Answer
The x component of the net electrostatic force on particle 3 is $~~0.17~N$
Work Step by Step
The x component of the electrostatic force on particle 3 due to particle 2 is in the +x direction.
The electrostatic force on particle 3 due to particle 4 is in the +x direction.
We can find the x component of the net electrostatic force on particle 3:
$F_x = \frac{\vert q_3 \vert~\vert~q_2 \vert}{4\pi~\epsilon_0~(\sqrt{2}~a)^2}~cos~45^{\circ}+\frac{\vert q_3 \vert~\vert~q_4 \vert}{4\pi~\epsilon_0~a^2}$
$F_x = \frac{\vert q_3 \vert}{4\pi~\epsilon_0~a^2}~(\frac{\vert~q_2 \vert cos~45^{\circ}}{2}+\vert~q_4 \vert)$
$F_x = \frac{(200\times 10^{-9}~C)}{(4\pi)~(8.854\times 10^{-12}~F/m)~(0.050~m)^2}~[\frac{(100\times 10^{-9}~C) cos~45^{\circ}}{2}+(200\times 10^{-9}~C)]$
$F_x = 0.17~N$
The x component of the net electrostatic force on particle 3 is $~~0.17~N$
