Question

Calculate the excess number of electrons on each of two similar charged spheres that are located 5 cm apart (in air),such that the force of repulsion between them is 36 x 10${}^{-19}$N

• A. 3262
• B. 1625
• C. 6250
• D. 1547

Answer 1

Answer: (C)

6250

Using Coulomb's Law,

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}$

We get, $q^2=\frac{36\times {10}^{-19}}{9\times {10}^9}\times (5\times {10}^{-2}{)}^2$

$q^2=\frac{36}{9}\times (5{)}^2\times {10}^{-32}$

$q=\frac{6}{3}\times 5\times {10}^{-16}$

$=10\times {10}^{-16}$

$={10}^{-15}C$

As $q=ne$

$n=\frac{{10}^{-15}}{1.6\times {10}^{-19}}$

$=0.625\times {10}^4$

$=6250$ electrons.