A capacitor of capacitance $C_1$ charged up to $V$ volt and then connected to an uncharged capacitor $C_2$.…
A capacitor of capacitance $C_1$ charged up to $V$ volt and then connected to an uncharged capacitor $C_2$. Then, the final potential difference across each will be:
$\frac{C_2 V}{C_1+C_2}$
$\frac{C_1 V}{C_1+C_2}$
$\left(1+\frac{C_2}{C_1}\right) V$
$\left(1-\frac{C_2}{C_1}\right) V$
Solution
$\begin{aligned}
& V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\left(\because V_2=0\right) \\
& \Rightarrow V=\frac{C_1 V_1}{C_1+C_2}
\end{aligned}$
Alternative Solution:
Sure, I'd be happy to explain.
When the two capacitors are connected, they will share the charge until they both reach an equilibrium, which means they will both have the same voltage across them. Since no external voltage source is connected, the total charge in the system remains conserved.
The initial charge on the first capacitor is $Q_1 = C_1V$. After the connection, let's assume the final voltage across both capacitors is $V'$.
The final charge on the first capacitor is $Q_1' = C_1V'$. The final charge on the second capacitor is $Q_2' = C_2V'$.
Since the total charge is conserved, the initial charge on the first capacitor must equal the total final charge on both capacitors. So, we have:
$Q_1 = Q_1' + Q_2'$
or
$C_1V = C_1V' + C_2V'$
This simplifies to
$V = (C_1 + C_2)V'$
And solving for $V'$, you get:
$V' = \frac{C_1V}{C_1 + C_2}$
So, the final potential difference across each capacitor is $\frac{C_1V}{C_1 + C_2}$, which corresponds to option B in your question.