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# Two large conducting plates are placed parallel to each other with a separation of 2.00 cm 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 :

Given:

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

Time taken by the electron =2μs=2× 10^{-6}s=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^{-31}kg

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/m^{2}

**C/m ^{2}**

**Therefore the surface charge density of the plate is given by**

**0.503× 10 ^{-12}C/m^{2}**

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