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Q.
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 bound charge $q_{b}$ can be expressed as :
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Solution:
When a dielectric is inserted in a capacitor Due to free charge
$\vec{E}=\vec{E}_{0}$ only After dielectric $E'=\frac{E_{0}}{k}$
$q_{B}=q_{f}\left(1-\frac{1}{k}\right)$