Question

An electron is released in a uniform electric field, and it experiences an electric force of 2.2 ✕ 10^-14 N downward. What are the magnitude and direction of the electric field?

Magnitude
____________ N/C
Direction

upward, to the left, to the right or downward?

Answer 1

Use F=EQ, where F is the force experienced in newtons, E is the magnitude of the electric field strength (in N/C)

and Q is the charge on the electron.

1. Magnitude = F/Q = (2.2*10^-14) / (1.6x10^-19) = 137500 N/C

2. Direction: direction of an electric field is always given by the direction of motion of a POSITIVELY-CHARGED

particle in the field. If the electron moves downward, then a positive particle will move upward. Hence direction of

field is upward.