Two identical capacitors 1 and 2 connected in series to a battery as shown in figure. Capacitor 2 contains a dielectric slab of dielectric constant K as shown. If $Q_{1}$ is the charge on each capacitor before removing the slab and $Q_{2}$ is the charge on each capacitor after
moving the slab, then the correct relation between $Q_{1}$ and $Q_{2}$
is.

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

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

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

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

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Solution

The correct option is C $\frac{Q_{1}}{Q_{2}} = \frac{2 K}{K + 1}$
Since the two capacitors are in series both have equal charges. Let $Q_{1}$ be the charge on the system.Then charge on the system is
$Q_{1} = (\frac{C \times K C}{C + K C}) V_{1} = (\frac{K}{K + 1}) C V_{1}$
$∴ V_{1} = \frac{Q_{1} (K + 1)}{K C}$ ....(1)
On removing the dielectric slab the charge on the system is
$Q_{2} = \frac{C \times C}{C + C} V_{2} = \frac{C V_{2}}{2}$
$∴ V_{2} = \frac{2 Q_{2}}{C} . . . . (2)$
The battery being the same $V_{1} = V_{2}$ from
Eqs. (1) and (2) we get $\frac{Q_{1}}{Q_{2}} = \frac{2 K}{(K + 1)}$

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