Answer:
Electric field, [tex]E=1.24\times 10^{12}\ N/C[/tex]
Explanation:
It is given that,
Charge, Q = +6.1 C
Distance, r = 21 cm = 0.21 m
We need to find the electric field. It is given by :
[tex]E=k\dfrac{Q}{r^2}[/tex]
[tex]E=9\times 10^9\times \dfrac{6.1}{(0.21)^2}[/tex]
[tex]E=1.24\times 10^{12}\ N/C[/tex]
So, the electric field at this distance is [tex]1.24\times 10^{12}\ N/C[/tex]. Hence, this is the required solution.
