Question

Two equal charges are 2m2m apart. If the charges and the distance are divided by two, how is the force between the charges affected? g

Answer 1

The force between the charges are not affected.

Step-by-step explanation:

Given;

distance between two equal charges, R = 2m

The force between the charges is given by;

`F = (kq^2)/(R^2)\\\\F_1 = (kq_1^2)/(R_1^2)\\\\When\ the \ charges \ and \ the \ distance \ are \ divided \ by \ two \ (q_2 = (q_1)/(2), \ R_2 = (R_1)/(2) )\\\\ F_2 = (kq_2^2)/(R_2^2)\\\\F_2 = (k(q_1/2)^2)/((R_1/2)^2)\\\\F_2= (4k*q_1^2)/(4*R_1^2)\\\\F_2 = (k*q_1^2)/(R_1^2)\\\\F_2 = F_1`

Therefore, the force between the charges are not affected.