Part A Part A what is the energy now stored if the capacitor was disconnected fr

Question

Part A Part A what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed? Part B What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed? Part A what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed? Part B What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed? Part B What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed? Part B What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed?

Explanation / Answer

a)

Charge remains constant ,so Energy stored in capacitor

U=(1/2)Q^2/C =(1/2)*Q^2/(eoA/d)

U=(1/2)*Q^2*d/eo*A

so

U/Uo =d/do

8.38/Uo =3.1/1.15

Uo=3.11 J

b)

Now Voltage is constant ,so energy stored in capacitor is

U=(1/2)*C*V^2 =(1/2)*(eoA/d)*V^2

so

U/Uo =do/d

8.38/Uo=1.15/3.1

Uo=22.59 J

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