A capacitor stores 60 mu C charge when conne | vedclass

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(C) Let the initial capacitance be $C_i$ and the potential difference across the battery be $V$. The initial charge is $q_i = C_i V = 60 \ \mu C$.When

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$3$

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(C) Let the initial capacitance be $C_i$ and the potential difference across the battery be $V$. The initial charge is $q_i = C_i V = 60 \ \mu C$.When a dielectric of constant $K$ is inserted,the new capacitance becomes $C_f = K C_i$.The new charge on the capacitor is $q_f = C_f V = K C_i V = K q_i$.The additional charge that flows through the battery is $\Delta q = q_f - q_i = 120 \ \mu C$.Substituting the values,we get $K q_i - q_i = 120 \ \mu C$.$(K - 1) 	imes 60 \ \mu C = 120 \ \mu C$.$K - 1 = \frac{120}{60} = 2$.$K = 3$.

$A$ capacitor stores $60 \ \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,an additional charge of $120 \ \mu C$ flows through the battery. The dielectric constant of the material inserted between the plates of the capacitor is

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