Answer:
[tex]2.8\cdot 10^{-11} C[/tex]
Explanation:
The electric field produced by a point charge is given by
[tex]E=k\frac{q}{r^2}[/tex]
where
k is the Coulomb's constant
q is the charge
r is the distance from the charge
In this problem, we have
[tex]E=100.0 N/C[/tex] is the strength of the electric field
[tex]r=5.0 cm=0.05 m[/tex] is the distance from the charge source of the field
Solving the formula for q, we find
[tex]q=\frac{Er^2}{k}=\frac{(100 N/C)(0.05 m)^2}{9\cdot 10^9 Nm^2C^{-2}}=2.8\cdot 10^{-11} C[/tex]
