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Capacitor question

2021 · 25 Jul · Shift 2 · Q54
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  5. /2021 · 25 Jul · Shift 2 · Q54

Capacitor question

2021 · 25 Jul · Shift 2 · Q54

JEE MainPhysicsCapacitorMCQ+4 / −1
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 :
  1. A
    qb=qf(1−1k){q_b} = {q_f}\left( {1 - {1 \over {\sqrt k }}} \right)qb​=qf​(1−k​1​)
  2. B
    qb=qf(1−1k){q_b} = {q_f}\left( {1 - {1 \over k}} \right)qb​=qf​(1−k1​)
  3. C
    qb=qf(1+1k){q_b} = {q_f}\left( {1 + {1 \over {\sqrt k }}} \right)qb​=qf​(1+k​1​)
  4. D
    qb=qf(1+1k){q_b} = {q_f}\left( {1 + {1 \over k}} \right)qb​=qf​(1+k1​)
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