Q106 / 200 · NEET 2024 (Re-NEET) · Physics

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

  • Correct answer
    A
    15
  • B
    7.5
  • C
    0.3
  • D
    150

Solution

Electrostatic energy stored $U=\frac{1}{2} C V^2$ $=\frac{1}{2} \times 12 \times 10^{-12} \times(50)^2$ $=6 \times 25 \times 10^{-10}$ $=15 \times 10^{-9} \mathrm{~J}$ $=15 \mathrm{~nJ}$

Hence, the correct answer is option: 'A'.

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