Physics

By symmetry, the net force on each charge has the same magnitude, and is directed away from the center.

First, calculate the forces from the charges along the sides, nearest our right hand, upper corner:

Now calculate the force from the opposite corner, in the direction away from the center. Since the charges are equal, this force differs only because it is further away. $$F_3=\frac{9.0\ast {10}^9\frac{N{m}^2}{C^2}\ast {(5.0\ast {10}^{-3}C)}^2}{{(\sqrt{2\ast 1.0}m\ast )}^2}=\frac{2.25\ast {10}^{5}N}{2}=1.125\ast {10}^{5}N$$

Now add all the forces together in the direction away from the center. F₁ contributes only to the "up" component of the resulting vector, and F₂ contributes only the the "right direction" compopnent of R. $$R=F_3+F_1\cos 45°+F_2\sin 45°=1.125\ast {10}^{5}N+2(2.25\ast {10}^{5}N)(\cos 45°)=5.3\ast {10}^{5}N$$

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