Question

Two unchanged spheres are rubbed together and then separated by a distance of 2cm in air. If the force of attraction between them is 22.5N. Then the number of electrons lost by positively charged sphere is

• A. $6.25\times {10}^{18}$
• B. $6.25\times {10}^{12}$
• C. $225$
• D. $625$

Answer 1

The correct option is B $6.25\times {10}^{12}$

$F=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}=\frac{1}{4\pi {ϵ}_0}\frac{\left(ne\right)\left(ne\right)}{r^2}$
$22.5=9\times {10}^9\times \frac{n^2{e}^2}{4\times {10}^{-4}}$
$n^2=\frac{22.5\times 4\times {10}^{-4}}{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}=\frac{90\times {10}^{-13}}{9\times (1.6\times {10}^{-19}{)}^2}$
$n=\frac{1\times {10}^{-6}}{1.6\times {10}^{-19}}=6.25\times {10}^{12}$