Let place the charge Q on the line between +4q and q charges as shown on the figure.
Then, according to the Coulomb's low, the interaction force between +4q and Q charges is:
"F_1 = k\\dfrac{4q\\cdot Q}{x^2}.\\\\"
The interaction force between +q and Q charges is:
"F_2 = k\\dfrac{q\\cdot Q}{(r - x)^2}, \\\\" where "r" is the distance between +4q and +q and "k" is the Coulomb's constant.
In the equilibrium state both forces are equal, thus obtain the equation:
"k\\dfrac{4q\\cdot Q}{x^2} = k\\dfrac{q\\cdot Q}{(r - x)^2}" .
Solving it for "x" , obtain:
"\\dfrac{4}{x^2} = \\dfrac{1}{(r - x)^2} \\Rightarrow 4(r-x)^2 = x^2 \\Rightarrow 2(r-x) = x \\\\\nx = \\dfrac{2}{3}r.\\\\"
Answer. The charge Q should be placed at the distance 2/3r from the charge +4q.
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