A 12 pF capacitor is connected to a 50 V battery. The electrostatic energy stored in the capacitor is

$1.5 × 10^{-8} J$

The correct answer is Option (4) → $1.5 × 10^{-8} J$

Given:

Capacitance $C = 12 \, pF = 12 \times 10^{-12} \, F$

Potential difference $V = 50 \, V$

Electrostatic energy stored:

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

Substitute values:

$U = \frac{1}{2} \cdot (12 \times 10^{-12}) \cdot (50)^{2}$

$U = 6 \times 10^{-12} \cdot 2500$

$U = 1.5 \times 10^{-8} \, J$

Final Answer: $1.5 \times 10^{-8} \, J$