The magnitude of the net force that is acting on particle q₃ is equal to 6.2 Newton.
Given the following data:
Charge = [tex]-2.35 \times 10^{-6}[/tex] C.
Distance = 0.100 m.
Scientific data:
Coulomb's constant = [tex]8.988\times 10^9 \;Nm^2/C^2[/tex]
How to calculate the net force.
In this scenario, the magnitude of the net force that is acting on particle q₃ is given by:
F₃ = F₁₃ + F₂₃
Mathematically, the electrostatic force between two (2) charges is given by this formula:
[tex]F = k\frac{q_1q_2}{r^2}[/tex]
Where:
- r is the distance between two charges.
Note: d₁₃ = 2d₂₃ = 2(0.100) = 0.200 meter.
For electrostatic force (F₁₃);
[tex]F_{13} = 8.988\times 10^9 \times \frac{(-2.35 \times 10^{-6} \times [-2.35 \times 10^{-6}])}{0.200^2}\\\\F_{13} = \frac{0.0496}{0.04}[/tex]
F₁₃ = 1.24 Newton.
For electrostatic force (F₂₃);
[tex]F_{13} = 8.988\times 10^9 \times \frac{(-2.35 \times 10^{-6} \times [-2.35 \times 10^{-6}])}{0.100^2}\\\\F_{13} = \frac{0.0496}{0.01}[/tex]
F₂₃ = 4.96 Newton.
Therefore, the magnitude of the net force that is acting on particle q₃ is given by:
F₃ = 1.24 + 4.96
F₃ = 6.2 Newton.
Read more on charges here: brainly.com/question/14372859
