Question

Two large conducting plates are placed parallel to each other with a separation of 2.00cm between them. An electron starting from rest near one of the plates reaches the other plate in 2.00 microseconds. Find the surface charge density on the inner surfaces.

Answer 1

s = 2cm = 2 × 10^–2m, 

u = 0, a = ? 

t = 2μs = 2 × 10^–6s

Acceleration of the electron, 

s= (1/2)at²

2 × 10^–2 = (1/2) × a × (2 × 10^–6)² => a = (2 x 2 x 10-²)/(4 x 10^-12) 

=> a = 10¹⁰m/s²

The electric field due to charge plate = σ/ε₀

= 50.334 × 10^–14 = 0.50334 × 10^–12c/m²