Question

A negative charge of -3.0 x 10^-5 C is placed 0.35 m from a positive charge of 9.0 x 10^-5 C. What is the force between the two charges

Answer 1

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;

$F = \frac{kQ_1Q_2}{r^2}$

where k is Coulomb's constant = 9 x 10⁹ Nm²/C²

$F = \frac{(9\times 10^9)(3.0\times 10^{-5})(9\times10^{-5} )}{0.35^2} \\\\F = 198.37 \ N$

Therefore, the force between the two charges is 198.37 N