A) zero
B) \[4\times {{10}^{-6}}J\]
C) \[6\times {{10}^{-6}}J\]
D) \[8\times {{10}^{-6}}J\]
Correct Answer: C
Solution :
Energy stored in capacitor \[{{E}_{1}}=\frac{1}{2}C{{V}^{2}}\] \[=\frac{1}{2}\times 600\times {{10}^{-12}}\times {{(200)}^{2}}\] \[=12\times {{10}^{-6}}J\] Again \[{{E}_{2}}=\frac{1}{2}(600+300)\times {{10}^{-12}}\times {{(200)}^{2}}\] \[=\frac{1}{2}\times 900\times {{10}^{-12}}\times 4\times {{10}^{4}}\] \[=18\times {{10}^{6}}J\] Energy lost \[={{E}_{1}}=-{{E}_{2}}\] \[=18\times {{10}^{-6}}-12\times {{10}^{-6}}\] \[=6\times {{10}^{-6}}J\]
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