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

A 12 pF capacitor is connected to a 50 V battery, the electrostatic energy stored in the capacitor in nJ is
  1. 15
  2. 7.5
  3. 0.3
  4. 150

Solution

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

Asked in: NEET 2024 (Re-NEET)

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