Two particles having charge `Q_(1)` and `Q_(2)` , when kept at a certain distance exert a force `F` on each other . If the distance between the two particles is reduced to half and the charge on each particle is doubled , the force between the particles would be

Answer

16 F

Explanation

`because F = (kq_(1)q_(2))/(r^(2))`
If `q'_(1) = = 2q_(1), " "q'_(2) = 2q_(2) " "r' = (r )/(2)`,
then `F' = (kq'_(1)q'_(2))/(r'_(2)) = (k(2q_(1))(2q_(2)))/((r )/(2))^(2)`
`F' = (16kq_(1)q_(2))/(r^(2))`
`F' = 16F`