Question

A parallel-plate vacuum capacitor has 7.92 J of energy stored in it. The separation between the plates is 3.50 mm . If the separation is decreased to 1.90 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?

U =?

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

energy stored in the capacitor = CV^{2} /2 =
Q^{2}/2C

When the source is disconnected the charge on the capcitore remains const.

C = A_{o}/d

seperation is decreased from 3.5 to 1.9

new capcitance C' = C *3.5/1.9 = 1.84C

Energy stored = Q^{2}/2*1.84C

energy stroed dcreases by a factor of 1.84

inital energy = 7.92 J

energy after seperation changed = 7.92/1.84 = 4.3 J

b) if the capcitor remain connected to the power source the voltage across the capictor remains const.

new energy = C'V^{2}/2 = 1.84CV^{2}/2

= 1.84 * 7.92 = 14.57 J

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