Answer
$F = (1.0\times 10^{-3}~N)~\hat{i}$
Work Step by Step
By symmetry the vertical components of the four forces cancel out.
By symmetry, the rightward component of the force due to each of the four charges on the corners are equal.
We can find the force on the charge in the middle:
$F = 4\times \frac{k~\vert q_1 \vert~\vert q_2 \vert}{r^2}~cos~45^{\circ}$
$F = 4\times \frac{(9.0\times 10^9~N~m^2/C^2)(1.0\times 10^{-9}~C)(2.0\times 10^{-9}~C)}{[(\sqrt{2})(0.0050~m)]^2}~cos~45^{\circ}$
$F = 1.0\times 10^{-3}~N$
We can express this force in component form:
$F = (1.0\times 10^{-3}~N)~\hat{i}$
