Question

Five identical equispaced conducting plates are given charges as shown in the figure. List -1 gives the labes of the sides of the plates and List -2 gives the charges

$\begin{array}{cccc}& \text{List- 1}& & \text{List - 2}\\ \text{I)}& \text{a}& \text{P)}& \text{Q}\\ \text{II)}& \text{d}& \text{Q)}& \text{2Q}\\ \text{III)}& \text{f}& \text{R)}& \text{3Q}\\ \text{IV)}& \text{h}& \text{S)}& \text{5Q/3}\\ & & \text{T)}& \text{-2Q}\\ & & \text{U)}& \text{-4Q/3}\end{array}$

When switch $S_1$ is open and $S_2$ is closed, match List-1 with List -2

• A. I$\to$ P II $\to$ S III $\to$ U IV $\to$ R
• B. I$\to$ P II $\to$ T III $\to$ R IV $\to$ Q
• C. I$\to$ P II $\to$ Q III $\to$ T IV $\to$ Q
• D. I$\to$ P II $\to$ S III $\to$ R IV $\to$ T

Answer 1

The correct option is A I$\to$ P II $\to$ S III $\to$ U IV $\to$ R



By balancing charge on A and D,
$Q+q-q^{{}^{\prime }}+q^{{}^{\prime }}+2Q+q=5Q$ -- (1)
By equating potentials on A and D,
$\frac{Q-q^{{}^{\prime }}}{C}+\frac{3Q-q^{{}^{\prime }}}{C}+\frac{-q^{{}^{\prime }}}{C}=0$ -- (2)
Solving the equations, we get
$q=Q$ and $q^{{}^{\prime }}=\frac{4Q}{3}$
Charge on a $=q=Q$
Charge on d $=3Q-q^{{}^{\prime }}=\frac{5Q}{3}$
Charge on f $=-q^{{}^{\prime }}=-\frac{4Q}{3}$
Charge on h $=2Q+q=3Q$