How strong is the repulsive force exerted on two point charges that each carry 1.0 E-6 C of negative charge and are 0.30 meters apart?

Answer 1

The repulsive force between two point charges can be calculated using Coulomb's Law, which is given by the formula ( F = k \frac{|q_1 q_2|}{r^2} ), where ( k ) is Coulomb's constant (( 8.99 \times 10^9 , \text{N m}^2/\text{C}^2 )), ( q_1 ) and ( q_2 ) are the magnitudes of the charges, and ( r ) is the distance between them. For two charges of ( 1.0 \times 10^{-6} , \text{C} ) each, and separated by ( 0.30 , \text{m} ), the force is calculated as ( F = 8.99 \times 10^9 \frac{(1.0 \times 10^{-6})^2}{(0.30)^2} \approx 9.99 , \text{N} ). Thus, the repulsive force exerted on the charges is approximately 10 N.