Solve this:

Dear Student ,
Let distance between charges 4q and q is a.
Let charge q^' is placed at a distance x from charge 4q.
Distance between charge q^' and q is (a-x)

At equilibrium , net force on the charge 4q is zero, therefore

$$\begin{gathered} \frac{k4q{q}^{\text{'}}}{x^2} = \frac{k4q^2}{a^2} \\ \frac{q^{\text{'}}}{q} = \frac{x^2}{a^2} ---- \left(1\right) \\ Now net force on charge q^{\text{'}} is also zero, therefore \\ \frac{k4q{q}^{\text{'}}}{x^2} = \frac{kq{q}^{\text{'}}}{{\left(a-x\right)}^2} \\ on solving we get x = (2/3)a and x = 2a \\ \sin ce charge is lying inside the two point so we take x = \frac{2}{3}a \\ From equation no\left(1\right) we get \\ \frac{q^{\text{'}}}{q} = \frac{4}{9} \end{gathered}$$
Regards