# Density of charge induced on a hollow sphere due to eccentric charge inside

Suppose we have a lone hollow metal sphere with net charge equal to $0$. If we were to put a point charge $Q$ inside of the sphere and move it, let's say, away from the sphere center at some distance $d$, determine the distribution of induced charge on the sphere. With no further explanation, the answer in my textbook states that the induced charge will be evenly distributed across inner surface of hollow sphere, no matter where the point charge is, as long as it is inside and not touching the sphere.

Can someone explain in detail or provide information on why this is true?

• Charge induced on the inner surface of the hollow sphere is not uniformly distributed in the charge inside is not at the centre. It is the charge on the outer surface which is always uniformly distributed, regardless of the distance $d$. – sammy gerbil Mar 21 '20 at 22:16

The explanation for this is partly Gauss' law and partly the nature of metal.

Metals have free electrons that move to try to compensate for electric fields. The free electrons near the surface on the inside of the sphere will have an uneven distribution if the point charge is off-centre. These electrons will counteract the effect of the positive charge inside the sphere so that there will be no electric field inside the metal shell. Let us say that the charge inside is positive and $+q$ - then there will be extra electrons on the inside of the sphere to counterbalance it - their total charge will be $-q$. Because the metal is neutral overall there will be a lack of electrons on the outer surface and a net positive charge of $+q$ on the outer surface because of the extra electrons. which are on the inside of the sphere.

Now because there is no electric field inside the metal the charge on the outside of the metal sphere will spread itself for the lowest energy configuration by spacing the charges as far apart as possible, which will give an even distribution.