A 60 mu F parallel plate capacitor whose pla | vedclass

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(B) Initial capacitance $C = 60 \ \mu F$. Initial potential $V = 250 \ V$. Charge $q = CV = 60 \ \mu F 	imes 250 \ V = 15000 \ \mu C$.When a dielectr

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(B) Initial capacitance $C = 60 \ \mu F$. Initial potential $V = 250 \ V$. Charge $q = CV = 60 \ \mu F 	imes 250 \ V = 15000 \ \mu C$.When a dielectric slab of thickness $t = 3 \ mm$ and dielectric constant $K = 5$ is inserted between plates separated by $d = 6 \ mm$,the new capacitance $C'$ is given by $C' = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.Since $C = \frac{\epsilon_0 A}{d}$,we have $C' = C \left[ \frac{d}{d - t + \frac{t}{K}} ight] = 60 \ \mu F \left[ \frac{6}{6 - 3 + \frac{3}{5}} ight] = 60 \ \mu F \left[ \frac{6}{3 + 0.6} ight] = 60 \ \mu F \left[ \frac{6}{3.6} ight] = 60 \ \mu F 	imes \frac{60}{36} = 60 \ \mu F 	imes \frac{5}{3} = 100 \ \mu F$.Since the charging source is removed,the charge $q$ remains constant. Thus,$q = C'V' \Rightarrow 15000 \ \mu C = 100 \ \mu F 	imes V' \Rightarrow V' = 150 \ V$.The change in potential difference is $\Delta V = V - V' = 250 \ V - 150 \ V = 100 \ V$.

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