Question

An electric field of 234,000 N/C points due west at a certain spot. What are the magnitude and direction of the force that acts on a charge of -7.25 µC at this spot?

Answer 1

F = 1.69 N

Step-by-step explanation:

Given that,

Electric field, E = 234,000 N/C

Charge, Q = -7.25 µC

We need to find the electric force acting on the charge. It can be given as follows :

`F=qE\\\\F=7.25* 10^(-6)* 234000\\\\F=1.69\ N`

As the charge is negative, the force will act in the opposite direction of electric field. Hence, the electric force is 1.69 N.