Answer:
d = 0.71 meters
Explanation:
It is given that,
Charge 1, [tex]q_1=16\ \mu C=16\times 10^{-6}\ C[/tex]
Charge 2, [tex]q_2=32\ \mu C=32\times 10^{-6}\ C[/tex]
Electrostatic force between charges, F = 9 N
Let d is the distance between the charges. The electrostatic force between the charges is given by the product of charges and divided by square of distance between them. Mathematically, it is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
[tex]d=\sqrt{\dfrac{kq_1q_2}{F}}[/tex]
[tex]d=\sqrt{\dfrac{9\times 10^9\times 16\times 10^{-6}\times 32\times 10^{-6}}{9}}[/tex]
d = 0.71 meters
So, the distance between the charges is 0.71 meters. Hence, this is the required solution.
