The intensity of the electric field is 30,000 N/C
Explanation:
The strength of the electric field produced by a single-point charge is given by the equation
[tex]E=k\frac{q}{r^2}[/tex]
where:
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
[tex]q=3\cdot 10^{-9}C[/tex] is the magnitude of the charge
r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity
Substituting, we find:
[tex]E=(8.99\cdot 10^9)\frac{3\cdot 10^{-9}}{(0.03)^2}=30,000 N/C[/tex]
Learn more about electric field:
brainly.com/question/8960054
brainly.com/question/4273177
#LearnwithBrainly
