Question

An isolated parallel plate capacitor of capacitance $C$ has four surfaces with charges $Q_1,Q_2,Q_3$ and $Q_4$ as shown in the figure below. The potential difference across plates is (In the given diagram, $Q_1$ and $Q_4$ are the charges on the outer surfaces of the left & right plates respectively)

• A. $\left|\frac{Q_2}{C}\right|$
• B. $\left|\frac{Q_3}{C}\right|$
• C. $\frac{[Q_1+Q_2]}{C}$
• D. $\left|\frac{Q_2+Q_3}{C}\right|$

Answer 1

Answer: (A), (B)

$\left|\frac{Q_3}{C}\right|$

For the parallel plate capacitor, $$Q=CV$$ Where,
$Q$ is the magnitude of the charges on the inner surface of the plates.

$C$ is the capacitance and $V$ is the potential difference across the plate.

Due to induction, charges on inner surfaces facing each other are equal and opposite.

So, $Q_2=-Q_3$

Thus, $V=\left|\frac{Q_2}{C}\right|=\left|\frac{Q_3}{C}\right|$

Hence, options (b) and (c) are the correct answers.