A 12 pF capacitor is connected to a 50 V battery. The electrostatic energy stored in the capacitor is
Answer & explanation
Correct answer: option 4
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: $1.5 \times 10^{-8} \, J$