Question

Figure shows three concentric spherical conducting shells A, B, and C of radii $R,2R$ and $4R$ respectively. Shells $A$ and $C$ are connected by a conducting wire and B is given charge $Q$. $[\text{Consider}\frac{1}{4\pi {ϵ}_0}=K]$
Match the following columns:

$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \text{A. Charge on shell A}& \left(p\right)\frac{Q}{3}\\ \text{B. Charge on shell C}& \left(q\right)-\frac{Q}{3}\\ \text{C. Potential of shell A}& \left(r\right)Q\\ \text{D. Potential of shell C}& \left(s\right)\frac{KQ}{4R}\\ & \left(t\right)\text{None}\end{array}$

• A. $A-q,B-p,C-s,D-s$
• B. $A-p,B-s,C-q,D-s$
• C. $A-s,B-t,C-q,D-p$
• D. $A-p,B-q,C-s,D-t$

Answer 1

The correct option is A $A-q,B-p,C-s,D-s$

Let $Q_A$ be charge on shell A
$Q_C$ be charge on shell C
$V_A$ be potential of shell A
$V_C$ be potential of shell C
Due to conservation of charge
$Q_A+Q_C=0$
Since A and C are connected by a conducting wire, they will have equal potential.
$V_A=V_C$
$\frac{K{Q}_A}{R}+\frac{KQ}{2R}+\frac{K{Q}_C}{4R}=\frac{K{Q}_A}{4R}+\frac{KQ}{4R}+\frac{K{Q}_C}{4R}$
$4Q_A+2Q=Q_A+Q$
$\Rightarrow 3Q_A=-Q$
$\Rightarrow Q_A=-\frac{Q}{3}$
$\Rightarrow Q_C=-Q_A=\frac{Q}{3}$

Potential of shell A
$\Rightarrow V_A=\frac{K}{R}(-\frac{Q}{3})+\frac{KQ}{2R}+\frac{K}{4R}\left(\frac{Q}{3}\right)$
$\Rightarrow V_A=\frac{K(-4+6+1)}{12R}$
$\Rightarrow V_A=\frac{KQ}{4R}=V_C$