Question

Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu F$ are individually charged by a $100 V D C$ source Capacitor $C_1$ is kept connected to the source and a dielectric slab is inserted between it plates Capacitor $C_2$ is disconnected from the source and then a dielectric slab is inserted in it Afterwards the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination The common potential of the combination will be ___$V$
(Assuming Dielectric constant $=10$ )

Answer 1

Given : Charge on C1 = KCE
Charge on C2 = CE
When they are connected in parallel charge will be equally divided so charge on one capacitor is
$q=\frac{K+1}{2}CV$
$\therefore$ $V=\frac{q}{KC}$
$=\frac{K+1}{2K}=55 V$
so, the correct answer is 55.