Question

A small plastic ball with a mass of 5.85 10-3 kg and with a charge of +0.147 µC is suspended from an insulating thread and hangs between the plates of a capacitor (see the drawing). The ball is in equilibrium, with the thread making an angle of 30.0° with respect to the vertical. The area of each plate is 0.0157 m2. What is the magnitude of the charge on each plate?

Answer 1

given

m = 5.85x10^-3Kg

Q = 0.147C

AREA A = 0.0157m^2

the following formula to be used are

F = Eq-------(1)

here F = m x g

         = (5.85x10^-3)(9.8)

      F = 0.057N

FROM EQN (1) WE HAVE

the magnitude of the charge q = F / E---------(2)

TO CALCULATE E WE MAKE USE OF THE FORMULA

E = / 0--------(3)

WE HAVE = Q /A---------(4)

                   = (0.147x10^-6) / 0.0157

              = 9.36x10^-6C/m^2------(5)

substituting the values of eqn (5) in eqn(3) we have

E = (9.36x10^-6) / (8.854x10^-12)

E = 1.057x10^6N/C---------(6)

SUBSTITUTING THE VALUES OF F AND E IN EQN (2) WE HAVE

q = (0.057) / (1.057x10^6)

q = 0.053C