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

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2. Electric Potential and 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.......$n $ $C$

A

$0.3$

B

$2.4$

C

$0.9$

D

$1.2$

(JEE MAIN-2018)

Solution

Charge on Capacitor, $\mathrm{Q}_{i}=\mathrm{CV}$

After inserting dielectric of dielectric constant = $\mathrm{K}$

$\mathrm{Q}_{\mathrm{f}}=(\mathrm{k} \mathrm{C}) \mathrm{V}$

Induced charges on dielectric

$ \mathrm{Q}_{\mathrm{ind}} =\mathrm{Q}_{\mathrm{f}}-\mathrm{Q}_{\mathrm{i}}=\mathrm{KCV}-\mathrm{CV} $

$=(\mathrm{K}-1) \mathrm{CV}=\left(\frac{5}{3}-1\right) \times 90 \,\mathrm{pF} \times 2\, \mathrm{V}=1.2\, \mathrm{nc} $

Standard 12

Physics

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(IIT-1998)