What happens to the force between two charges if the distance between them is a doubled and b halved?

Text Solution

Solution : From Coulombs law, `F prop 1/d^(2)`. So <br> (a) When distance is reduced to half, force increases by four times. `[ :' F_(2)=(F_(1)d_(1)^(2))/((d_(1)/2)^(2))=4F_(1)]` <br> (b) When distance is doubled, then force is reduced by four times. `[ :' F_(2)=(F_(1)d^(2))/(|2d_(1)|^(2))=1/4 F_(1)]`