Question

Two charges attract each other with a force of 1.5 N. What will be the force if the distance between them is reduced to one-ninth of its original value?

Answer 1

Answer:

Fnew=121.5N

Explanation:

the coulombs law

f=k×q1.q2÷r.r

the force is same on the both charges

let distance between the two charges be r.

f=k×q1.q2÷r.r

1.5=k.q1.q2÷r.r

when distance is reduced

Fnew=121.5N