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

Question

Two identical parallel-plate capacitors, each with capacitance C, are charged to potential difference ?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. (Use any variable or symbol stated above as necessary.)

Two identical parallel - plate capacitors, each with capacitance C, are charged to potential difference ?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. (Use any variable or symbol stated above as necessary.) Find the total energy of the system of two capacitors before the plate separation is doubled. Find the potential difference across each capacitor after the plate separation is doubled. Find the total energy of the system after the plate separation is doubled. Reconcile the difference in the answers to parts (a) and (c) with the law of conservation of energy

Explanation / Answer

a)

U=(1/2)CV^2 +(1/2)CV^2

U=CV^2

b)

d'=2d

C=eoA/d' =eoA/2d =(1/2)(eoA/d)

C'=C/2

The total charge is same as before

C*V+C*V =C*V' +V'*(C/2)

V' =4V/3

c)

U' =(1/2)C*(4V/3)^2+(1/2)(C/2)(4V/3)^2

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

d)

the extra energy comes from work put into the system by pulling the capacitor plates apart

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