An electrostatic force of attraction between two point charges A and B is 1000 N.If the charge on A is increased by 25% and that on B is reduced by 25% and the initial distance them is decreased by `25%`, find the new force of attraction between them.

Explanation

(i) Given the force of attraction between two point charges `q_(1)` and `q_(2)` is `= F_(1)`.
`= (1)/(4piin_(0)) (q_(1)q_(2))/(d^(2)) = 1000 N`
When charge on `q_(1)` increase by `25%` then it becomes `(125)/(100) q_(1)`
When change on `q_(2)` decreases by `25%`, then it will be `= 75/100 q_(2)`
The initial distance (d) has reduced by `25%` then the present distance force would
`= F_(2) = (1)/(4piin_(0))((125)/(100) q_(1)'(75)/(100)q_(2))/((75/100d)^(2))`
Find the value of `F_(2)` in terms of `F_(1)`.
(ii) `1666.67 N`.