Question

There are $'n'$ electrons of charge $'e'$ in a drop of oil of density $\rho$ . It is in equilibrium in an electric field $E$ . Then the radius of drop is:
(A) ${\left( {\dfrac{{2neE}}{{4\pi \rho g}}} \right)^{1/2}}$
(B) ${\left( {\dfrac{{neE}}{{\rho g}}} \right)^{1/2}}$
(C) ${\left( {\dfrac{{3neE}}{{4\pi \rho g}}} \right)^{1/3}}$
(D) ${\left( {\dfrac{{2neE}}{{\pi \rho g}}} \right)^{1/3}}$

Answer 1

To solve this question, we need to consider the vertical equilibrium of the drop, which is acted upon by its weight and the electric field. In the equation thus formed, writing the mass in terms of its radius will give the value of the radius.

Formula used: The formulae which are used to solve this question are given by
$F = qE$ , where $F$ is the electric force on a charge $q$ due to an electric field $E$ .
$q = ne$ , where $q$ is the total charge of $n$ electrons, and $e$ is the charge of an electron.
$V = \dfrac{4}{3}\pi {R^3}$ , where $V$ is the volume of a sphere of radius $R$ .

Complete step by step solution:
Let the radius of the drop be $R$ .
We first calculate the weight of the drop. If the mass of the drop is $m$ , then its weight must be
$W = mg$ (1)
We know that the mass of a body is related to its density and volume as
$m = \rho V$
The drop is a sphere of radius $R$ . So its volume is $V = \dfrac{4}{3}\pi {R^3}$ . So we have
$m = \dfrac{4}{3}\pi {R^3}\rho$
Substituting in (1), we get
$W = \dfrac{4}{3}\pi {R^3}\rho g$ (2)
Now we calculate the electric force on the drop. We know that the electric field force is given by
$F = qE$ (3)
Since the drop contains $n$ electrons, so it has a charge of
$q = ne$
Substituting in (3) we have
$F = neE$ (4)
We know that the weight of a body always acts vertically downwards. According to the question, the drop is in equilibrium with the electric field. So, the electric force of the field must be acting vertically upwards on drop, so as to balance its weight. So from the vertical equilibrium of the drop, we have
$W = F$
From (2) and (4)
$\dfrac{4}{3}\pi {R^3}\rho g = neE$
Separating the radius, we have
${R^3} = \dfrac{{neE}}{{\dfrac{4}{3}\pi \rho g}}$
${R^3} = \dfrac{{3neE}}{{4\pi \rho g}}$
Finally taking the cube root on both the sides, we get
$R = {\left( {\dfrac{{3neE}}{{4\pi \rho g}}} \right)^{1/3}}$
This is the required expression for the radius of the drop.
Hence the correct answer is option C.

Note:
We should not take the direction of the electric field in an upward direction, so as to oppose the weight which acts in the downward direction. This is because the drop of oil given in the question is negatively charged due to the presence of the electrons. So the direction of the electric force will be opposite to that of the electric field, meaning in downward direction which will not balance the drop. So, the electric field must be in the downward direction only.