Question

A neutral sphere of radius $r$ and density $\rho $ is placed in a uniform electric field $E$ that exists on the earth's surface in the vertically upward direction. If the atomic number and the mass number of the material of the sphere are $Z$ and $A$ respectively, then the fraction of electrons that should be removed from the sphere for it to remain in equilibrium is [Assume that the sphere is able to hold the necessary charge without any leakage. Here $N_{A}$ -- Avogadro number]

• A. $\frac{n}{n_{total}}=\frac{\rho g A}{e E N_{A} Z}$
• B. $\frac{n}{n_{total}}=\frac{4 g A}{\pi e E N_{A} Z}$
• C. $\frac{n}{n_{total}}=\frac{g A}{e E N_{A} Z}$
• D. $\frac{n}{n_{total}}=\frac{\pi \rho g A}{3 e E N_{A} Z}$

Answer 1

Answer: (C)

Let us assume that we remove $n$ electrons from the sphere due to which the net charge on the sphere is $q=ne$
For the sphere to be in equilibrium
$mg=qE=neE$
$\frac{4}{3}\pi r^{3}\rho g=neE$
$\Rightarrow n=\frac{4 \pi r^{3} \rho g}{3 e E}$
The number of moles of the metal in the sphere is
$n_{mol}=\frac{m}{A}=\frac{4 \pi r^{3} \rho }{3 A}$
So the number of metal atoms will be
$n_{metal atoms}=\frac{4 \pi r^{3} \rho }{3 A}\times N_{A}$
This means the total number of electrons in the sphere is
$n_{total}=\frac{4 \pi r^{3} \rho N_{A}}{3 A}\times Z$
Hence, the fraction is
$\frac{n}{n_{total}}=\frac{4 \pi r^{3} \rho g}{3 e E}\times \frac{3 A}{4 \pi r^{3} \rho N_{A} Z}$
$\Rightarrow \frac{n}{n_{total}}=\frac{g A}{e E N_{A} Z}$