Question

suppose a charge q=+50 uC experiences a force of 0.040 N which points to the right. What is the magnitude and direction of the electric field that causes this force?

Answer 1

800 N/C to the right

Step-by-step explanation:

The relationship between electric force, electric field and charge is

`E=(F)/(q)`

where

E is the electric field

F is the force acting on the charge

q is the charge

In this problem, we have

`F = 0.040 N` is the force

`q=+50 \mu C=+50 \cdot 10^(-6)C` is the charge

Substituting into the formula, we find the electric field magnitude

`E=(0.040 N)/(+50 \cdot 10^(-6)C)=800 N/C`

Concerning the direction:

- The electric field and the force have same directions if the charge is positive

- The electric field and the force have opposite directions if the charge is negative

Here the charge is positive, so the electric field has same direction as the force: therefore, to the right.