Answered

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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?

Sagot :

Trancescient

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

Explanation:

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

[tex]F = k\dfrac{q_aq_b}{r^2}[/tex]

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

[tex]F' = k\dfrac{q_aq_b}{(2r)^2} = k\dfrac{q_aq_b}{4r^2}[/tex]

[tex]\;\;\;\;\;= \frac{1}{4}\left(k\dfrac{q_aq_b}{r^2}\right) = \frac{1}{4}F[/tex]

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

Hope you could understand.

If you have any query, feel free to ask.

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