Question

A parallel-plate vacuum capacitor has 8.76 J of energy stored in it. The separation between the plates is 3.70 mm . If the separation is decreased to 1.35 mm ,

a)what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed?

b)What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed?

Answer 1

a)

inital energy in the capacitor

U1 = 8.76 = 0.5 Q^2 / C

0.5 Q^2 / ( eo A / d) = U1

0.5* Q^2 d / eo A = U1

here, for charge to be constant d is directly proportional to Energy so

U2 / U1 = d2 / d1

U2 = 8.76 * 1.35 / 3.7

U2 = 3.196 J

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b)

now, for constant voltage

U1 = 0.5 CV^2 = 0.5 eo A V^2 / d

so U1 is inversely proportional to d

U2 / U1 = d1/ d2

U2 = 8.76*3.7 / 1.35

U2 = 24 J

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