Question

2 pts.) An electron is placed in an electric field of intensity 700 N/Cj. What are themagnitude and direction of the acceleration of this electron due to this field? (melectron=9.1 × 10-31 kg, qe= 1.60 × 10-19 C)

Answer 1

Step-by-step explanation:

We have,

Electric field intensity is 700 N/Cj

It is required to find the magnitude and direction of the acceleration of this electron due to this field.

The electric force is balanced by the force due to its motion i.e.

qE =ma

a is acceleration of this electron

`a=(qE)/(m)\\\\a=(1.6* 10^(-19)* 700)/(9.1* 10^(-31))\\\\a=1.23* 10^(14)\ m/s^2`

So, the acceleration of the electron is
`1.23* 10^(14)\ m/s^2` and it is moving in +y direction.