Question

Three charged particles of charges 3 μC, -2 μC, and 4 μC are placed on the X-Y plane at (1 cm, 0), (2.5 cm, 0), and (1 cm, 2 cm) respectively. Determine the magnitude and direction of the force acting on a -2 μC charge.

Answer 1

Answer:

The total force is $F_{tot}=4.29\: N$

The direction is $\omega=32.43^{\circ}$

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.

$tan(\alpha)=\frac{2}{1.5}$

$\alpha=53.13^{\circ}$  

Now, let's find the magnitude of the force F(12).

$|F_{12}|=k\frac{q_{1}q_{2}}{d_{12}}$

Where:

• k is the Coulomb constant (9*10⁹ NC²/m²)
• q1 is 3 μC
• q2 -2 μC
• d(12) is the distance between q1 and q2 ( 1.5 cm = 0.015 m)

$|F_{12}|=9*10^{9}\frac{3*10^{-6}*2*10^{-6}}{0.015}$

$|F_{12}|=3.6\: N$

The magnitude of the force F(23) will be:

$|F_{23}|=k\frac{q_{2}q_{3}}{d_{23}}$

The distance between these charges is:

$d_{23}=\sqrt{1.5^{2}+2^{2}}$

$d_{23}=2.5\: m$

$|F_{23}|=9*10^{9}\frac{2*10^{-6}*4*10^{-6}}{0.025}$

$|F_{23}|=2.88\: N$

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:

$F_{tot-x}=-F_{12}-F_{23}cos(53.13)$

$F_{tot-x}=-3.6-2.88cos(53.13)$

$F_{tot-x}=-5.33\: N$

Total force in y-direction:

$F_{tot-y}=F_{23}sin(53.13)$

$F_{tot-y}=2.88sin(53.13)$

$F_{tot-y}=2.3\: N$

Therefore, the magnitude of the total force will be:

$|F_{tot}|=\sqrt{(-3.62)^{2}+(2.3)^{2}}$

$|F_{tot}|=4.29\: N$

and the direction is:

$tan(\omega)=\frac{2.30}{3.62}$

$\omega=32.43^{\circ}$

I hope it helps you!