A parallel plate capacitor having capacitance | vedclass

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Question

(B) Initial capacitance $C = 12 \, pF$,Initial potential $V = 10 \, V$.Initial charge $Q = CV = 12 \, pF 	imes 10 \, V = 120 \, pC$.Initial energy $U

Vedclass · Accepted Answer

$508$

Vedclass · Answer

(B) Initial capacitance $C = 12 \, pF$,Initial potential $V = 10 \, V$.Initial charge $Q = CV = 12 \, pF 	imes 10 \, V = 120 \, pC$.Initial energy $U_i = \frac{Q^2}{2C} = \frac{(120)^2}{2 	imes 12} = \frac{14400}{24} = 600 \, pJ$.When the dielectric slab is inserted,the new capacitance becomes $C' = kC = 6.5 	imes 12 = 78 \, pF$.Since the battery is disconnected,the charge $Q$ remains constant.Final energy $U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2kC} = \frac{U_i}{k} = \frac{600}{6.5} \approx 92.31 \, pJ$.The work done by the capacitor on the slab is the decrease in potential energy: $W = U_i - U_f = 600 - 92.31 = 507.69 \, pJ$.Rounding to the nearest integer,we get $508 \, pJ$.

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