Answer:
the force between the two charges is 198.37 N
Explanation:
Given;
charge of the first particle, Q₁ = -3 x 10⁻⁵ C
charge of the second particle, Q₂ = 9.0 x 10⁻⁵ C
distance between the two charges, r = 0.35 m
The force between the two charges is calculated from Coulomb's law;
![F = \frac{kQ_1Q_2}{r^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7BkQ_1Q_2%7D%7Br%5E2%7D)
where k is Coulomb's constant = 9 x 10⁹ Nm²/C²
![F = \frac{(9\times 10^9)(3.0\times 10^{-5})(9\times10^{-5} )}{0.35^2} \\\\F = 198.37 \ N](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%289%5Ctimes%2010%5E9%29%283.0%5Ctimes%2010%5E%7B-5%7D%29%289%5Ctimes10%5E%7B-5%7D%20%29%7D%7B0.35%5E2%7D%20%5C%5C%5C%5CF%20%3D%20198.37%20%5C%20N)
Therefore, the force between the two charges is 198.37 N