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?

E= 3.4893 N/C

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

now puting the values we get

therefore, E= 3.4865 N/C

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge

0.55x745.7j and a calculator