Question

An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electron is 4.52×105 m/s pointed toward the east. What is the speed of the electron when it reaches point B, which is a distance of 0.395 m east of point A?

Answer 1

Answer:

Final velocity of electron, $v=6.45\times 10^5\ m/s$    

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, $u=4.52\times 10^5\ m/s$

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

$v^2=u^2+2as$........(1)

a is the acceleration, $a=\dfrac{F}{m}$

We know that electric force, F = qE

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

Use above equation in equation (1) as:

$v^2=u^2+\dfrac{2qEs}{m}$

$v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m$

v = 647302.09 m/s

or

$v=6.45\times 10^5\ m/s$

So, the final velocity of the electron when it reaches point B is $6.45\times 10^5\ m/s$. Hence, this is the required solution.