Answer
$F_x = 1.31\times 10^{-22}~N$
Work Step by Step
We can find the distance between the particles:
$r = \sqrt{(2.00~mm)^2+(6.00~mm)^2}$
$r = \sqrt{40}~mm$
We can find the x component of the electrostatic force on particle 2 due to particle 1:
$F_x = \frac{q_1~q_2}{4\pi~\epsilon_0~r^2}~cos~\theta$
$F_x = \frac{(4e)(6e)}{4\pi~\epsilon_0~r^2}~cos~\theta$
$F_x = \frac{(4)(1.6\times 10^{-19}~C)(6)(1.6\times 10^{-19}~C)}{(4\pi)~(8.854\times 10^{-12}~F/m)~(\sqrt{40}\times 10^{-3}~m)^2}~(\frac{6.00~mm}{\sqrt{40}~mm})$
$F_x = 1.31\times 10^{-22}~N$
