Question

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

• A. $600$
• B. $625$
• C. $525$
• D. $125$

Answer 1

The correct option is B $625$

$F=\frac{q_1{q}_2}{4\pi {ϵ}_o{r}^2}$

Given here that- $q_1=q_2=q\left(say\right)$

$r=3cm=3\times {10}^{-2}m$

$\frac{1}{4\pi {ϵ}_o}=9\times {10}^9$

And, $F={10}^{-19}N$

Hence $9\times {10}^9\times \frac{q^2}{9\times {10}^{-4}}={10}^{-19}$

$⟹q^2\times {10}^{13}={10}^{=-19}$

$⟹q^2={10}^{-32}$

$⟹q={10}^{-16}C$

Let N=number of electrons

$q=Ne⟹N=\frac{q}{e}$

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

$⟹N=625$

Answer-(B)