Answer:
The total force is [tex]F_{tot}=4.29\: N[/tex]
The direction is [tex]\omega=32.43^{\circ}[/tex]
Explanation:
First, we need to find the angle with respect to the horizontal, of the force between q2 (-2 μC) and q3 (3 μC).
Let's use the tangent function.
[tex]tan(\alpha)=\frac{2}{1.5}[/tex]
[tex]\alpha=53.13^{\circ}[/tex]
Now, let's find the magnitude of the force F(12).
[tex]|F_{12}|=k\frac{q_{1}q_{2}}{d_{12}}[/tex]
Where:
[tex]|F_{12}|=9*10^{9}\frac{3*10^{-6}*2*10^{-6}}{0.015}[/tex]
[tex]|F_{12}|=3.6\: N[/tex]
The magnitude of the force F(23) will be:
[tex]|F_{23}|=k\frac{q_{2}q_{3}}{d_{23}}[/tex]
The distance between these charges is:
[tex]d_{23}=\sqrt{1.5^{2}+2^{2}}[/tex]
[tex]d_{23}=2.5\: m[/tex]
[tex]|F_{23}|=9*10^{9}\frac{2*10^{-6}*4*10^{-6}}{0.025}[/tex]
[tex]|F_{23}|=2.88\: N[/tex]
So, we have the force F(12) in the second quadrant and F(23) in the second quadrant too but with 53.13 ° with respect to the horizontal.
We just need to add these two forces (vectors) and get the total force acting on q2.
Total force in x-direction:
[tex]F_{tot-x}=-F_{12}-F_{23}cos(53.13)[/tex]
[tex]F_{tot-x}=-3.6-2.88cos(53.13)[/tex]
[tex]F_{tot-x}=-5.33\: N[/tex]
Total force in y-direction:
[tex]F_{tot-y}=F_{23}sin(53.13)[/tex]
[tex]F_{tot-y}=2.88sin(53.13)[/tex]
[tex]F_{tot-y}=2.3\: N[/tex]
Therefore, the magnitude of the total force will be:
[tex]|F_{tot}|=\sqrt{(-3.62)^{2}+(2.3)^{2}}[/tex]
[tex]|F_{tot}|=4.29\: N[/tex]
and the direction is:
[tex]tan(\omega)=\frac{2.30}{3.62}[/tex]
[tex]\omega=32.43^{\circ}[/tex]
I hope it helps you!
