Problem 1) Two identical parallel-plate capacitors, each with capacitance 18.0 µ

Question

Problem 1) Two identical parallel-plate capacitors, each with capacitance 18.0 µF, are charged to potential difference 53.5 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 tripled.

a) Find total energy before the plate separation is tripled.

b) Find potential difference across plates after the plate separation is tripled.

c) Find total energy of the system after the plate separation is tripled.

d) Why is the answer to part (a) and part (c) different? Does it obey the law of conservation of energy?

Explanation / Answer

C = Capacitance = 18 uF

V = potential difference = 53.5

charge stored in each capacitor = Q = 53.5 x 18 uF = 963 uC

Common potential = V' = total charge / total capacitance = 2 x 963 / (18 + 18) = 53.5 volts

a)

total energy stored is given as

E = (0.5) (C1 + C2) V2

E = (0.5) (18 + 18) x 10-6 (53.5)2 = 0.0515 J

b)

since the capacitors are connected in parallel , the potential difference across the plates remains same

C)

after plate saperation is tripled , capacitance of one of the capacitor becomes 1/3rd

so C1 = 18 uF   and C2 = 6uF

total energy stored is given as

E = (0.5) (C1 + C2) V2

E = (0.5) (18 + 6) x 10-6 (53.5)2 = 0.0343 J

d)

yes the law of conservation of energy is still obeyed . the difference in energy is due to the fact that some energy is used to saperate the plates.. so initial and final energy are still equal

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