The intensity of the electric field is 30,000 N/C
Explanation:
The intensity of the electric field around 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 from the charge
Substituting and solving, we find the strength of the electric field:
[tex]E=\frac{(8.99\cdot 10^9)(3\cdot 10^{-9})}{(0.03)^2}=30,000 N/C[/tex]
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