The magnitude of the net force that is acting on particle q₃ is equal to 6.2 Newton.
<u>Given the following data:</u>
Charge = 
C.
Distance = 0.100 m.
<u>Scientific data:</u>
Coulomb's constant = 
<h3>How to calculate the net force.</h3>
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:
<u>Where:</u>
- r is the distance between two charges.
<u>Note:</u> d₁₃ = 2d₂₃ = 2(0.100) = 0.200 meter.
For electrostatic force (F₁₃);
![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}](https://tex.z-dn.net/?f=F_%7B13%7D%20%3D%208.988%5Ctimes%2010%5E9%20%5Ctimes%20%5Cfrac%7B%28-2.35%20%5Ctimes%2010%5E%7B-6%7D%20%5Ctimes%20%5B-2.35%20%5Ctimes%2010%5E%7B-6%7D%5D%29%7D%7B0.200%5E2%7D%5C%5C%5C%5CF_%7B13%7D%20%3D%20%5Cfrac%7B0.0496%7D%7B0.04%7D)
F₁₃ = 1.24 Newton.
For electrostatic force (F₂₃);
![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}](https://tex.z-dn.net/?f=F_%7B13%7D%20%3D%208.988%5Ctimes%2010%5E9%20%5Ctimes%20%5Cfrac%7B%28-2.35%20%5Ctimes%2010%5E%7B-6%7D%20%5Ctimes%20%5B-2.35%20%5Ctimes%2010%5E%7B-6%7D%5D%29%7D%7B0.100%5E2%7D%5C%5C%5C%5CF_%7B13%7D%20%3D%20%5Cfrac%7B0.0496%7D%7B0.01%7D)
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.
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Answer:
Newtons second law of motion relates to force, inversely related to its <em><u>mass</u></em><em> </em>and acceleration.
C
Atomic radius is the distance between the center of the nucleus to the outermost orbital shell of the atom. Assume the atom is like a football stadium and the nucleus of the atom is a ball placed at the center of the pitch. The atomic radius is from the center of the ball to the edge of the football stadium.
Explanation:
This atomic radius decreases from left to right of a periodic table because of increases in protons in the nucleus along the periodic table. This increased proton count has a higher attractive force on the electron orbitals of the atom. This decreases the atomic radius
The atomic radius of atoms down a column of the periodic table increase because an extra orbital shell is added to the atoms with every period down the column.
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