Magnitude of potential difference

Question

A small object of mass 2.92 g and charge 10.5 uC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground created by two plates separated by 10.0cm. What is the magnitude of potential difference?

Answer 1

Let

q = 10.5uC be the object's charge,

d = 10cm be the distance between the plates,

V be the potential to be calculated,

m = 2.92g be the object's mass,

g = 9.8m/s² be gravitational acceleration,

F = mg be the object's weight,

E be the electric field between the plates.

We are told that the force of the electric field E is exactly equal the object's weight F,

therefore

E = F/q

In general in a capacitor

E = V/d

Therefore

V/d = F/q

V = Fd/q

Substituting actual numbers:

V = 2.92g x 9.8m/s² x 10cm / 10.5uC

= 27.25 g m/s² cm / uC

= 27.25 10^-3kg m/s² 10^-2m / 10^-6 As

= 272.5V