WBJEE · Physics · Capacitance
If the charge on a capacitor is increased by 2 coulomb, the energy stored in it increases by \(21 \%\). The original charge on the capacitor is
- A \(10 \mathrm{C}\)
- B \(20 \mathrm{C}\)
- C \(30 \mathrm{C}\)
- D \(40 \mathrm{C}\)
Answer & Solution
Correct Answer
(B) \(20 \mathrm{C}\)
Step-by-step Solution
Detailed explanation
\(\frac{\frac{q_f^2}{2 C}-\frac{q_i^2}{2 C}}{\frac{q_i^2}{2 C}} \times 100=21\) and \(q_f-q_i=2\) solving we get \(q_i=20\) coulomb
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