Question

Two conducting large plates $P_1$ and $P_2$ are placed near to each other at very small separation 'd'. The plate area of either face of plate is A. Now a charge $+2Q$ is given to plate $P_1$ and $–Q$ to the plate $P_2$ (neglect ends effects). Potential difference between the plates $P_1$ and $P_2$

• A. $\frac{Qd}{2{ϵ}_{0}A}$
• B. $\frac{3Qd}{2{ϵ}_{0}A}$
• C. $\frac{Qd}{3{ϵ}_{0}A}$
• D. $\frac{Qd}{4{ϵ}_{0}A}$

Answer 1

The correct option is B $\frac{3Qd}{2{ϵ}_{0}A}$

Magnitude of charge on each inner face
$q=\frac{2Q-(-Q)}{2}=\frac{3Q}{2}$
Potential difference $=\frac{q}{C}=\frac{3Qd}{2{ϵ}_{0}A}$

Why this question? Tip: The potential difference is only due to the charges on the inner surfaces, as the fields due to charges on outer surfaces cancel each other in the region between the plates.