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Q.
The charge flowing through the cell on closing the key $k$ is equal to
Electrostatic Potential and Capacitance
Solution:
When switch is open, the charge given to equivalent capacitor.
$q_{i}=\left(\frac{3 C \times C}{3 C+C}\right) \cdot V=\frac{3}{4} C V$
When switch is closed, $q_{f}=C V$
$\Delta q=q_{f}-q_{i}=C V-\frac{3}{4} C V=\frac{1}{4} C V$
Hence, charge passed through switch will be $\frac{1}{4} C V$.
