A 10 pF capacitor is connected to a 24 V battery. The electrostatic energy stored in the capacitor is

$2.88 × 10^{-9} J$

The correct answer is Option (4) → $2.88 × 10^{-9} J$

Given:

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

$V = 24\,V$

Formula:

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

Substitute:

$U = \frac{1}{2} \times 10 \times 10^{-12} \times (24)^2$

$U = 5 \times 10^{-12} \times 576$

$U = 2.88 \times 10^{-9}\,J$

Final Answer: Electrostatic energy stored = 2.88 × 10⁻⁹ J