Calculate the excess number of electrons that must be placed on each of the two similar uncharged spheres that are located 5 cm part of (in air), such that the force of repulsion between them is `36 xx 10^(-19) N`.

Explanation

(i) The uncharged shphere are neutral .
Use Coulomb's law
`F = (1)/(4piin_(0)) xx (q_(1) q_(2))/(r^(2))`
Substitute the value of `F, in_(0)` and
`1/(4pi)` and determine the value of `q_(1)q_(2)`.
But `q_(1) = q_(2) = "ne"`
Where m is the number of electrons and e is the charge on each electrons.
`(e = 1.602 xx 10^(-19)C)`
Find out the value of n from the above relation.
`(ii) n = 6250`.