Question 164 of 180

A capacitor of capacitance $50pF$ is charged by $100V$ source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is____ $nJ$.

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Answer: 125

Initial charge on first capacitor will be, $Q=CV$
As the second capacitor is identical to the first one, hence potential drop across both capacitor will be equal to $\dfrac{V}{2}$.
Now, loss of energy, $\Delta H={U}_{i}-{U}_{f}$
$\Rightarrow \Delta H=\dfrac{1}{2}C{V}^{2}-\dfrac{1}{2}\left(2C\right)\times {\left(\dfrac{V}{2}\right)}^{2}$
$\Rightarrow \Delta H=\dfrac{1}{2}C{V}^{2}-\dfrac{1}{4}C{V}^{2}$ $\Rightarrow \Delta H=\dfrac{1}{4}C{V}^{2}=\dfrac{1}{4}\times 50\times {10}^{-12}\times {100}^{2}=125\times {10}^{-9}J=125nJ$

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