Question

A positive charge of 4.0 x (10^-5) C experiences a force

of 0.36 N when located at a certain point. What is the
electric field intensity at that point? (Show your work:
Equation, steps and units)

Answer 1

9000N/C

Step-by-step explanation:

Given parameters:

Charge = 4 x 10⁻⁵C

Magnitude of force = 0.36N

Unknown:

Electric field intensity = ?

Solution:

This is the force that would be experienced by a unit test charge at a point.

Electric field intensity =
`(Force)/(Charge)`

Insert parameters and solve;

Electric field intensity =
`(0.36)/(4 x 10^(-5) )`

Electric field intensity = 9000N/C