Question

The excess (equal in number) of electrons that must be placed on each of two small spheres spaced $3 \,cm$ 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(n e)(n e)}{r^{2}}\left(\text { where } k=\frac{1}{4 \pi \varepsilon_{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 $