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 :

Given capacitor \[C=400\times {{10}^{-12}}F\] and V= 100 V Initially energy stored \[=\frac{1}{2}\,C{{V}^{2}}\] \[=\frac{1}{2}\,[400\times {{10}^{-12}}\,\times {{({{10}^{2}})}^{2}}]\] \[=\frac{1}{2}\times 4\times {{10}^{-10}}\,\times {{10}^{4}}\,=2\times {{10}^{-6}}\,J\] \[\Rightarrow \,\]  Charge \[Q=CV=4\times {{10}^{-10}}\,\times 100\,=4\times \,{{10}^{-8}}C\] By again connecting the charged capacitor with a uncharged ideal capacitor Now the charge Q will divide into two capacitors. As V and C are same for both, Q' will be same for both. So, \[Q'=\frac{Q}{2}=2\times {{10}^{-8}}C\] So, energy stores in each capacitors = \[\left( \frac{{{(Q')}^{2}}}{2C} \right)\,=\frac{{{(Q')}^{2}}}{2C}\] \[=\frac{{{(2\times {{10}^{-8}})}^{2}}}{8\times {{10}^{-10}}}\,={{10}^{-6}}J\,=0.5\,\times {{10}^{-6}}J\] Hence total final energy \[=2\times 0.5\,\times {{10}^{-6}}\] \[=1\times {{10}^{6}}\,J={{10}^{-6}}J\] So energy lost \[=2\times {{10}^{-6}},\,\,-{{10}^{-6}}\,={{10}^{-6}}J\]