## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$F = \frac{0.586~k~Q~q}{L^2}$
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: $F = \sqrt{(\frac{0.414~k~Q~q}{L^2})^2+(\frac{0.414~k~Q~q}{L^2})^2}$ $F = \frac{0.586~k~Q~q}{L^2}$