Answer
$F = 34.5~N$
Work Step by Step
We can find the magnitude of the electrostatic force on particle 2 due to particle 1:
$F = \frac{\vert q_1 \vert~\vert~q_2 \vert}{4\pi~\epsilon_0~r^2}$
$F = \frac{(4.0~\mu C)(3.0~\mu C)}{(4\pi)~(8.854\times 10^{-12}~F/m)~(\sqrt{(0.055~m)^2+()0.010~m^2})^2}$
$F = \frac{(4.0~\mu C)(3.0~\mu C)}{(4\pi)~(8.854\times 10^{-12}~F/m)~(0.0559~m)^2}$
$F = 34.5~N$
