A parallel-plate capacitor has capacitance 4.00 mu F. How much energy is stored

Question

A parallel-plate capacitor has capacitance 4.00 mu F. How much energy is stored in the capacitor if it is connected to a 18.00 V battery? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully, mu j If the battery is disconnected and the distance between the charged plates doubled, what is the energy stored? The battery is subsequently reattached to the capacitor, but the plate separation remains as in part (b). How much energy is stored?

Explanation / Answer

(a)

energy stored in capacitor is,

U = (1/2)*C*V^2

U = (1/2)*4*10^(-6)*18^2 = 648 uJ

(b)

lf battery is disconnected and distance b/w plates doubled,

C = A*e0 / d

C' = A*e0 / 2d

C' = C / 2

energy stored in capacitor,

U' = Q^2 / 2*C' = Q^2 / 2*(C / 2)

U' = 2*U = 1296 uJ

(c)

when battery is reattached and plate separation is doubled,

U' = (1/2)*C'*V^2

U' = (1/2)*(C / 2)*V^2 = (1 / 2)*(C*V^2 / 2)

U' = U / 2

U' = 324 uJ

answer

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