A capacitor of capacitance $C_1$ is charged upto V volt and then connected to an uncharged capacitor of capacity $C_2$. The final potential difference across each will be:

$\frac{C_1 V}{\left(C_1+C_2\right)}$

The correct answer is Option (3) → $\frac{C_1 V}{\left(C_1+C_2\right)}$

The total charge before the connection is the charge on $C_1$ and after the connection, the total charge is distributed on both the capacitor -

$Q_{total}=Q_1=C_1×V$

Total capacitance of a system when $C_1$ and $C_2$ are in parallel.

$C_{total}=C_1+C_2$

Since, the total charge is conserved and both capacitor end up having the same value of $V_f$.

$Q_{total}=(C_1+C_2)×V_f$

$∴(C_1+C_2)×V_f=C_1V$

$V_f=\frac{C_1V}{C_1+C_2}$