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If $q_{f}$ is the free charge on the capacitor plates and $q_{b}$ is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then the magnitude of bound charge $q_{b}$ can be expressed as :

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$q_{b} = q_{f} (1 - \frac{1}{\sqrt{k}})$

$q_{b} = q_{f} (1 - \frac{1}{k})$

$q_{b} = q_{f} (1 + \frac{1}{\sqrt{k}})$

$q_{b} = q_{f} (1 + \frac{1}{k})$

answer is B.

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Detailed Solution

When a dielectric is inserted in a capacitor Due to free charge $\vec{E} = {\vec{E}}_{0}$ only

After dielectric $E^{'} = \frac{E_{0}}{k}$

$q_{B} = q_{f} (1 - \frac{1}{k})$

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