A capacitor consists of two 6.4-cm -diameter circular plates separated by 1.0 mm . The plates are charged to 130 V , then the battery is removed. How much energy is stored in the capacitor? How much work must be done to pull the plates apart to where the distance between them is 2.0 mm ?

### Answer

(a)

d = diameter of the circular plates of the capacitor = 6.4 cm = 0.064 m

A = Area of each plate of the capacitor = (0.25) \pid2 = (0.25) (3.14)(0.064)2 = 0.00322 m2

d = separation of the plates = 1 mm = 1 x 10-3 m

Capacitance of the capacitor is given as

C = \epsilon _{o}A/d

C = (8.85 x 10-12)(0.00322)/(1 x 10-3)

C = 2.85 x 10-11 F

V = potential difference between the plates = 130 Volts

Energy stored by the capacitor is given as

U = (0.5) CV2

U = (0.5) (2.85 x 10-11)(130)2

U = 2.40825 x 10-7 J

b)

Q = charge stored by the capacitor

Charge stored by the capacitor is given as

Q = C V

Q = (2.85 x 10-11)(130)

Q = 3.705 x 10-9 C

d' = new distance between the plates of capacitor = 2 mm = 0.002 m

New Capacitance of the capacitor is given as

C' = \epsilon _{o}A/d'

C' = (8.85 x 10-12)(0.00322)/(2 x 10-3)

C' = 1.42485 x 10-11 F

Energy stored by the new capacitor is given as

U' = (0.5) Q2/C'

U' = (0.5) (3.705 x 10-9)2/(1.42485 x 10-11)

U' = 4.82 x 10-7 J

W = work done in pulling the plates apart

work done in pulling the plates apart is same as the difference in the energy stored in capacitor before and after pulling the plates

W = U' - U

W = ( 4.82 x 10-7 ) - (2.40825 x 10-7)

W' = 2.41 x 10-7 J