Answer:
Electric field, E = 5 N/C
Explanation:
It is given that,
Charge on the point charge, [tex]Q=6\ \mu C=6\times 10^{-6}\ C[/tex]
Distance at which the point charge is placed, d = 100 m
Electric field is given by :
[tex]E=\dfrac{kQ}{r^2}[/tex]
[tex]E=\dfrac{9\times 10^9\times 6\times 10^{-6}}{(100)^2}[/tex]
E = 5.4 N/C
E = 5.4 N/C
or E = 5 N/C
Hence, this is the required solution.
