Question

An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after traveling a distance d.
a. What is the magnitude of the electric field? (Use any variable or symbol stated above along with the following as necessary: e for the charge of the electron; Apply the work-kinetic energy theorem, and relate the force to the electric field.)
b. What is the direction of the electric field?
i. in the direction of the electron's motion
ii. opposite to the direction of the electron's motion
iii. perpendicular to the direction of the electron's motion

Answer 1

Answer:

a) E = $\frac{K}{e*d}$  b) i.

Explanation:

Assuming no other forces acting on the electron, as the field is pushing the electron back till it comes to rest momentarily, applying the work-energy theorem, the loss in kinetic energy, must be equal to the work done by the field on the electron.

The force that does work on the electron is the one due to the electric field E, and by definition of electric field, can be expressed as follows:

W = q*E = -e*E (1)

This work must be equal to the change in Kinetic Energy:

ΔK = Kf -K₀ = 0 - K = -K (2)

From (1) and (2), and solving for E, we have:

E =$\frac{K}{e*d}$

b) As the electric field (by convention) has the direction that it would have a positive test charge, the electrons (being negative charges),when starting from rest),have the direction opposite to the field.

In this case, as the electron has an initial velocity, and the field is opposing to the electron movement, we conclude that the electric field is in the direction of the electron´s motion, due to the electron slows down, and then comes momentarily to an stop, before changing direction and move opposite to the field as it is its natural behavior.