Answer:
9000N/C
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 = [tex]\frac{Force}{Charge}[/tex]
Insert parameters and solve;
Electric field intensity = [tex]\frac{0.36}{4 x 10^{-5} }[/tex]
Electric field intensity = 9000N/C
