Question

Twolargeparallelconductingplatescarryingequalandoppositechargesontheir 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 magnitude of the electric field at the position of the electron between the plates E = F/e = 4.8*10^-16/1.6*10^-19 = 3(kV/m). The potential difference between the plates U = El = 3000*0.75 = 2.25 9 (kV). So, E = 3 kV/m, U = 2.25 kV.