A 12 μF parallel plate capacitor is charged to 200 V and then connected to the 7.0 μF capacitor. The plates are physically pulled apart to double the original separation. a)What is the stored energy? b)Explain the difference in stored energy between the two capacitors. c)Calculate the work performed by the outside agent pulling the plates apart
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A 12 μF parallel plate capacitor is charged to 200 V and then connected to the 7.0 μF capacitor. The plates are physically pulled apart to double the original separation.
a)What is the stored energy?
b)Explain the difference in stored energy between the two capacitors.
c)Calculate the work performed by the outside agent pulling the plates apart
d)The average force exerted on the plates
e)The average power if the separation occurs over 10 s.
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It's a question of Energy conservation :
Given :
C1= Capacitance of a charged capacitor=12µF = 12 x 10-6 F
C2= Capacitance of the uncharged capacitor =2µF = 2 x 10-6 F
Supply voltage, V1 = 200 V
Electrostatic energy stored in C1 is given by,
When C2 is connected to the circuit, the potential acquired by it is V2.
According to the conservation of charge, initial charge on capacitor C1 is equal to the final charge on capacitors, C1 and C2.
Now, plates are pulled apart, so the new capacitance will be the half of it, i.e.
a) Electrostatic energy for the combination of two capacitors is given by,
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