Two identical parallel-plate capacitors, each with capacitance 19.5 F, are charg

Question

Two identical parallel-plate capacitors, each with capacitance 19.5 F, 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 doubled.

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

(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.
J

(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. Energy Stored in a Capacitor = .5 C V2

Total Energy Stored = C V2 = 2.8 * 10-2 J

b. When Plate seperation is doubled. Capacitance is halved. Therfore Voltage would be Doubled

Therefore one will have Voltage 53.5 V. While other will have Voltage of 107 V

c. Energy Stored in the smaller capacitor will be same while that of larger is doubled. Therefore Total energy = 3/2 * Initial energy stored

=4.2 * 10-2 J

d. The energy stored is increased due to the Positive work done by the agent pulling the plates apart.

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