Question

In Millikan's oil drop experiment an oil drop carrying a charge $Q$ is held stationary by a potential difference 2400V between the plates. To keep a drop of half the radius stationary the potential difference had to be made 600V. What is the charge on the second drop?
A: $\dfrac{Q}{4}$
B: $\dfrac{Q}{2}$
C: $Q$
D: $\dfrac{{3Q}}{2}$

Answer 1

In Millikan’s oil drop experiment, the gravitational force experienced by the oil drop is balanced by the electrostatic force created by the given potential. By finding the relationship between these two forces under the given conditions, and simplifying those relations, we can find the charge on the second drop.

Complete step by step answer:
In the Millikan’s oil drop experiment it’s been given in the question that the oil droplet was held stationary by a potential.
Let’s find the gravitational force,
Gravitational force = $mg$, where $m$represents the mass of the oil drop.
Since, $m$denotes the mass it can be represented as $density \times volume$.
Assume the density of the oil drop to be $\rho$ and the radius of the oil drop to be $R$
Volume of oil drop will be $\dfrac{{4\prod {R^3}}}{3}$
Hence the gravitational force on oil drop will be,
$mg = density \times volume \times g = \rho \times \dfrac{{4\prod {R^3}}}{3} \times g$……………………. (1)
it is necessary to do this step as in the question nothing was said about the mass of the drop but the radius, therefore we must express mass in terms of the radius.
Now, we’ll have a look at the electrostatic force
Electrostatic force on a charge $Q$can be expressed as $QE$, where $E$denotes the electric field.
As the question gives information only about the electric potential and not the electric field, we need to express the electric field in terms of electric potential.
For a spherical charge, $E = \dfrac{V}{L}$
Thus, $QE = \dfrac{{QV}}{L}$ ……………………. (2)
Since both forces balance each other we can equate equation (1) and equation (2),
$QE = mg$
i.e. $\dfrac{{QV}}{L} = \rho \times \dfrac{{4\prod {R^3}}}{3} \times g$
or $Q = \rho \times \dfrac{{4\prod {R^3}}}{3} \times g \times \dfrac{L}{V}$
Since $\rho \times \dfrac{{4Lg\prod }}{3}$ is a constant,
$Q \propto \dfrac{{{R^3}}}{V}$
We have successfully obtained the relation between charge, radius, and potential now we can have the answer by taking the ratio and simplifying the equation.
i.e. $\dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{{R_2^3 \times {V_1}}}{{R_1^3 \times {V_2}}}$
Rearranging and substituting the values,
${Q_2} = \dfrac{{{Q_1} \times {{\left( {\dfrac{1}{2}} \right)}^3}}}{{\left( {\dfrac{{2400}}{{600}}} \right)}}$
Hence we get, ${Q_2} = \dfrac{{{Q_1}}}{2}$, where ${Q_1}$is our initial charge $Q$
Therefore, the charge of the second drop ${Q_2}$is $\dfrac{{{Q_{}}}}{2}$.

So, the correct answer is “Option B”.

Note:
The key to solving this question lies in the understanding of the basic relations between radius and mass, as well as potential and electric field to rearrange the expressions of forces to get the desired relations between the variables. The electric potential and not the electric field, we need to express the electric field in terms of electric potential.