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

Force experienced by the electron F= 4.8×10⁻⁶N

Charge of an electron q= 1.6×10⁻¹⁹C

Electric field experienced by the charge is,

E= F/q= 4.8×10⁻¹⁶/1.6×10⁻¹⁹= 3×10³N/C

Potential difference between the plates, V= E. dS

Where, ds= distance between the plates

E= electric field due to the charged plates

 V= E.ds= 3×10³×0.75= 2.25×10³V