Question

In Millikan’s experiment, an oil drop of mass $4.9 \times {10^{ - 14}}\;{\rm{kg}}$ is balanced by applying a potential difference of $9.8\;{\rm{kV}}$ between the two plates which are $12.8\;{\rm{mm}}$ apart. Calculate the number of elementary charges on the drop. $\left( {{\rm{Take}}\;g = 10\;{\rm{m}}{{\rm{s}}^{ - 2}}} \right)$.

Answer 1

To solve this question first we must derive the formula for the number of elementary charges on the drop by using some other formulas, then substitute the given values into the obtained formula to get the final value.

Complete step by step answer: Given:
The mass of the oil drop is $m = 4.9 \times {10^{ - 14}}\;{\rm{kg}}$.
The potential difference is $V = 9.8\;{\rm{kV}}$.
The distance between the plates is $d = 12.8\;{\rm{mm}}$.
The acceleration due to gravity is $g = 10\;{\rm{m}}{{\rm{s}}^{ - 2}}$.

The oil drop experiment was conducted by Robert Millikan and Harvey Fletcher in year $1909$. This experiment was conducted to find the charge of an electron.

The value of charge calculated by them was $1.5924 \times {10^{ - 19}}\;{\rm{C}}$. They further described that the electrons are the subatomic particle having a negative charge and orbiting around the nucleus. The flow of these electrons in the conductor is responsible for electricity.

First consider the following reaction.
$\begin{array}{c} Eq = mg\\\ q = \dfrac{{mg}}{E} \end{array}$

Here, $q$ is charge, $m$ is mass, $g$ is acceleration due to gravity, and $E$ is electric field.

As we know that $E = \dfrac{V}{d}$.
Here, $V$ is potential difference, and $d$ is separation.

By substituting $E = \dfrac{V}{d}$ in the equation $q = \dfrac{{mg}}{E}$, we get,
$\begin{array}{c} q = \dfrac{{mg}}{{\dfrac{V}{d}}}\\\ = \dfrac{{mgd}}{V} \end{array}$

As we know that $q = ne$.
By substituting, $q = ne$ in the equation $q = \dfrac{{mgd}}{V}$, we get
$\begin{array}{c} ne = \dfrac{{mgd}}{V}\\\ n = \dfrac{{mgd}}{{Ve}} \end{array}$

By substituting $4.9 \times {10^{ - 14}}\;{\rm{kg}}$ for $m$, $10\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ for $g$, $12.8 \times {10^{ - 3}}\;{\rm{m}}$ for $d$, $9.8 \times {10^3}\;{\rm{V}}$ for $V$, and $1.6 \times {10^{ - 19}}\;{\rm{C}}$ for $e$ in the equation $n = \dfrac{{mgd}}{{Ve}}$, we get

n = \dfrac{{\left( {4.9 \times {{10}^{ - 14}}} \right)\left( {10} \right)\left( {12.8 \times {{10}^{ - 3}}} \right)}}{{\left( {9.8 \times {{10}^3}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)}}\\\ \Rightarrow n= 4 \end{array}$$ **Therefore, the number of elementary charges on the drop are $4$.** **Note:** Make sure to use correct relations to get the final relation to find the number of elementary charges on the drop, and finally do not forget to substitute all the values in the same unit to get the correct answer.