Question

three charged particles lie on a straight line and are separated by distances d. Charges q1 and q2 are held fixed. Charge q3 is free to move but happens to be in equilibrium (no net electrostatic force acts on it)

Answer 1

Answer:

$\boxed {q_1=-4q_2}$

Explanation:

Using the attached figure

Considering that the distance of separation is 2d then  

$F_1=\frac {q_1q_3}{4\pi\epsilon_o(2d)^{2}}$

Also, considering that distance of separation between  and  is d then

$F_2=\frac {q_1q_3}{4\pi\epsilon_o(d)^{2}}$

The net force acting on  is

$F=F_1+F_2=0\\F=\frac {q_1q_3}{4\pi\epsilon_o(2d)^{2}}+ \frac {q_1q_3}{4\pi\epsilon_o(d)^{2}}=0\\F=\frac {q_3}{4\pi \epsilon_o d^{2}}(q_2+0.25q_1)=0\\F=0.25q_1+q_2=0$

Therefore

$\boxed {q_1=-4q_2}$