If the charge on a capacitor is increased by 2 C, the energy stored in it increases by 44%. The original charge on the capacitor is (in C)

Solution

Physics sir · 4 steps

Same capacitor, so energy goes as charge squared: a 44% rise in U means a 20% rise in Q.

1. Energy and charge. The capacitor itself does not change, so C is fixed. Use U = Q²/2C, which shows U ∝ Q².
2. Set up the ratio. The new energy is 1.44 times the old one. In the ratio of energies, 2C cancels.

   U'/U = ((Q+2)/Q)² = 1.44 ⇒ (Q+2)/Q = 1.2

3. Solve for Q. Q + 2 = 1.2Q, so 0.2Q = 2 and Q = 10 C. Check: (12/10)² = 1.44, a 44% increase.
4. So. The original charge is 10 C, option (a).

The idea to keep. For a fixed capacitor, U ∝ Q². Take the square root of the energy ratio to get the charge ratio, then solve one simple linear equation.

Answer from the NTA final answer key. Solution by PrepWiser.