Respuesta :
Answer:
The force magnitude is 1.75 μN acting outward from the origin towards x₂
Explanation:
The given parameters, are;
The location of q₁ = The origin
The location of q₂ = 14.7 mm from q₁
The repulsive force exerted on q₂ by q₁ = 2.62 μN
The location the particle q₂ is located to = 18.0 mm from q₁
By Coulomb's law, we have;
[tex]F = k\dfrac{q_1 \times q_2 }{r^2}[/tex]
Where;
k = Coulomb constant ≈ 8.99 × 10⁹ kg·m³/(s²·C²)
r = The distance between the particles
F = The force acting between the particles
When r = 14.7 mm F = 2.62 μN
∴ q₂ × q₁ = r² × F/k = (14.7×10^(-3))²×2.62×10^(-6)/(8.9875517923^9) ≈ 1.48 × 10⁻¹⁸
q₂ × q₁ = 1.48 × 10⁻¹⁸ C²
When the distance is increased to 18.0 mm, we have;
F = (8.9875517923^9) × 1.4796647 × 10^(-18)/((18×10^(-3))²) = 1.75 × 10⁻⁶ N
∴ F = 1.75 μN.
Therefore;
The force magnitude is 1.75 μN outward from the origin towards x₂.
To magnitude of the force that [tex]q_2[/tex] exerts on [tex]q_1[/tex] at this new location is 1.75 µN.
Given the following data:
- Initial distance of [tex]q_2[/tex] = 14.7 mm to m = 0.0147 meter.
- Repulsive force = 2.62 µN
- Final distance of [tex]q_2[/tex] = 18.0 mm to m = 0.018 meter.
Scientific data:
- Coulomb's constant = [tex]8.99 \times 10^9\; Nm^2/C^2[/tex]
To calculate the force (magnitude and direction) that [tex]q_2[/tex] exerts on [tex]q_1[/tex] at this new location, we would apply Coulomb's law:
The force in an electric field.
Mathematically, the force in an electric field is given by this formula:
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Where:
- k is Coulomb's constant.
- q is the charge.
- r is the distance.
[tex]q_1q_2 =\frac{F}{k} \times r^2\\\\q_1q_2 =\frac{2.62 \times 10^{-6}}{8.99 \times 10^9} \times 0.0147^2\\\\q_1q_2 = 2.914 \times 10^{-16} \times 0.00021609\\\\q_1q_2 = 6.30 \times 10^{-20}\;C^2[/tex]
For the force, when r = 0.018:
[tex]F =\frac{8.99 \times 10^9 \times 6.30 \times 10^{-20}}{0.018^2} \\\\F =\frac{5.66 \times 10^{-10}}{0.000324} \\\\F = 1.75 \times 10^{-6}[/tex]
Note: 1 µN = [tex]1 \times 10^{-6}\;N[/tex]
Force, F = 1.75 µN.
In conclusion, the direction of the force at the new location is outward from the origin along [tex]X_2[/tex].
Read more on electric field here: https://brainly.com/question/14372859
