A 10 pF capacitor is connected to a 24 V battery. The electrostatic energy stored in the capacitor is
Answer & explanation
Correct answer: option 4
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