Question

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

• A. 25
• B. 225
• C. 625
• D. 1250

Answer 1

Answer: (C)

From Coulomb's law
$F=\frac{k{q}_1{q}_2}{r^2}=\frac{k(ne)(ne)}{r^2}\left(\text{ where }k=\frac{1}{4\pi {\epsilon }_0}\right)$
or $n^2=\frac{F{r}^2}{k{e}^2}=\frac{10^{-19}\times {\left(3\times 10^{-2}\right)}^2}{9\times 10^9\times {\left(1.6\times 10^{-19}\right)}^2}$
$=\frac{1}{9\times 10^{28}}\times \frac{{\left(3\times 10^{-2}\right)}^2}{{\left(1.6\times 10^{-19}\right)}^2}$
$=\frac{10^6}{(1.6{)}^2}\therefore n=\frac{10^3}{1.6}=625$