JEE Main · 2018 · 08 Apr 2018 · Offline · Capacitance

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A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant

K = \frac{5}{3}

is inserted between the plates, the magnitude of the induced charge will be:

(A)0.9 nC
(B)1.2 nC✓ Correct
(C)0.3 nC
(D)2.4 nC

Answer: (B)

Solution

With the battery still connected the voltage is fixed, so inserting the dielectric raises the stored charge to

Q = K C V = \frac{5}{3} \times 90 \times 1 0^{- 12} \times 20 = 3

nC. The induced bound charge that appears on the dielectric faces is a fraction of this,

Q_{in d} = Q (1 - \frac{1}{K}) = 3 (1 - \frac{3}{5}) = 3 \times \frac{2}{5} = 1.2

nC.

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option

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