Answer:
F = 10.788 N
Explanation:
Given that,
Charge 1, [tex]q_1=6\ \mu C=6\times 10^{-6}\ C[/tex]
Charge 2, [tex]q_2=2\ \mu C=2\times 10^{-6}\ C[/tex]
Distance between charges, d = 0.1 m
We know that there is a force between charges. It is called electrostatic force. It is given by :
[tex]F=\dfrac{kq_1q_2}{d^2 }\\\\F=\dfrac{8.99\times 10^9\times 6\times 10^{-6}\times 2\times 10^{-6}}{(0.1)^2 }\\\\F=10.788\ N[/tex]
So, the force applied between charges is 10.788 N.
