A) 12 ´ 10?6 J
B) 9 ´ 10?4 J
C) 4 ´ 10?6 J
D) 4 ´ 10?9 J
Correct Answer: B
Solution :
The increase in energy of the capacitor \[\Delta U=\frac{1}{2}C(V_{2}^{2}-V_{1}^{2})=\frac{1}{2}(6\times {{10}^{-6}})\,({{20}^{2}}-{{10}^{2}})\] \[=3\times {{10}^{-6}}\times 300=9\times {{10}^{-4}}J\]
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