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A capacitor of capacity $C_1$ charged upto V volt and then connected to an uncharged capacitor $C_2$. Then final P.D. across each will be
A
$\dfrac{C_2V}{C_1 + C_2}$
B
$\dfrac{C_1V}{C_1 + C_2}$
C
$\left(1 + \dfrac{C_2}{C_1}\right)$
D
$\left(1 - \dfrac{C_2}{C_1}\right)V$
Detailed Solution
When two capacitors are connected, charge flows until both are at the same (common) potential. The total charge is conserved.
Common potential: $V_{common} = \dfrac{\text{total charge}}{\text{total capacitance}} = \dfrac{C_1V_1 + C_2V_2}{C_1 + C_2}$
Here $V_1 = V$ and $V_2 = 0$ (uncharged):
$V_{common} = \dfrac{C_1V}{C_1 + C_2}$
So the final p.d. across each capacitor is $\dfrac{C_1V}{C_1 + C_2}$.
Common potential: $V_{common} = \dfrac{\text{total charge}}{\text{total capacitance}} = \dfrac{C_1V_1 + C_2V_2}{C_1 + C_2}$
Here $V_1 = V$ and $V_2 = 0$ (uncharged):
$V_{common} = \dfrac{C_1V}{C_1 + C_2}$
So the final p.d. across each capacitor is $\dfrac{C_1V}{C_1 + C_2}$.