Answer
The y component of the net electrostatic force on particle 3 is $~~-0.046~N$
Work Step by Step
The electrostatic force on particle 3 due to particle 1 is in the -y direction.
The y component of the electrostatic force on particle 3 due to particle 2 is in the +y direction.
We can find the y component of the net electrostatic force on particle 3:
$F_y = \frac{\vert q_3 \vert~\vert~q_2 \vert}{4\pi~\epsilon_0~(\sqrt{2}~a)^2}~sin~45^{\circ}-\frac{\vert q_3 \vert~\vert~q_1 \vert}{4\pi~\epsilon_0~a^2}$
$F_y = \frac{\vert q_3 \vert}{4\pi~\epsilon_0~a^2}~(\frac{\vert~q_2 \vert sin~45^{\circ}}{2}-\vert~q_1 \vert)$
$F_y = \frac{(200\times 10^{-9}~C)}{(4\pi)~(8.854\times 10^{-12}~F/m)~(0.050~m)^2}~[\frac{(100\times 10^{-9}~C) sin~45^{\circ}}{2}-(100\times 10^{-9}~C)]$
$F_y = -0.046~N$
The y component of the net electrostatic force on particle 3 is $~~-0.046~N$
