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

Question

TestHub · Accepted Answer

Initial charge on first capacitor will be, $Q = C V$

As the second capacitor is identical to the first one, hence potential drop across both capacitor will be equal to $\frac{V}{2}$ .

Now, loss of energy, $Δ H = U_{i} - U_{f}$

$\Rightarrow Δ H = \frac{1}{2} C V^{2} - \frac{1}{2} (2 C) \times {(\frac{V}{2})}^{2}$

$\Rightarrow Δ H = \frac{1}{2} C V^{2} - \frac{1}{4} C V^{2}$ $\Rightarrow Δ H = \frac{1}{4} C V^{2} = \frac{1}{4} \times 50 \times 10^{- 12} \times 100^{2} = 125 \times 10^{- 9} J = 125 nJ$

Physics - Capacitor Question with Solution | TestHub

Doubts & Discussion