AP EAMCET · PHYSICS · Capacitance
The energy stored in a capacitor is W. To double the charge on the plates of the capacitor, the additional work to be done is
- A W
- B 4W
- C \(\frac{4}{3} \mathrm{~W}\)
- D 3W
Answer & Solution
Correct Answer
(D) 3W
Step-by-step Solution
Detailed explanation
\( W_1 = \frac{1}{2} \frac{Q^2}{C} \) \( W_2 = \frac{1}{2} \frac{(2Q)^2}{C} = 4 \left( \frac{1}{2} \frac{Q^2}{C} \right) = 4W \) \( \Delta W = W_2 - W_1 = 4W - W = 3W \)
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