A parallel plate capacitor of capacitance 2 μF has energy $10×10^{-5} J$. The potential difference and the charge on the capacitor are:

10 V, 20 μC

The correct answer is Option (3) → 10 V, 20 μC

Given:

Capacitance, $ C = 2\ \mu F = 2 \times 10^{-6}\ \text{F} $

Energy stored, $ U = 10 \times 10^{-5}\ \text{J} = 1 \times 10^{-4}\ \text{J} $

Formula for energy stored in a capacitor:

$ U = \frac{1}{2} C V^2 $

Substituting values:

$ 1 \times 10^{-4} = \frac{1}{2} \times 2 \times 10^{-6} \times V^2 $

$ 1 \times 10^{-4} = 1 \times 10^{-6} \times V^2 $

$ V^2 = 100 $

$ V = 10\ \text{V} $

Charge on the capacitor:

$ Q = C V = 2 \times 10^{-6} \times 10 = 2 \times 10^{-5}\ \text{C} $

Therefore,

Potential difference $ = 10\ \text{V} $

Charge on the capacitor $ = 2 \times 10^{-5}\ \text{C} $