1) Let's put [tex]q_1 [/tex] and [tex]q_2[/tex] both on the x-axis such that [tex]q_2[/tex] is on the right side of [tex]q_1[/tex]. The distance between the two charges is [tex]d=2~cm[/tex].
2) Let's put [tex]q_3[/tex] on the x-axis and let's call [tex]r[/tex] the distance between charge 1 and charge 3. As a consequence, the distance between charge 2 and 3 will be [tex]d+r[/tex]: if r is positive, then charge 3 will be located on the left of charge 1, if r is negative, charge 3 will be located between charge 1 and charge 2.
3) The electrostatic force between charge 1 and charge 3 is
[tex]F_{13} = \frac{q_1 q_3}{4 \pi \epsilon_0 r^2} [/tex]
while the force between charge 2 and charge 3 is
[tex]F_{23} = \frac{q_2 q_3}{4 \pi \epsilon_0 (r+d)^2} [/tex]
4) The problem says the two forces are equal in intensity. Therefore we can write [tex]F_{13} = F_{23}[/tex]. Using the equations written at step 3), and substituting [tex]q_1=q[/tex] and [tex]q_2=4q[/tex] as mentioned in the problem, [tex]F_{13} = F_{23}[/tex] becomes
[tex] \frac{q}{r^2}= \frac{4q}{(r+d)^2} [/tex]
Using [tex]d=2~cm[/tex], we can solve the equation, and we get two solutions:
[tex]r=2~cm[/tex]
[tex]r=-0.67~cm[/tex]
Both solutions are correct. With the former, the charge 3 is located 2 cm on the left of charge 1, while with the latter, charge 3 is located 0.67 cm on the right of charge 1, between charge 1 and 2.
