Question

A simple and common technique for accelerating electrons is shown in the figure, where there is a uniform electric field between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to pass through.
Calculate the horizontal component of the acceleration of the electron if the field strength is 2.55*10^4 N/C. Express your answer in m/s^2 and assume the electric field is pointing in the negative x-direction.

Answer 1

Answer : $4.483\times 10^{15}\ m/s^2$.

Explanation:

It is given that,

Electric field strength, $E=2.55\times 10^{4}\ N/C$

We know that,

Charge of electron, $q=1.6\times 10^{-19}\ C$

Mass of electron, $m=9.1\times 10^{-31}\ kg$

From the definition of electric field, $F=qE$...............(1)

According to Newton's second law, F = ma..........(2)

From equation (1) and (2)

$ma=qE$

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

$a=\dfrac{1.6\times 10^{-19}\ C\times 2.55\times 10^{4}\ N/C }{9.1\times 10^{-31}\ kg}$

$a=0.4483\times 10^{16}\ m/s^2$

or

$a=4.483\times 10^{15}\ m/s^2$

So, the horizontal component of acceleration of an electron is $4.483\times 10^{15}\ m/s^2$.

Hence, it is the required solution.