Question

A parallel-plate vacuum capacitor has 7.22J of energy stored in it. The separation between the plates is 2.70mm . If the separation is decreased to 1.85mm ,

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

Energy stored in capacitor = U = 7.22 J

U =(1/2)CV^2

C =(eo)A/d

C*d=(eo)A=constant

C2d2=C1d1

C2=C1d1/d2

The separation between the plates is 2.70mm . The separation is decreased to 1.85 mm.

Initial separation between the plates =d1= 2.70mm .

Final separation = d2 = 1.85 mm
(A)

if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same

Energy=U =(1/2)q^2/C

U2C2 = U1C1

U2 =U1C1 /C2

U2 =U1d2/d1

Final energy = Uf = initial energy *d2/d1

Final energy = Uf =7.22 *1.85/2.70

Final energy = Uf = 4.947J

B) If the capacitor remained connected to the potential source while the separation of the plates was changed, potential remains same.

Final energy = Uf =(1/2)C!V^2

Final energy = Uf =(1/2)[C1d1/d2]V^2

Final energy = Uf =(1/2)[C1V^2]d1/d2

Final energy = Uf = initial energy *d1/d2

Final energy = Uf =7.22 *2.70/1.85

Final energy = Uf =10.537 J