In (Fig. 3.73), if the inner shell has charge `+q_0` and the outer shell is earthed, then
.
i. determine the charge on the outer shell, and
ii. find the potential of the inner shell.

Explanation

(i) we know that charges on facing surfaces are equal and opposite. If the charge on the inner sphere is `q_0`, then the charge on the inner surface of the shell should be `-q_0`. Let the charge on the outer surface of the shell be `q_(2)`. As the shell is earthed, so its potential should be zero. Thus,
`V_("shell") = (k q_0)/r_(2) + (k (- q_2))/r_(2) + (k q_2)/r_(2) = 0` or `q_2 = 0`
Hence, the charge on the outer surface of the shell is zero. Final charges appearing are shown in (Fig. 3.74) (b)
a. ,
(ii) Potential of inner sphere is
`V_i = (k q_0)/r_(1) + (k(- q_0))/r_(2) = (q_0)/(4 pi epsilon_0)[(1)/r_(1) - (1)/r_(2)]`.