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 :
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
[tex]q_1\ and\ q_2[/tex] 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 :
[tex]F'=\dfrac{kq^2}{r'^2}[/tex]
[tex]F'=\dfrac{kq^2}{(2r)^2}[/tex]
[tex]F'=\dfrac{1}{4}\dfrac{kq^2}{r^2}[/tex]
[tex]F'=\dfrac{1}{4}\times 7.35[/tex]
F' = 1.83 N
So, the new force between charges will be 1.83 N. Therefore, this is the required solution.
