Answer:
Explanation:
Given
[tex]q_1=8.01\ nC[/tex]
[tex]q_2=4.50\ nC[/tex]
distance between the charges is given by [tex]d=1.87\ m[/tex]
Electrostatic force between the charges is given by
[tex]F_{12}=F_{21}=\frac{kq_1q_2}{d^2}[/tex]
where k=constant [tex]=9\times 10^{9}[/tex]
[tex]F_{12}=F_{21}=\frac{9\times 10^9\times 8.01\times 4.50\times 10^{-18}}{(1.87)^2}[/tex]
[tex]F_{12}=F_{21}=92.769\times 10^{-9}\ N[/tex]
as both the charges are of similar nature therefore they repel each other
