Given:

Distance travelled by the electron=2cm=2× 10^-2m=s

Time taken by the electron =2μs=2× 10^-6s=t

Let the surface charge density of the plate be σ

Let the acceleration of electron be a

Using 2^nd law of motion

Where,

u=initial velocity

t=time taken to travel

a=acceleration of particle

s=displacement of particle

it is given that electron starts from rest ∴ u=0

This acceleration is provided by the force due to electric field between plates

Force applied to the particle is given by

…(i)

Where,

m=mass of electron=9.1× 10^-31kg

this force is provided by the electric field (E) between the plates which is given by

…..(ii)

Equating eqns.(i) and (ii)

..(iii)

We know that

Electric field due to a conducting plate of surface charge density σ is given by

Putting this value of E in eqn.(iii)

Putting values of s, m, t and q

C/m²

C/m²

Therefore the surface charge density of the plate is given by

0.503× 10^-12C/m²