Question

2. A parallel plate capacitor has an area of 3x10-4m2 and a plate separation of 1mm. Calculate the (a) capacitance, (b) charge on each plate if a 9V battery is connected, (c) surface charge density and (d) the electric field between the plates. Show the equations used in the calculation and any derivation.

Answer 1

given
surface area, A = 3*10^-4 m^2
plate sepration, d = 1 mm = 1*10^-3 m

a) Capacitance, C = A*epsilon/d

= 3*10^-4*8.854*10^-12/(1*10^-3)

= 2.66*10^-12 F <<<<<<<<------------Answer

b) charge on each plate, Q = C*V

= 2.66*10^-12*9

= 2.39*10^-11 C <<<<<<<<------------Answer

c) surface charge density, sigma = Q/A

= 2.39*10^-11/(3*10^-4)

= 7.97*10^-8 C/m^2 <<<<<<<<------------Answer

d) electric field between the plates, E = V/d

= 9/(1*10^-3)

= 9*10^3 N/C <<<<<<<<------------Answer

(or) E = sigma/epsilon

= 7.97*10^-8/(8.854*10^-12)

= 9*10^3 N/C <<<<<<<<------------Answer