Answer:
So the repulsive force between the pith ball will be [tex]1.44\times 10^{-3}N[/tex]
Explanation:
We have given that the pith ball have the equal charge q = -40 nC =[tex]-40\times 10^{-9}C[/tex]
Distance between the charges = 10 cm =0.1 m
According to coulombs law [tex]F=\frac{KQ_1Q_2}{R*2}[/tex]
[tex]F=\frac{9\times 10^9\times -40\times 10^{-9}\times -40\times 10^{-9}}{0.1^2}=1440000\times 10^{-9}=1.44\times 10^{-3}N[/tex]
So the repulsive force between the pith ball will be [tex]1.44\times 10^{-3}N[/tex]
