question_answer

A 400 pF capacitor is charged by a 100 V supply. How much electrostatic energy is lost in the process of disconnecting from the supply and connecting another uncharged 400 pF capacitor?

A)  \[{{10}^{-5}}J\]

B)  \[{{10}^{-6}}J\]

C)  \[{{10}^{-7}}J\]

D)  \[{{10}^{-4}}J\]

Correct Answer: B

Solution :

 \[C=400\times {{10}^{-12}}F\] \[V=100\,V\] Initially energy stored\[=\frac{1}{2}C{{V}^{2}}\] \[=\frac{1}{2}400\times {{10}^{-12}}\times {{({{10}^{2}})}^{2}}\] \[=\frac{1}{2}\times 4\times {{10}^{-10}}\times {{10}^{4}}\] \[\Rightarrow \] \[Q=CV=4\times {{10}^{-10}}\times 100\] \[=4\times {{10}^{-8}}C\] \[\Rightarrow \]By again connecting the charged capacitor with a uncharged ideal capacitor. \[\Rightarrow \] \[\Rightarrow \] Now, the charge Q will divide into two capacitors. \[\Rightarrow \]As V and C are same for both, 0 will be same for both, \[\Rightarrow \]So   \[Q=\frac{Q}{2}=2\times {{10}^{-8}}C\] \[\Rightarrow \]So, energy stored in both the capacitors \[=\left( \frac{{{(Q)}^{2}}}{2C} \right)=\frac{{{(Q)}^{2}}}{2C}\] \[=\frac{{{(2\times {{10}^{-8}})}^{2}}}{8\times {{10}^{-10}}}\] \[={{10}^{-6}}J\] So energy lost \[=2\times {{10}^{-6}}-0.25\times {{10}^{-6}}\] \[=0.75\times {{10}^{-6}}J\approx {{10}^{-6}}J\]