Question

A capacitor consists of two 42 cm-diameter circular plates separated by 1.0 mm. The plates are charged to 170 V,then the battery is removed (= 8 85 102 m) How much energy is stored in the capacitor? Express your answer in joules. ΑΣΦ. o 2 Submit Best A Part B How much work must be done to pull the lates apart to where the distance between them is 20 mm Express your answer in joules |||| th 2 IN

Answer 1

Part A.

Energy stored in the capacitor is given by:

Uc = (1/2)*C*V^2 = (1/2)*C*(Q/C)^2 = Q^2/(2C)

C = Capacitance of capacitor = e0*A/d

A = Area of plate = pi*D^2/4

D = diameter of plate = 4.2 cm = 0.042 m

d = plate separation = 1.0 mm = 1.0*10^-3 m

So,

C = 8.854*10^-12*pi*0.042^2/(4*1.0*10^-3) = 12.27*10^-12 F

Q = charge stored on capacitor = C*V = 12.27*10^-12*170 = 2.086*10^-9 C

Now since battery is removed, So charge on capacitor will remain constant. And

Uc = (2.086*10^-9)^2/(2*12.27*10^-12)

Uc = 17.73*10^-8 J

Part B.

Work-done = Change in potential energy stored

W = dU = Uf - Ui

Ui = 17.73*10^-8 J

Uf = Q^2/(2*Cf)

Cf = e0*A/d1 = 8.854*10^-12*pi*0.042^2/(4*2.0*10^-3) = 6.133*10^-12 F

So,

Uf = (2.086*10^-9)^2/(2*6.13*10^-12) = 35.46*10^-8 J

And

W = 35.46*10^-8 - 17.73*10^-8

W = 17.73*10^-8 J

Let me know if you've any query.