Question

An electron is accelerqated in the uniform field betwen two parallel charged oplates. The separation of the plates is 1.20 cm

Answer 1

Answer:

The speed of electron is $8.7\times10^{6}\ m/s$

Explanation:

Given that,

Separation of the plate = 1.20 cm

Suppose the field is $E=1.80\times10^{4}\ N/C$.

If the electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate.

What the speed does it leave the hole?

We need to calculate the acceleration

Using formula of electric force

$F = qE$

$ma=qE$

$a=\dfrac{qE}{m}$

We need to calculate the speed of electron

Using equation of motion

$v^2=u^2+2as$

$v^2=2as$

Put the value of acceleration in the formula

$v^2=2\times\dfrac{qE}{m}\times s$

Put the value into the formula

$v^2=2\times\dfrac{1.6\times10^{-19}\times1.80\times10^{4}\times1.20\times10^{-2}}{9.11\times10^{-31}}$

$v=\sqrt{2\times\dfrac{1.6\times10^{-19}\times1.80\times10^{4}\times1.20\times10^{-2}}{9.11\times10^{-31}}}$

$v=8.7\times10^{6}\ m/s$

Hence, The speed of electron is $8.7\times10^{6}\ m/s$