Answer
The magnitude of the net force is
$\frac{9.05\times10^{9}\times Q^{2}}{l^{2}}N=10.1\frac{kQ^{2}}{l^{2}}N$
The direction of the net force is
$61.3^{\circ}$
Work Step by Step
To solve this problem, we must first find the electrostatic force of each of the other forces (Q, 3Q, and 4Q) on charge 2Q using Coulomb's law. Then, we need to find the vector sum to find the total magnitude and direction of the 3 forces.
Coulomb's Law states $F=\frac{kQ_{1}Q_{2}}{r^{2}}$
1. Charge of Q on 2Q
$F_{Q}=\frac{(8.99\times10^{9}\frac{N\times m^{2}}{C})\times Q\times2Q}{l^{2}}=\frac{17.98\times10^{9}\times Q^{2}}{l^{2}}N$
2. Charge of 3Q on 2Q
$F_{3Q}=\frac{(8.99\times10^{9}\frac{N\times m^{2}}{C})\times3Q\times2Q}{l^{2}}=\frac{53.94\times10^{9}\times Q^{2}}{l^{2}}N$
3. Charge of 4Q on 2Q
Because the charge 4Q and charge 2Q are on opposite vertices of a square, the distance between them is
$\sqrt (2l^{2})=\sqrt2\times l$ according to the Pythagorean Theorem.
$F_{4Q}=\frac{(8.99\times10^{9}\frac{N\times m^{2}}{C})\times4Q\times2Q}{2\times l^{2}}=\frac{71.92\times10^{9}\times Q^{2}}{2\times l^{2}}N$
Now that we have found the magnitude of the 3 electrostatic forces, we now need to add them using vector addition.
4. Resolve force of 4Q into its x and y components
Because the four charges are on the four vertices of a square,
$F_{4Qx}=\frac{71.92\times10^{9}\times Q^{2}}{2\times l^{2}}\times cos(45^{\circ})N=\frac{71.92\times10^{9}\times Q^{2}}{2\times l^{2}}\times \frac{\sqrt 2}{2}N=\frac{25.43\times10^{9}\times Q^{2}}{l^{2}}N$
Because $cos(45^{\circ})=sin(45^{\circ})$, $F_{4Qx}=F_{4Qy}$
To find the net horizontal and vertical forces, simply add all the vertical and horizontal components
$F_{x}=\frac{17.98\times10^{9}\times Q^{2}}{l^{2}}N+\frac{25.43\times10^{9}\times Q^{2}}{l^{2}}N=\frac{43.41\times10^{9}\times Q^{2}}{l^{2}}N$
$F_{y}=\frac{53.94\times10^{9}\times Q^{2}}{l^{2}}N+\frac{25.43\times10^{9}\times Q^{2}}{l^{2}}N=\frac{79.37\times10^{9}\times Q^{2}}{l^{2}}N$
The magnitude of the net force is
$F=\sqrt(\frac{79.37\times10^{9}\times Q^{2}}{l^{2}}N)^{2}+(\frac{43.41\times10^{9}\times Q^{2}}{l^{2}}N)^{2})=\frac{9.05\times10^{9}\times Q^{2}}{l^{2}}N=10.1\frac{kQ^{2}}{l^{2}}N$
The direction is
$arctan(\frac{F_{y}}{F_{x}})=arctan(\frac{\frac{79.37\times10^{9}\times Q^{2}}{l^{2}}N}{\frac{43.41\times10^{9}\times Q^{2}}{l^{2}}N})=61.3^{\circ}$
