question_answer

A \[400\text{ }pF\]capacitor is charged by a' \[100\text{ }V\] supply. How much electrostatic energy is lost in the process of disconnecting from the supply and connecting another uncharged \[400\text{ }pF\] 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 \[=\frac{1}{2}C{{V}^{2}}\] \[=\frac{1}{2}400\times {{10}^{-12}}\times {{({{10}^{2}})}^{2}}\] \[=\frac{1}{2}\times 4\times {{10}^{-10}}\times {{10}^{4}}\] \[=2\times {{10}^{-6}}J\] \[\Rightarrow \] \[Q=CV=4\times {{10}^{-10}}\times 100\] \[=4\times {{10}^{-8}}C\] \[\Rightarrow \] 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, Q' will be same for both, \[\Rightarrow \] So   \[Q'=\frac{Q}{2}=2\times {{10}^{-8}}C\] \[\Rightarrow \]So, energy stored in both the capacitors \[=\left( \frac{{{(Q')}^{2}}}{2C} \right)=\frac{{{(Q')}^{2}}}{2C}\] \[=\frac{(2\times {{10}^{-8}})}{8\times {{10}^{-10}}}={{10}^{-6}}J\] So energy lost \[=2\times {{10}^{-6}}-0.25\times {{10}^{-6}}\] \[=0.75\times {{10}^{-6}}J\approx {{10}^{-6}}J\]