Two identical parallel-plate capacitors, each with capacitance 17.5 Mu F, are ch

Question

Two identical parallel-plate capacitors, each with capacitance 17.5 Mu F, are charged to potential difference 56.0 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 tripled. (a) Find the total energy of the system of two capacitors before the plate separation s tripled. 0.05488 J (b) Find the potential difference across each capacitor after the plate separation is tripled. 74.7 Your response differs from the correct answer by more than 10%. Double check your calculations. V (c) Find the total energy of the system after the plate separation is tripled. 0.0732 Your response differs from the correct answer by more than 10%. Double check your calculations. J (d) Reconcile the difference in the answers to parts (a) and (c) with the law of conservation of energy. Negative work is done by the agent pulling the plates apart. Positive work is done by the agent pulling the plates apart. No work is done by the agent pulling the plates apart.

Explanation / Answer

C = 17.5 X 10-6 F and V = 56v

a)

E = 1 / 2 C V2 = 1 / 2 C V2 X 3

= 3 / 2 C V2

= 3 / 2 X 17.5 X 10-6 X 562

= 82320 X 10-6

E = 0.082320 J

b)

Q = CV0 X 3 = 3 CV0

triple the spacing and the capacitance is cutting in one third

C + 1 / 3 C = 1.33 C

3 CV0 = V X 1.33C

V = 3 / 1.33V0 = 2.25 V0

c)

total energy of the system E = 1 / 2 CV2

E = 1/2 X 1.33 C X ( 2.25 V0 )2

E = 3.36 C V02

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