Answer: [tex]30.09\times 10^{-15}\ C[/tex]
Explanation:
Given
The first charge is [tex]q_1=-6\ C[/tex]
Attraction force [tex]F=65\ N[/tex]
the second charge is at [tex]0.005\ m[/tex] away
Suppose the second charge is [tex]q_2[/tex]
The electrostatic force is given by
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
Insert the values
[tex]65=\dfrac{9\times 10^9\times 6\times q_2}{[5\times 10^{-3}]^2}\\\\\Rightarrow q_2=\dfrac{65\times 25\times 10^{-6}}{54\times 10^9}\\\\\Rightarrow q_2=\dfrac{1625\times 10^{-6-9}}{54}\\\\\Rightarrow q_2=30.09\times 10^{-15}\ C[/tex]
