Question

Two electric charges, held a distance, dd, apart experience an electric force of magnitude, FF, between them. If one of the charges is doubled in magnitude while maintaining the same separation between the charges, what is the new magnitude of the force between them?

Answer 1

Answer:

$F'=2F$

Explanation:

The Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them:

$F=\frac{kq_1q_2}{d^2}$

In this case, we have $q_1'=2q_1$:

$F'=\frac{kq'_1q_2}{d^2}\\F'=\frac{k(2q_1)q_2}{d^2}\\F'=2\frac{kq_1q_2}{d^2}\\F'=2F$