Two identical capacitors 1 and 2 are connected in series to a battery as shown in figure. Capacitor 2 contains a dielectric slab of dielectric constant K as shown. $Q_{1}$ and $Q_{2}$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $q_{1}$ and $q_{2}$ . Then :

$\frac{q_{1}}{Q_{1}} = \frac{K + 1}{K}$
$\frac{q_{2}}{Q_{2}} = \frac{K + 1}{2}$
$\frac{q_{2}}{Q_{2}} = \frac{K + 1}{2 K}$
$\frac{q_{2}}{Q_{2}} = \frac{K}{2}$

Question

Two identical capacitors 1 and 2 are connected in series to a battery as shown in figure. Capacitor 2 contains a dielectric slab of dielectric constant K as shown. $Q_{1}$ and $Q_{2}$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $q_{1}$ and $q_{2}$ . Then :

$\frac{q_{1}}{Q_{1}} = \frac{K + 1}{K}$
$\frac{q_{2}}{Q_{2}} = \frac{K + 1}{2}$
$\frac{q_{2}}{Q_{2}} = \frac{K + 1}{2 K}$
$\frac{q_{2}}{Q_{2}} = \frac{K}{2}$

Toppr Admin · Accepted Answer