Answer:
[tex]q_2=9.59\times 10^{-7}\ C[/tex]
Explanation:
It is given that,
Charge at origin, [tex]q_1=6.6\ \mu C=6.6\times 10^{-6}\ C[/tex]
Let second charge is [tex]q_2[/tex] and it is placed at x = 0.1 m
Resulting force on the second charge, [tex]F_2=5.7\ N[/tex]
The electric force acting on the charged particles is given by the following formula as :
[tex]F=\dfrac{kq_1q_2}{x^2}[/tex]
[tex]q_2=\dfrac{Fx^2}{kq_1}[/tex]
[tex]q_2=\dfrac{5.7\times (0.1)^2}{9\times 10^9\times 6.6\times 10^{-6}}[/tex]
[tex]q_2=9.59\times 10^{-7}\ C[/tex]
So, the second charge is [tex]9.59\times 10^{-7}\ C[/tex]. Hence, this is the required solution.
