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?
B. What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed?
energy stored in the capacitor = CV2 /2 = Q2/2C
When the source is disconnected the charge on the capcitore remains const.
C = Ao/d
seperation is decreased from 3.5 to 1.9
new capcitance C' = C *3.5/1.9 = 1.84C
Energy stored = Q2/2*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'V2/2 = 1.84CV2/2
= 1.84 * 7.92 = 14.57 J
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