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.

The total force is

The direction is

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.

 

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

Where:

k is the Coulomb constant (9*10⁹ NC²/m²)q1 is 3 μCq2 -2 μCd(12) is the distance between q1 and q2 ( 1.5 cm = 0.015 m)

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

The distance between these charges is:

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:

Total force in y-direction:

Therefore, the magnitude of the total force will be:

and the direction is:

I hope it helps you!

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