Answer
$
\begin{aligned}
E_{\text {net }, x} & =20.6 \mathrm{~N} / \mathrm{C} .
\end{aligned}
$
Work Step by Step
The magnitude of the net field is
$
\begin{aligned}
E_{\text {net }, x} & =2\left(\frac{1}{4 \pi \varepsilon_0} \frac{2 \sqrt{2}|q|}{\pi r^2}\right) \cos 45^{\circ}\\&=\frac{1}{4 \pi \Sigma_0} \frac{4|q|}{\pi r^2} \\
& =\frac{\left(8.99 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{C}^2\right) 4\left(4.50 \times 10^{-12} \mathrm{C}\right)}{\pi\left(5.00 \times 10^{-2} \mathrm{~m}\right)^2}\\&=20.6 \mathrm{~N} / \mathrm{C} .
\end{aligned}
$
