Question

Two charged particles exert an electrostatic force of 7 N on each other. What will the magnitude of the electrostatic force be if the distance between the two charges is reduced to one-fourth of the original distance?

Answer 1

We know force between two charge particles

F=k Q ₁* Q ₂/ R ² =7 N ..1

according to the question R'=R/4

F'=k Q ₁*Q ₂ /R'²

so F'=k Q ₁ *Q ₂ /(R/4)²

F'=16*k Q ₁*Q ₂ /R ² ..2

using equation 1 in 2

F'=16*7

F'= 112 N Answer.