# Charges and Conducting Shells

I need a little help with this problem. Its from the 2004 physics bowl.

50. A solid spherical conducting shell has inner radius a and outer radius 2a. At the center of the shell is located a point charge +Q. What must the excess charge on the shell be in order for the charge density on the inner and outer surfaces of the shell to be exactly equal?

I reasoned that the charge on the inner surfaces of the shell is -Q because the electrons are attracted towards the point charge. Thus the charge on the outer shell is +Q. The surface area of the outer shell is $$16\pi a^2 [/itex] and the surface are of the inner shell is [tex] 4\pi a^2 [/itex]. So the charge density on the outer is shell is [tex]\frac{Q}{16\pi a^2} [/itex] and the charge denisty on the inner shell is [tex]\frac{-Q}{4\pi a^2} [/itex]. So in order to make the charge density equal you need to add -5Q to the shell so that that charge accumulates outside and yields a charge density on the outer surface of [tex]\frac{-Q}{4\pi a^2} [/itex]. So the answer is -5Q which is also the answer according to the answer key, however my physics teacher does not agree. Is my reasoning correct cause I have to agrue this with him tommorow. Thanks ## Answers and Replies Related Introductory Physics Homework Help News on Phys.org Any help people, I am not very familiar with E&M SpaceTiger Staff Emeritus Science Advisor Gold Member thechunk said: Any help people, I am not very familiar with E&M Your solution looks right to me. The electric field inside the conductor must be zero, so the inner wall of the shell must have a total charge of -Q to balance the point charge. This of course gives a density of [tex]\kappa=-\frac{Q}{4\pi a^2}$$

When charge is added to the conductor, the inner wall must retain the same charge density if the field is remain zero inside, so all added charge goes to the outer surface. The outer surface has a larger radius, so it'll need a larger charge to get the same density:

$$\kappa=\frac{Q_{outer}}{16\pi a^2}=-\frac{Q}{4\pi a^2}$$

The solution to this equation is, of course,

$$Q_{outer}=-4Q$$

Giving a total charge on the conductor of

$$Q_{tot}=Q_{outer}+Q_{inner}=-4Q+(-Q)=-5Q$$