Answer:
Electric field, E = 3.0 N/C
Step-by-step explanation:
It is given that,
Charge particles,

Distance from charge, r = 3.7 m
We need to find the magnitude of the electric field. The formula is given by :


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.