Answer
$F = \frac{0.586~k~Q~q}{L^2}$
Work Step by Step
We can find the component of the force in the +x direction:
$F_x = \frac{k~(4Q)(q)}{(\sqrt{2}~L)^2}~sin~45^{\circ} - \frac{k~(Q)(q)}{L^2}$
$F_x = (0.414)~(\frac{k~Q~q}{L^2})$
We can find the component of the force in the +y direction:
$F_y = \frac{k~(4Q)(q)}{(\sqrt{2}~L)^2}~cos~45^{\circ} - \frac{k~(Q)(q)}{L^2}$
$F_y = (0.414)~(\frac{k~Q~q}{L^2})$
We can find the magnitude of the net force:
$F = \sqrt{(\frac{0.414~k~Q~q}{L^2})^2+(\frac{0.414~k~Q~q}{L^2})^2}$
$F = \frac{0.586~k~Q~q}{L^2}$
