Answer:
The magnitude of the electric field is 1.124 X 10⁷ N/C
Explanation:
Magnitude of electric field is given as;
[tex]E = \frac{kq}{r^2} , N/C[/tex]
where;
E is the magnitude of the electric field, N/C
q is the point charge, C
k is coulomb's constant, Nm²/C²
r is the distance of the point charge, m
Given;
q = 5mC = 5×10⁻³ C
r = 2m
k = 8.99 × 10⁹ Nm²/C²
Substitute these values and solve for magnitude of electric field E
[tex]E = \frac{kq}{r^2} = \frac{(8.99 X10^9)(5X10^{-3})}{2^2} = 1.124 X10^7 N/C[/tex]
Therefore, the magnitude of the electric field is 1.124 X 10⁷ N/C
