The strength of the electric field is 5 N/C
Explanation:
The magnitude of the electric field produced by a single-point charge is given by:
[tex]E=\frac{kQ}{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=6 \mu C = 6\cdot 10^{-6}C[/tex] is the charge producing the field
r = 100 m is the distance from the charge at which we want to calculate the field
Substituting into the equation, we find the s trength of the electric field:
[tex]E=\frac{(8.99\cdot 10^9)(6\cdot 10^{-6})}{(100)^2}=5.4 N/C \sim 5 N/C[/tex]
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