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JEE Main 2021 (25 Jul, Shift 2) · Dielectrics

Dielectrics question from JEE Main 2021 (25 Jul, Shift 2)

Q76JEE Main 2021 · 25 JUL · SHIFT 2Easy

Dielectrics

If qf is the free charge on the capacitor plates and qb is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge qb an be expressed as:

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Answer: (b)

Field inside the dielectric is $\frac{1}{k}$ of the free-charge field: $q_f - q_b = \frac{q_f}{k}$ .
$q_b = q_f\left(1 - \frac{1}{k}\right)$ .
So the answer is (b).

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