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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…

JEE Main 2021 — Physics Electromagnetism

2021mcqmedium

If qf{q}_{f}qf​ is the free charge on the capacitor plates and qb{q}_{b}qb​ is the bound charge on the dielectric slab of dielectric constant kkk placed between the capacitor plates, then bound charge qb{q}_{b}qb​ can be expressed as :

Official previous-year question

Held on 25 Jul 2021 · Verified 6 Jul 2026.

Options

  1. A

    qb=qf(1−1k){q}_{b}={q}_{f}(1-\frac{1}{\sqrt{k}})qb​=qf​(1−k​1​)

  2. B

    qb=qi(1−1k){q}_{b}={q}_{i}(1-\frac{1}{k})qb​=qi​(1−k1​)

  3. C

    qb=qf(1+1k){q}_{b}={q}_{f}(1+\frac{1}{\sqrt{k}})qb​=qf​(1+k​1​)

  4. D

    qb=qf(1+1k){q}_{b}={q}_{f}(1+\frac{1}{k})qb​=qf​(1+k1​)

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