Given 3 point charges are located in a Cartesian Coordinate system where Charge 1 (having a charge of 2nC) is placed at the origin, Charge 2 is placed at $(0,5)$ and has a charge of 2nC as well, and Charge 3 is placed at $(8,0)$ and has a charge of 3nC. I am wondering what is the overall force being applied to the Charge 3. The problem is to find the net charges acting on charge 3. My premise to solving the problem, was to use Coulomb's law, and the summation of forces.
Equations to use
$$\vec{F}=k\frac{q_1 q_2}{r^2}$$
$$\sum\vec{F}=\sqrt{{(\sum\vec{F_y}})^2+(\sum\vec{F_x})^2}$$
My thought was that the force exerted by the charge 1 on 3 is horizontal repulsion, and then the charge 2 on 3 would be horizontal and vertical repulsion meaning that the net acting force on Charge 3 is south of east of where point Charge 3 is.
Best Answer
I suggest you start by setting the problem up just like in the picture I have attached.
ASSUMING ALL CHARGES ARE POSITIVE AS IN THE PROBLEM STATEMENT
You can calculate each of the forces separately:
$$ F_1 = k\frac{q1q3}{r_{13}^2} = k\frac{(2)(3)}{8^2}\hat{i} $$
$F_2$ is slightly more complicated
$$ F_2 = k\frac{q2q3}{r_{23}^2} = k\frac{(2)(3)}{(\sqrt{8^2+5^2})^2} $$
Now you have to break down $F_2$ into two different vectors $F_{2x}$ and $F_{2y}$, which are the projections of $F_2$ on the x,y axis.
You can do this by figuring out the angle $\theta$ between $F_1$ and $F_2$.
Doing some simple trig:
$$ \theta = \arctan\left(\frac{5}{8}\right) $$
Then you can say:
$$ F_{2x} = F_2\cos(\theta)\hat{i} $$
$$ F_{2y} = F_2\cos(\theta)\hat{j} $$
Now that all your forces are represented as vectors on the $x-y$ plane, you can find the resultant by just adding the vectors.
Can you proceed from here?