A) \[18\times {{10}^{-4}}J\]
B) \[9\times {{10}^{-4}}J\]
C) \[4.5\times {{10}^{-4}}J\]
D) \[9\times {{10}^{-6}}J\]
Correct Answer: B
Solution :
\[\Delta E={{E}_{Final}}-{{E}_{Initial}}=\frac{1}{2}C(V_{Final}^{2}-V_{Initial}^{2})\] \[=\frac{1}{2}\times 6\times ({{20}^{2}}-{{10}^{2}})\times {{10}^{-6}}\] \[=3\times (400-100)\times {{10}^{-6}}=3\times 300\times {{10}^{-6}}=9\times {{10}^{-4}}J\]
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