Answer:
Electric field, E = 3.0 N/C
Explanation:
It is given that,
Charge particles, [tex]q=4.66\ nC=4.66\times 10^{-9}\ C[/tex]
Distance from charge, r = 3.7 m
We need to find the magnitude of the electric field. The formula is given by :
[tex]E=k\dfrac{q}{r^2}[/tex]
[tex]E=9\times 10^9\times \dfrac{4.66\times 10^{-9}}{(3.7)^2}[/tex]
E = 3.06 N/C
or
E = 3.0 N/C
So, the magnitude of the electric field produced by a point charge is 3.0 N/C. Hence, this is the required solution.
