Question

) a charge of 6.15 mc is placed at each corner of a square 0.100 m on a side. determine the magnitude and direction of the force on each charg

Answer 1

Because charges are positioned on a square the force acting on one charge is the same as the force acting on all others. 
We will use superposition principle. This means that force acting on the charge is the sum of individual forces. I have attached the sketch that you should take a look at.
We will break down forces on their x and y components:
$F_x=F_3+F_2cos(45^{\circ})$
$F_y=F_1+F_2sin(45^{\circ})$
Let's figure out each component:
$F_1=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\ F_3=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}$
$F_2=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}$
Total force acting on the charge would be:
$F=\sqrt{F_x^2+F_y^2}$
We need to calculate forces along x and y axis first( I will assume you meant micro coulombs, because otherwise we get forces that are huge).
$F_x=F_3+F_2cos(45^{\circ})=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}+\frac{1} {4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot\cos(45)=46N$
$F_y=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot sin(45)+\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}=46N$
Now we can find the total force acting on a single charge:
$F=\sqrt{F_x^2+F_y^2}=\sqrt{46^2+46^2}=65N$
As said before, intensity of the force acting on charges is the same for all of them.