Question

Five identical equispaced conducting plates are given charges as shown in figure. List -1 gives the labels 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 closed and $S_2$ is open, match List-1 with List -2

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

Answer 1

The correct option is C I$\to$ P II $\to$ Q III $\to$ T IV $\to$ Q

The charges will get distributed as follows


Since the potentials of plates C & E are same
$\frac{q}{C}=\frac{4Q-q}{C}\Rightarrow q=2Q$
$\therefore \text{a}\to Q,d\to 2Q,f\to -2Q,h\to 2Q$