Question

Two point charges exert a 7.35 N force on each other. What will the force become if the distance between them is increased by a factor of 2

Answer 1

Answer :

New force becomes, F' = 1.83 N

Explanation:

Let two point charges exert a force of 7.35 N force on each other. The electric force between two charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

$q_1\ and\ q_2$ are charges

r is the distance between charges if the distance between them is increased by a factor of 2, r' = 2r

New force is given by :

$F'=\dfrac{kq^2}{r'^2}$

$F'=\dfrac{kq^2}{(2r)^2}$

$F'=\dfrac{1}{4}\dfrac{kq^2}{r^2}$

$F'=\dfrac{1}{4}\times 7.35$

F' = 1.83 N

So, the new force between charges will be 1.83 N. Therefore, this is the required solution.