A particle `A` having a charge `10 mu C` is held fixed on a horizontal surface. A block of mass `80 g` and having charge stays in equilibrium on the surface at a distance `3 cm` from the first charge. The coefficient of friction between the surface and the block is `mu = 0.5`. Find the range within which the charge on the block may lie.

Explanation

The block is in equilibrium because electric force is balanced by friction force. If `q is + ve` , friction will act towards left , if `q is - ve` , friction will act towards right. We know friction force develops according to need from `0` to `mu N`. To get maximum value of charge that can be given to the second particle , we should equate repulsive force by maximum friction force.
If `q is + ve`
`F_(e ) = mu mg`
`(1)/( 4 pi in_(0)) ((10 xx 10^(-6))q)/((3 xx 10^(-2))^(2)) = 0.5 xx 80 xx 10^(-3) xx 10`
`q = 4 xx 10^(-9) C =` maximum `+ ve` charge
If `q is -ve`
`F_(e ) = mu mg`
`q = 4 xx 10^(-9) C =` maximum `-ve` charge
Range of charge is from `-4 xx 10^(-9) C` to `4 xx 10^(-9) C`.
If charge smaller than `+- 4 xx 10^(-9) C` to `4 xx 10^(-9) C` is given , friction force develops up to that extent so that it balances electric e.g. if `q = 0`, friction is not developed.