Two identical parallel-plate capacitors, each with capacitance 15.5 uF, are char

Question

Two identical parallel-plate capacitors, each with capacitance 15.5 uF, are charged to potential difference 49.0 V and then disconnected from the battery. They are then connected to each other in parallel with plates of like sign connected. Finally, the plate separation in one of the capacitors is doubled (a) Find the total energy of the system of two capacitors before the plate separation is doubled. (b) Find the potential difference across each capacitor after the plate separation is doubled (c) Find the total energy of the system after the plate separation is doubled (d) Reconcile the difference in the answers to parts (a) and (c) with the law of conservation of energy. Positive work is done by the agent pulling the plates apart. Negative work is done by the agent pulling the plates apart. No work is done by pulling the agent pulling the plates apart.

Explanation / Answer

a) Total energy,

U = 0.5 C (?V)^2 + 0.5 C (?V)^2 = C(?V)^2

U = 15.5 x 10^-6 x 49^2 = 0.037 J

(b) Initial total charge = Qtoti = 2 C V

final total capacitance = Ctotf = C + C/2 = 3C/2

Qtotf = Qtoti = 2 C V

?V' = Qtotf/Ctotf = 2 C V/(3C/2) = (4/3) ?V

= (4/3) x 49^2 = 3201.33 V

(c)

U' = 0.5 (3C/2) (4/3)^2(?V)^2 = (4/3) C (?V)^2

U' = (4/3) 15.5 x 10^-6 x 49^2 = 0.05 J

(d) ?U = (4/3) C (?V)^2 - C (?V)^2 = (1/3) C (?V)^2

We had to do some work to double the plate separation. This work stored as potential energy.

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