A small charged ball lies within the hollow of a metallic spherical shell of radius R. For three situations, the net charges on the ball and shell, respectively, are (1) +4q, 0; (2) -6q, +10q; (3) +16q, -12q. Rank the situations according to the charge on (a) the inner surface of the shell and (b) the outer surface, most positive first.

a) (2) > (1) > (3)

b) (1) = (2) = (3)

Explanation:

a)

As the electric field inside a conductor, in electrotatic conditions, must be zero, if we apply the Gauss law to a spherical surface which radius falls witthin the conductor, total electric flux through this surface, must be zero also.Now, as the electric  flux is proportional to the total charge enclosed by the surface, if  the flux is zero, this means that the total charge must be zero.So, for the three cases, as total charge must be zero, there must be a charge on the inner surface of the shell, equal and opposite to the charge of the ball: q = +4q ⇒ qin = - 4 q (1) q = -6q ⇒ qin = +6q q = +16 q ⇒ qin = -16qSo, ranking the situations according to the charge on the inner surface of the shell, most positive first, we have:+6q > -4q > -16q⇒(2) > (1) > (3)

b)

Once we know the charges on the inner surface, as total charge on the shell must be conserved, the charge on the outer surface must meet the following condition:qin + qou = q₀⇒ qou = qo-qinq₀ = 0 , qin = -4q ⇒ qou = + 4qq₀ = +10q, qin = +6q ⇒ qou = +4qq₀ =-12 q, qin =-16 q ⇒ qou = +4qAs it can be seen, the charge on the outer surface is the same for the three cases, so (1) =(2) = (3).