Answer:
188.7 N
Explanation:
We are given that
[tex]q_1=4.3\times 10^{-6} C[/tex]
[tex]q_2=-5\times 10^{-6} C[/tex]
[tex]r=3.2 cm=\frac{3.2}{100}=0.032 m[/tex]
[tex]1 m=100 cm[/tex]
[tex]k=8.98755\times 10^9Nm^2/C^2[/tex]
We know that Coulomb's force
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Using the formula
[tex]F=\frac{8.98755\times 10^9\times 4.3\times 10^{-6}\times 5\times 10^{-6}}{(0.032)^2}[/tex]
[tex]F=188.7 N[/tex]