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?
b. What is the direction of the electric field?

1. in the direction of the electron's motion
2. opposite to the direction of the electron's motion
3. perpendicular to the direction of the electron's motion

Answer 1

Answer:

The answer is below

Explanation:

a) The work done is equal to the loss in kinetic energy (KE)

Change in kinetic energy (ΔKE) = Final kinetic energy - initial kinetic energy

Final KE = $\frac{1}{2}mv_f^2$

But the final velocity is 0 (at rest). Hence:

Final KE = $\frac{1}{2}mv_f^2=\frac{1}{2}m(0)^2=0$

ΔKE = 0 - K = -K

W = ΔKE = -K

Also, the work done (W) = charge (q) * distance (d) * electric field intensity (E)

W = qEd

but q = -e, hence:

W = -e * E * d

Using:

W = ΔKE

-e * E * d = -K

E= K / (e * d)

b) The electric field is in the direction of the electrons motion