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) respectivel

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.

in progress 0
Thanh Hà 1 year 2021-09-03T21:03:03+00:00 1 Answers 1 views 0

Answers ( )

    0
    2021-09-03T21:04:49+00:00

    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:

    • 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)

    [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!

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )