Question

What is the magnitude of the acceleration experienced by an electron in an electric field of 1200 N/C? How does the direction of the acceleration depend on the direction of the field at that point?

Answer 1

Step-by-step explanation:

It is given that,

Electric field experienced by an electron,
`E=1200\ N/C`

Charge on electron,
`q=-1.6* 10^(-19)\ C`

Mass of electron,
`m=9.1* 10^(-31)\ C`

The electric force is balanced by the force acting on electron as,

ma = qE

`a=(qE)/(m)`

`a=(-1.6* 10^(-19)* 1200)/(9.1* 10^(-31))`

`a=-2.1* 10^(14)\ m/s^2`

So, the acceleration of the electron is
`-2.1* 10^(14)\ m/s^2`. As the electron is a negatively charged particle, the field lines is from negative to positive charge.