Answer
$y = 2.7~cm$
Work Step by Step
If the net electrostatic force on particle 2 due to particles 1 and 3 is zero, then the forces on particle 2 due to the other two particles must be equal in magnitude and opposite in direction.
We can find $r_2$, the distance between particle 2 and particle 3:
$\frac{\vert q_1 \vert~\vert q_2 \vert}{4\pi~\epsilon_0~r_1^2}= \frac{\vert q_2 \vert~\vert q_3 \vert}{4\pi~\epsilon_0~r_2^2}$
$\frac{q_1}{r_1^2}= \frac{q_3}{r_2^2}$
$r_2^2= \frac{q_3~r_1^2}{q_1}$
$r_2= \sqrt{\frac{q_3}{q_1}}~r_1$
$r_2= \sqrt{\frac{4.0~\mu C}{3.0~\mu C}}~\sqrt{(0.055~m)^2+(0.010~m)^2}$
$r_2 = 6.455~cm$
Since particle 3 also has a positive charge, particle 3 should be placed on the other side of particle 2 in the opposite direction from particle 1.
In part (b), we found that relative to the positive direction of the x- axis, the direction of the electrostatic force on particle 2 due to particle 1 is $-10.3^{\circ}$
Then particle 3 should be placed at an angle of $169.7^{\circ}$ relative to the positive x axis.
We can find the y coordinate of particle 3:
$y = (1.5~cm)+(6.455~cm)~sin(169.7^{\circ})$
$y = 2.7~cm$
