Question

the distance between two charges q a and q b is r and the force between them is F. What is the force between them if the distance between them is doubled?

Answer 1

The force will be reduced to 1/4 of the original

Explanation:

According to Coulomb's law, the force between two charges $q_a\:\text{and}\:q_b,$ separated by a distance r is given by

$F = k\dfrac{q_aq_b}{r^2}$

where k is the Coulomb constant.

Now let F' be the force between the two charges when their separation distance is doubled. We can write this force as

$F' = k\dfrac{q_aq_b}{(2r)^2} = k\dfrac{q_aq_b}{4r^2}$

$\;\;\;\;\;= \frac{1}{4}\left(k\dfrac{q_aq_b}{r^2}\right) = \frac{1}{4}F$

Therefore, the force will be reduced to a quarter of its original value.

Answer 2

Hope you could understand.

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