Question

A parallel-plate capacitor is constructed of two horizontal 17.2-cm-diameter circular plates. A 1.8 g plastic bead with a charge of -4.8 nC is suspended between the two plates by the force of the electric field between them. What is the charge on the positive plate?

Answer 1

Given values:
diameter (d) = 0.172 m which means radius (r) = 0.086 m
mass of bead = 1.8 g = 0.0018 kg
charge of bead = -4.8 nC = -4.8 X 10^-9 C

Equation to use:
E = Q/(p*A), where p is permittivity constant which is 8.85 X 10^-12 C/(N*m^2), A is area of capacitor.
But you are solving for the charge on the positive plate (Q) so rearrange this formula:
Q = E*p*A

F=mg

F=0.01764N

E = F/q

E=3.675*10^6 N/C

Q = E*p*A

Q=7.55X10^-7 C