Answer:
The new force will become 1/4 times of the initial force.
Explanation:
The Coulomb force between two charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them.
It can be given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex] ....(1)
r is distance
If we decrease the r by a factor of 2, r'=2r.
New force will become,
[tex]F'=k\dfrac{q_1q_2}{r'^2}\\\\=k\dfrac{q_1q_2}{(2r)^2}\\\\=k\dfrac{q_1q_2}{4r^2}\\\\=\dfrac{1}{4}\times \dfrac{kq_1q_2}{r^2}\\\\F'=\dfrac{F}{4}[/tex]
So, the new force will become 1/4 times of the initial force.
