Question

A small conducting spherical shell with inner radius a and outer radius b is concentric with a larger conducting spherical shell with inner radius c and outer radius d. The inner shell has a total charge of -2q and the outer shell has a total charge of +4q.

Select True or False for the following statements.
True False  The radial component of the electric field in the region r > d is given by +4q/(4??₀r²).
True False  The radial component of the electric field in the region c < r < d is given by -2q/(4??₀r²).
True False  The total charge on the inner surface of the small shell is zero.
True False  The total charge on the outer surface of the large shell is +2q.
True False  The total charge on the inner surface of the large shell is +2q.
True False  The total charge on the outer surface of the small shell is -6q.
True False  The radial component of the electric field in the region r < a is given by +2q/(4??₀r²).

Answer 1

The all questions is based on Gauss' theorem and the fact that inside a conductor the electric field is zero.

So, the field inside the metal part of each shell is zero.

Now start with a spherical Gaussian surface centered in the inner shell and of radius just short of a.

There is no charge contained in this Gaussian surface, so the electric field is 0.

Next, expand the Gaussian to a radius between a and b: you are now inside the inner metal shell.

Since this is a conductor, the electric field field is 0.

Now let the Gaussian surface expand to a radius >b and < c, in between the two shells: you have a charge +1q, hence there is an electric field.

Now expand the Gaussian surface between c and d: the electric field is 0.

Wait a minute, you say, there is a charge +1q on the inner shell. If the field is 0, the total charge must be 0, where did the charge +1q go?
The answer is: it stayed where it has been all along, on the inner shell. What is happening is that the field that it generated attracted charges of the opposite sign in the conductor. Remember that in a conductor charges are free to move, that is why it is a conductor: it "conducts" electricity.

So, enough charges (-1q exactly) will move to the inner surface of the larger metal shell, so that the net charge enclosed by the Gaussian surface is 0: the field inside the second metal shell is now 0.