What will be the effect on the force between two charged particles, if the distance between them is halved and the charge on both particles is kept constant?

The correct answer is option 2):(Force will be four times)

Concept:

• The force between the two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
• Hence, if the distance between charges is halved (charges remaining kept constant), the force between the two charges is quadrupled. Force will be four times. F is the force between the charge q₁ and q2
• F = \(k q_1​q_2\over r^2\)

where

K  is the constant

q_1,q₂ are the charges

r is the distance between the charge 

• When the distance between the  q1 and q₂ is halved 
• F = \(k q_1​q_2\over (r/2)^2\)
• F =  4  \(k q_1​q_2\over r^2\)  Force will be four times