Answer:
800 N/C to the right
Explanation:
The relationship between electric force, electric field and charge is
[tex]E=\frac{F}{q}[/tex]
where
E is the electric field
F is the force acting on the charge
q is the charge
In this problem, we have
[tex]F = 0.040 N[/tex] is the force
[tex]q=+50 \mu C=+50 \cdot 10^{-6}C[/tex] is the charge
Substituting into the formula, we find the electric field magnitude
[tex]E=\frac{0.040 N}{+50 \cdot 10^{-6}C}=800 N/C[/tex]
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.
