Two identical parallel-plate capacitors, each with capacitance C, are charged to

Question

Two identical parallel-plate capacitors, each with capacitance C, are charged to potential difference ?V and connected in parallel. Then the plate separation in one of the capacitors is doubled. (Use DeltaV for ?V and C as necessary.) (a) Find the total energy of the system of two capacitors before the plate separation is doubled. U = (b) Find the potential difference across each capacitor after the plate separation is doubled. ?V ' = (c) Find the total energy of the system after the plate separation is doubled. U ' = (d) Reconcile the difference in the answers to parts (a) and (c) with the law of conservation of energy.

Explanation / Answer

(a) Find the total energy of the system of two capacitors before the plate separation is doubled.

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

(b) Find the potential difference across each capacitor after the plate separation is doubled. ?V ' =

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

(c) Find the total energy of the system after the plate separation is doubled. U ' =

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

(d) Reconcile the difference in the answers to parts (a) and (c) with the law of conservation of energy.

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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