Answer:
Distance between the charges, r = 0.8 meters
Explanation:
Given that,
Charge 1, [tex]q_1=+8.4\ \mu C=+8.4\times 10^{-6}\ C[/tex]
Charge 2, [tex]q_2=+5.6\ \mu C=+5.6\times 10^{-6}\ C[/tex]
Repulsive force between charges, F = 0.66 N
Let r is the distance between charges. The formula for the electrostatic force is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
[tex]r=\sqrt{\dfrac{kq_1q_2}{F}}[/tex]
[tex]r=\sqrt{\dfrac{9\times 10^9\times 8.4\times 10^{-6}\times 5.6\times 10^{-6}}{0.66}}[/tex]
r = 0.8009 meters
or
r = 0.8 meters
So, the distance between the charges i 0.8 meters. Hence, this is the required solution.
