JEE Main 2021 (25 Jul, Shift 2) Physics Q11: Dielectrics

Question

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 :

Accepted Answer

Option 1: $q_b = q_f\left(1 - \frac{1}{k}\right)$

Answer

$q_b = q_f\left(1 - \frac{1}{k}\right)$

Answer

$q_b = q_f\left(1 - \frac{1}{\sqrt{k}}\right)$

Answer

$q_b = q_f\left(1 + \frac{1}{k}\right)$

Answer

$q_b = q_f\left(1 + \frac{1}{\sqrt{k}}\right)$

JEE Main 25 July 2021, Shift 2, Physics Q11

Same idea in other shifts