Question

A parallel-plate vacuum capacitor has 7.28 J of energy stored in it. The separation between the plates is 2.90 mm . If the separation is decreased to 1.90 mm

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

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

U =		J

Answer 1

Charger remains constant, then energy stored in capacitor: U = 1/2 Q^2/C = Q^2 d/2 epsilon_0 A U = U_0 = d/d_0 7.28/U_0 = 2.90/1.90 U_0 = 4.76 J (b) Now voltage is constant then energy stored in capacitors, U = 1/2 CV^2 1/ epsilon_0 A/d V^2 U/U_0 = d_0/d 7.28/U_0 = 1.90/2.90 U_0 = 11.11 J