Two identical small metallic sphere having arbitrary charges are separated by a distance ' r ' in air. They are then touched by two more identical metal spheres respectively and seperated After- wards A and B are brought to `r/(2) ` distance. Show that initial interaction is same as final interaction-

Explanation

`F=( 1)/( 4piin _0) (q_1q_2)/( r^(2) ) ` where, `q_1` and `q_2` are the arbitary charges in A and B respectively . As and A and B are touched by identical sphere the charges are respectively reduced by ` (q_1)/( 2 ) and ( q_2)/( 2 ) ` Since
they are then placed at `(r)/(2) ` the new force is ` F.=(1)/( 4pi in _0) ((q_1)/(2) xx (q_2)/( 2) )/( (r)/(2) ^(2)) = (1)/( 4pi in _0) (q_1q_2)/( r^(2)) ((1//4)/(1//4)) `
` = (1)/( 4pi in _0) (q_1q_2)/( r^(2))` F, hence result.