Question

Two points charges are brought closer together,increasing the force between them by a factor of 25.By what factor wa their separation decreased?

Answer 1

Answer:

The separation between the charges was decreased by a factor of 0.2

Explanation:

The Coulomb's force between two charges is given by;

$F = \frac{kq^2}{r^2} \\\\let \ kq^2 \ be \ constant\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\increasing \ the \ force \ between \ them \ by \ factor \ of \ 25\\\\(F_2 = 25F_1)\\\\r_2^2 = \frac{F_1r_1^2}{25F_1}\\\\r_2^2 = \frac{r_1^2}{25}\\\\r_2 = \sqrt{\frac{r_1^2}{25} }\\\\ r_2 = \frac{r_1}{5}$

r₂ = 0.2r₁

Therefore, the separation between the charges was decreased by a factor of 0.2.