For the three-point charges fixed in a right triangle, we have:
1. The magnitude of the electric force on the +1.00 μC charge is 1.79 N.
2. The force does an angle of 21.68° with the x-axis.
1. Magnitude of the electric force
First, let's denote the charges as is shown in the picture below:
- q₁: is the charge 1 = +0.800 μC = +0.800x10⁻⁶ C
- q₂: is the charge 2 = +1.00 μC = +1.00x10⁻⁶ C
- q₃: is the charge 3 = -0.600 μC = -0.600x10⁻⁶ C
The magnitude of the electric force on the charge 2 (+1.00 μC) is given by:
[tex]|F_{net}| = \sqrt{(\Sigma\vec{F}_{x})^{2} + (\Sigma\vec{F}_{y})^{2}}[/tex] (1)
Where:
- [tex]\Sigma\vec{F}_{x}[/tex]: is the sum of the forces acting on the x-axis
- [tex]\Sigma\vec{F}_{y}[/tex]: is the sum of the forces acting on the y-axis
We can calculate the electrical forces with Coulomb's law:
[tex]\vec{F} = \frac{Kq_{1}q_{2}}{d^{2}}[/tex]
Where:
- K is the Coulomb's constant = 9.00x10⁹ Nm²/C²
- q₁ and q₂ are the charges
- d is the distance between the charges
Forces in the x-axis ([tex]\Sigma\vec{F}_{x}[/tex])
The forces in the x-component are given by:
[tex]\Sigma\vec{F}_{x} = \vec{F}_{21}_{x} + \vec{F}_{23}_{x}[/tex]
[tex]\Sigma\vec{F}_{x} = \vec{F}_{21}_{x}cos(\theta) + \vec{F}_{23}_{x}[/tex]
[tex]\Sigma\vec{F}_{x} = \frac{Kq_{1}q_{2}}{d_{12}^{2}}cos(\theta) + \frac{Kq_{2}q_{3}}{d_{23}^{2}}[/tex] (2)
Where:
- θ: is the angle of the force F₂₁ with the x-axis
- d₁₂ = z = 9.60 cm = 0.0960 m
We can calculate the angle θ with the following trigonometric function:
[tex]sin(\theta) = \frac{y}{z}[/tex]
[tex]\theta = sin^{-1}(\frac{y}{z}) = sin^{-1}(\frac{8.10 cm}{9.60 cm}) = 57.5 ^\circ[/tex]
To find the distance x (d₂₃), we need to use Pythagoras:
[tex]x = \sqrt{z^{2} - y^{2}} = \sqrt{(0.0960 m)^{2} - (0.0810 m)^{2}} = 0.051 m[/tex]
After entering θ and x (d₂₃) into equation 2, we have:
[tex]\Sigma\vec{F}_{x} = \frac{9.00\cdot 10^{9} Nm^{2}C^{-2}(0.800 \cdot 10^{-6} C)(1.00 \cdot 10^{-6} C)}{(0.0960 m)^{2}}cos(57.5) + \frac{9.00 \cdot 10^{9}Nm^{2}C^{-2}(1.00 \cdot 10^{-6} C)(-0.600 \cdot 10^{-6} C)}{(0.051 m)^{2}}[/tex]
[tex]\Sigma\vec{F}_{x} = [0.78*cos(57.5) + (-2.08) N] = -1.66 N[/tex]
Hence, the x-component of the force is -1.66 N.
Forces in the y-axis ([tex]\Sigma\vec{F}_{y}[/tex])
The only force acting on the y-axis is the y-component of the force F₂₁, so:
[tex]\Sigma\vec{F}_{y} = -\vec{F}_{21}_{y} = -\vec{F}_{21}sin(\theta)[/tex]
The minus sign is because the vector is pointing in the negative y-direction (see the picture below).
[tex]\Sigma \vec{F}_{y} = -\frac{9.00\cdot 10^{9} Nm^{2}C^{-2}(0.800 \cdot 10^{-6} C)(1.00 \cdot 10^{-6} C)}{(0.0960 m)^{2}}sin(57.5) = -0.66 N[/tex]
Hence, the y-component of the force is -0.66 N.
Finally, the magnitude of the electric force on the charge +1.00 μC is (eq 1):
[tex]|F_{net}| = \sqrt{(-1.66 N)^{2} + (-0.66 N)^{2}} = 1.79 N[/tex]
Therefore, the magnitude of the electric force on the +1.00 μC charge is 1.79 N.
2. Direction of the force with the x-axis
According to the picture below, the angle of the force with respect to the x-axis is given by:
[tex]tan(\beta) = \frac{\Sigma\vec{F}_{y}}{\Sigma\vec{F}_{x}}[/tex]
[tex]\beta = tan^{-1}(\frac{-0.66 N}{-1.66 N}) = 21.68 ^\circ[/tex]
Therefore, the force does an angle of 21.68° with the x-axis.
Find more about Coulomb's law here:
brainly.com/question/506926
I hope it helps you!
