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

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

Step-by-step explanation:

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

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

r₂ = 0.2r₁

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