The correct option is
D Charge distribution on shells is as shown in the figure below
Since, there is no charge present inside shell
A. The whole charge
q given to the shell
A will appear on its outer surface.
Charge on its inner surface will be zero.
Moreover if a Gaussian surface is drawn on shell
B, net charge enclosed by it should be zero.
Therefore, charge on its inner surface will be
−q.
Now let
q′ be the charge on its outer surface, then charge on the inner surface of
C will be
−q′ and on its outer surface will be,
2q−(−q′)=2q+q′ as total charge on
C is
2q.
Given that Shell
B is earthed. Hence, its potential should be zero. i.e,
VB=0
⇒14πε0[q2R−q2R+q′2R−q′3R+2q+q′3R]=0
Solving this equation, we get
q′=−43q
∴2q+q′=2q−43q=23q
Therefore, charges on different surfaces in tabular form are given below:
Hence, option (d) is the correct answer.