Question

The electrons in a particle beam each have a kinetic energy of 1.60×10−17 J. What are the magnitude and direction of the electric field that will stop these electrons in a distance of 10.0cm?

Answer 1

`E = F/q = 1.60×10^-16 N/1.60×10^-19 C = 1xx10^3 C`

Explanation:

Use the Work-Energy Theorem: `W_("net") = DeltaK`
As the electron slows to a halt, its change in kinetic energy is:
`DeltaK = K_f −K_i =0−(1.60×10^-17 J) = −1.60×10^-17 J`
So `W = −1.60×10^-17 J`
Let the electric force on the electron has magnitude `F`. The electron moves a distance `d = 10 .0 cm` opposite the direction of the force so that the work done is:
`W =−Fd; −1.60×10^-17 J= −F(10.0×10^-2 m)`
solving for, `F = 1.60×10^-16 N`
Now knowing the charge of the electron we can evaluate the electric field, E:
`E = F/q = 1.60×10^-16 N/1.60×10^-19 C = 1xx10^3 C`