Question

Two like charges exert a force of repulsion of 3.5 Non each other. What will be the force if the distance
between the charges increased to five times the original value ?

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Answer 1

Explanation:

As the force due to two point charges is equal to :

F= Kq1q2/r²

So from here we get to know that force is inversely proportional to r²

So the ans should be 3.5/25= 0.14

Here r: radius

q: charge

K: 9×10^9

F: force