Answer:
-11776.36 N
This force is attractive since both charges are of opposite sign
Explanation:
Given that
[tex]q_1=-18.4\mu C=-18.4\times 10^{-6}C\ and\\\ q_2=53\mu C=53\times 10^{-6}C[/tex]
Distance between the spheres = 2.73 cm =0.0273 m
where K is Coulomb's constant = [tex]9.10^ 9 [N.m^2 /C^2][/tex]
According to coulombs law we know that force between two charges is given by
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
[tex]F = \frac{(9\times10^9)\times (-18.4 \times 10^-^6)(53 \times 10^-^6}{0.0273^2)} \\= -11776.36N[/tex]
This force is attractive since both charges are of opposite sign