the outside surface. And that on the inside surface PLUS the point charge at the center, must add up to zero.

Call it Qi and Qo, for inner and outer charges.

We established that:
Qi + (+Q) = 0

Thus:
Qi = -Q


By conservation of charge, we know that:
Qi + Qo = -3*Q

With knowledge that Qi = -Q:
(-Q) + Qo = -3*Q

Thus:
Qo = -2*Q

Results:
A: Qi = -Q
B: Qo = -2*Q

C and D, verbal description.

For the electric field inside the outer sphere, the charges on the outer sphere ultimately don't matter because they are spherically symmetric. Any electric field due to them, will add up to zero, as per Gauss's law.

That said, only the inside point charge of +Q affects the electric field inside the inner sphere surface. And it behaves according to Coulomb's law, such that E = 1/(4*pi*epsilon0) * Q/r^2. This is true as long as r < R. End it with an open circle, to indicate an inequality.

Within the sphere body, the electric field is zero. For r = R, the electric field equals zero. Draw a filled circle on the horizontal axis accordingly.

Once beyond the sphere body, it now behaves according to Coulomb's law, with the 1/r^2 falloff. Although, now the charge that causes it, is +Q + (-3*Q) = -2*Q. So, for r >R, E=1/(4*pi*epislon0) * (-2*Q)/r^2.