Explanation:
Let [tex]q_1[/tex] is the first charge and [tex]q_2[/tex] is the second charge.
Force between them, F = 0.12 N
Distance between charges, d = 0.46 m
(a) Force acting between two point charges is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
[tex]q_1q_2=\dfrac{Fd^2}{k}[/tex]
[tex]q_1q_2=\dfrac{0.12\times (0.46)^2}{9\times 10^9}[/tex]
[tex]q_1q_2=2.82\times 10^{-12}[/tex]..............(1)
Also,
[tex]q_1+q_2=-8.9\ \mu C=-8.9\times 10^{-6}\ C[/tex]............(2) (both charges are negative)
On solving equation (1) and (2) :
[tex]q_1=-8.571\ C[/tex]
and
[tex]q_2=-0.329\ C[/tex]
(b) If the force is attractive, F = -0.12 N
[tex]q_1q_2=\dfrac{Fd^2}{k}[/tex]
[tex]q_1q_2=\dfrac{-0.12\times (0.46)^2}{9\times 10^9}[/tex]
[tex]q_1q_2=-2.82\times 10^{-12}[/tex]..............(3)
[tex]q_1+q_2=-8.9\ \mu C=-8.9\times 10^{-6}\ C[/tex]............(4)
Solving equation (3) and (4) we get :
[tex]q_1=-0.306\ C[/tex]
[tex]q_2=9.206\ C[/tex]
Hence, this is the required solution.
