Question

2.A parallel plate air capacitor has rectangular
plates each of length 20cm and breadth 10cm,
separated by a distance of 2mm.The potential
difference between the plates is
500V.Calculate:a)Capacitance of capacitor
b)Charge on each plate
c) Electric filed intensity between the two
plates.(Ans:C=8.85 x10-11,Q =4.425 x10-8C,E
=250x103V/m)​

Answer 1

a) The capacitance of the capacitor = 8.85×10^-12 Farad

b) The charge on each plate = 44.25 × 10^-10 Coulomb

c) The electric field intensity between the two plates = 2.5 × 10^3 V/m

The length of the capacitor = 20cm breadth = 10cm

d = the seperation between the capacitor plates = 2mm

V = potential difference = 500V

Area of the plate = A = 20 × 10 = 200 cm²

The capacitance of the capacitor = Aε0/d

= (200 × 10^-4 × 8.85 × 10^-11)/(2 × 10^-3)

C = 8.85×10^-12 Farad

Charge on each plate = CV

= 8.85×10^-12 × 500

= 44.25 × 10^-10 Coulomb

Electric field intensity between the plates = V/d = 500/(2×10^-3)

= 2.5 × 10^3 V/m