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. $\frac{n}{n_{total}}=\frac{\rho gA}{eE{N}_{A}Z}$
• B. $\frac{n}{n_{total}}=\frac{4gA}{\pi eE{N}_{A}Z}$
• C. $\frac{n}{n_{total}}=\frac{gA}{eE{N}_{A}Z}$
• D. $\frac{n}{n_{total}}=\frac{\pi \rho gA}{3eE{N}_{A}Z}$

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
$mg=qE=neE$
$\frac{4}{3}\pi r^3\rho g=neE$
$\Rightarrow n=\frac{4\pi r^3\rho g}{3eE}$
The number of moles of the metal in the sphere is
$n_{mol}=\frac{m}{A}=\frac{4\pi r^3\rho }{3A}$
So the number of metal atoms will be
$n_{metalatoms}=\frac{4\pi r^3\rho }{3A}\times N_A$
This means the total number of electrons in the sphere is
$n_{total}=\frac{4\pi r^3\rho N_A}{3A}\times Z$
Hence, the fraction is
$\frac{n}{n_{total}}=\frac{4\pi r^3\rho g}{3eE}\times \frac{3A}{4\pi r^3\rho N_{A}Z}$
$\Rightarrow \frac{n}{n_{total}}=\frac{gA}{eE{N}_{A}Z}$