The intensity of the electric field is 30 000 N/C
Explanation:
The magnitude of the electric field produced by a single-point charge is given by:
[tex]F=k\frac{q}{r^2}[/tex]
where:
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q[/tex] is the magnitude of the charge
r is 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 from the charge
Substituting into the equation, we find the intensity of the field:
[tex]E=(8.99\cdot 10^9) \frac{3\cdot 10^{-9}}{0.03^2}=30,000 N/C[/tex]
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