Question

Two large parallel conducting plates carrying equal and opposite charges on their
surfaces facing each other are placed at a distance of 0.75 m. An electron placed at
some point between the plates experiences an electrostatic force of 4.8 × 10-16 N.
Calculate the magnitude of the electric field at the position of the electron between
the plates. Also determine the potential difference between the plates.

Answer 1

The electric field magnitude is given by :

E = F/e

Where :

e = The elementary charge or rather the change if a single proton.

This is a constant which is given as :

e = 1.6 × 10^-19 C

E = Magnitude of the electric field.

F = Electrostatic force.

Doing the substitution we have :

E = (4.8 × 10^-16) / (1.6 × 10^-19)

= 3.0 × 10^3

The potential difference is given by :

Pd = Electric field × the distance between the plates.

= 3000 × 0.75 = 2250 V

= 2.25 × 10^3 V