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
JEE MAIN 2021, Diffcult
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When a dielectric is inserted in a capacitor
Due to free charge $\vec{E}=\vec{E}_{0}$ only
After dielectric $E^{\prime}=\frac{E_{0}}{k}$
$q_{B}=q_{f}\left(1-\frac{1}{k}\right)$
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