The excess (equal to number) number of electrons that must be placed on each of two small spheres spaced `3cm` apart with force of repulsion between the spheres of be `10^(-19)N` is

Options

A. 25
B. 225
C. 625
D. 1250

Answer

C

Explanation

From Coulomb.s law
`F = (kq_(1)q_(2))/(r^(2)) = (k(n e)(n e))/(r^(2)) ("where " k = (1)/(4pi epsilon_(0)))`
or `n^(2) = (Fr^(2))/(k e^(2)) = (10^(-19) xx (3 xx 10^(-2))^(2))/(9 xx 10^(9) xx (1.6 xx 10^(-19))^(2))`
`= (1)/(9 xx 10^(28)) xx ((3 xx 10^(-2))^(2))/((1.6 xx 10^(-19))^(2)) = (10^(6))/((1.6)^(2)) :. n = (10^(3))/(1.6) = 625`