Question

An electron is released from rest in a uniform electric field. The electron accelerates, travelling 6.20 m in 4.50 µs after it is released. What is the magnitude of the electric field in N/C?

Answer 1

E= 3.4893 N/C

Step-by-step explanation:

Given s=6.20 m , t=2.50μs, m=9.11*10^-31 Kg , q= 1.6*10^-19 C

the distance traveled by the electron in time t is

s=ut+0.5at^2

here, u is the initial velocity of the electron, t is time taken and

a is acceleration.

Since the electron is initially at rest u=0

now s=0.5at^2

Therefore a=2s/t^2

also. we know that strength of electric field is

E=ma/q

`E= (2ma)/(qt^2)`

now puting the values we get

`E=(9.11* 10^-31* 2* 6.20)/(1.6* 10^-19* (4.5* 10^-6)^2)`

therefore, E= 3.4865 N/C