Question

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

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

Answer 1

Answer: (C)

$625$

We need force of repulsion $={10}^{-19}N$
Let n no. of increase electrons are placed on each sphere
$\Rightarrow \frac{K\left(ne\right)\left(ne\right)}{(3\times {10}^{-2}{)}^2}={10}^{-19}$

$\Rightarrow n^2{e}^2=\frac{{10}^{-19}\times 9\times {10}^{-4}}{9\times {10}^9\times e^2}$

$\Rightarrow n^2=\frac{{10}^{-32}}{e^2}\Rightarrow n=\frac{{10}^{-16}}{e}$

$n=\frac{{10}^{-16}}{1.6\times {10}^{-19}}$

$n=\frac{1000}{1.6}=625$