Question

Three conducting bodies A(solid sphere), B(hollow sphere) and C(hollow sphere with thickness R) are arranged as shown in the figure. A charge $q$ is given to the inner sphere. Match the statements from List I with the values in List II and select the correct answer using the code given below the lists.

	Column-I		Column-II
(A)	When $S_1,S_2$ both are open, capacity of the system is $C_1$, then $C_1$ is	(p)	$16\pi {ϵ}_{0}R$
(B)	When $S_1$ is closed but $S_2$ is open, capacity is $C_2$, then $C_2$ is	(q)	$\frac{16\pi {ϵ}_{0}R}{3}$
(C)	When $S_2$ is closed but $S_1$ is opened, capacity is $C_3$, then $C_3$ is	(r)	$\frac{48\pi {ϵ}_{0}R}{5}$
(D)	When both $S_1$ and $S_2$ are closed, capacity is $C_4$, then $C_4$ is	(s)	$\frac{48\pi {ϵ}_{0}R}{11}$

Answer 1

For Option B, since s1 is closed and s2 is open, we simply add the potentials of each of the interior bodies to find the capacitance outside. Because of the connecting wire, the smaller and the medium sphere both have charge q/2.Thus, we get V=2x(kq/2*3R + kq/2*2R + kq/2*R).

Following this, to get capacitance we get C=q/V=48(pi)(epsilon knot)*R/11 so we get answer as S.

Solve the other parts similarly using self potential.