Q 12-02-140JEE MainJEE Main 2021 (25 Jul, Shift 2)Easy
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:
Answer: (B) $q_b = q_f\left(1 - \frac{1}{k}\right)$
The field inside the dielectric is reduced by the factor $k$: $\dfrac{q_f - q_b}{A\varepsilon_0} = \dfrac{q_f}{kA\varepsilon_0}$.
$$q_b = q_f\left(1 - \frac{1}{k}\right)$$
Solution by Sreeraj P, M.Sc Physics