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?
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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
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