Question

Two point charges attract each other with an electric force of magnitude F.a. If one charge is reduced to one-third its original value and the distance between the charges is doubled, what is the resulting magnitude of the electric force between them?b. If one charge is reduced to one-fourth its original value, how should the distance between the two charges change such that the force F remains constant?

Answer 1

Answer:

a. $F'=\frac{1}{12}F$

b. $d'=\frac{d}{2}$

Explanation:

a. The magnitude of the electric force is given by the Coulomb's law:

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

In this case, we have $q_1'=\frac{q_1}{3}$ and $d'=2d$:

$F'=\frac{kq_1'q_2}{d'^2}\\F'=\frac{k(\frac{q_1}{3})q_2}{(2d)^2}\\F'=\frac{kq_1q_2}{3*4d^2}\\F'=\frac{1}{12}\frac{kq_1q_2}{d^2}\\F'=\frac{1}{12}F$

b. In this case, we have $q_1'=\frac{q_1}{4}$:

$F=\frac{kq_1'q_2}{d'^2}\\\frac{kq_1q_2}{d^2}=\frac{k(\frac{q_1}{4})q_2}{d'^2}\\\frac{1}{d^2}=\frac{1}{4d'^2}\\d'^2=\frac{d^2}{4}\\d'=\frac{d}{2}$