A parallel plate capacitor of capacitance 90 | vedclass

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(D) Initial charge on the capacitor is $Q_i = CV = 90 \ pF 	imes 20 \ V = 1800 \ pC = 1.8 \ nC$.After inserting a dielectric of dielectric constant $

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$1.2$

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(D) Initial charge on the capacitor is $Q_i = CV = 90 \ pF 	imes 20 \ V = 1800 \ pC = 1.8 \ nC$.After inserting a dielectric of dielectric constant $K = \frac{5}{3}$,the new capacitance becomes $C' = KC = \frac{5}{3} 	imes 90 \ pF = 150 \ pF$.The new charge on the capacitor is $Q_f = C'V = 150 \ pF 	imes 20 \ V = 3000 \ pC = 3.0 \ nC$.The magnitude of the induced charge $Q_{ind}$ on the dielectric is given by the difference between the final charge and the initial charge:$Q_{ind} = Q_f - Q_i = (K-1)CV$$Q_{ind} = \left(\frac{5}{3} - 1ight) 	imes 90 \ pF 	imes 20 \ V$$Q_{ind} = \left(\frac{2}{3}ight) 	imes 1800 \ pC = 1200 \ pC = 1.2 \ nC$.

$A$ 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.......$nC$.

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