Question

Two charges are separated by a distance d and exert mutual attractive forces of F on each other. If the charges are separated by a distance of d/3, what are the new mutual forces

Answer 1

Answer:

Explanation:

Given

Two charges are separated by a distance d and exert a mutual attractive force F on each other.

Suppose $q_1$ and $q_2$ is the charge so force is given by

$F=\frac{kq_1q_2}{d^2}-----1$

When distance is decrease to $\frac{d}{2}$

Force is given by

$F'=\frac{kq_1q_2}{(\frac{d}{3})^2}$

$F'=\frac{9kq_1q_2}{d^2}-----2$

divide 1 and 2

$\frac{F}{F'}=\dfrac{\frac{kq_1q_2}{d^2}}{\frac{9kq_1q_2}{d^2}}$

$F'=9 F$

thus the new force become 9 times of previous force