A parallel-plate vacuum capacitor has 7.62 J of energy stored in it. The separat

Question

A parallel-plate vacuum capacitor has 7.62 J of energy stored in it. The separation between the plates is 3.70 mm . If the separation is decreased to 1.10 mm ,

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?

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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?

A parallel-plate vacuum capacitor has 7.62 J of energy stored in it. The separation between the plates is 3.70 mm . If the separation is decreased to 1.10 mm ,

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?

U =   J  

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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?

U =

Explanation / Answer

d1 = initial saperation of plates = 3.70 mm

d2 = final saperation of plates = 1.10 mm

C1 = initial Capacitance = 7.62 J

C2 = final Capacitance = ?

Since Capacitance is inversly related to saperation of plates

C1/C2 = d2/d1

C1/C2 = 1.10/3.70

C2 = 3.4 C1

when the capacitor is disconnected from the potential source

energy stored remain same

U2 = U1 = 7.62 J

as the saperation of plates is decreased , the capacitance increases. Due to increase in capacitance , the capacitor could store more energy , but since the capacitor is disconnected from the Voltage source, it does not gain any extra energy and hence the Potential difference across the capacitor decreases so as toe have the same energy as before.

b)

Given that : U1 = (0.5) C1 V2 = 7.62 J

U2 = (0.5) C2 V2 = (0.5) (3.4 C1 ) V2 = (3.4) (7.62) = 26 J

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