A tiny spherical drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength $\dfrac{81\pi}{7} \times 10^5 Vm^{-1}$. When the field is switched off, the drop is observed to fall terminal velocity $2\times 10^{-3}\ ms^{-1}$. Given $g = 9.8\ ms^{-2}$, viscosity of air is $1.8\times 10^{-5}\ Nsm^{-2}$and the density of oil $is\ 900\ kg m^{-3}$, the magnitude of ‘q’ is:
$\text{A.}\quad 1.6 \times 10^{-19}\ C$
$\text{B.}\quad 3.2 \times 10^{-19}\ C$
$\text{C.}\quad 4.8 \times 10^{-19}\ C$
$\text{D.}\quad 8 \times 10^{-19}\ C$

Hint: When any charge is placed in an external electric field, it experiences electric force. Terminal velocity is a velocity which is achieved in the presence of some resistance force after the net force on an object becomes zero.
Formula used:
$F = qE \ and \ F_{R} = 6\pi \eta rv$

Complete answer:
When the charge containing bubbles is placed in a vertical direction, it will experience a force in the direction of the field which is equal to $F = qE$. As it lies in gravity, it will experience a force of gravity either, equal to its weight, in a vertically downward direction. As viscosity of air (air friction) has to be considered, the bubble will experience a viscous force, opposite to the direction of motion. This force will be equal to $F_{R} = 6\pi\eta rv$ where ‘r’ is the radius of the bubble and $\eta$ is the viscosity of air.
Since it is given that the bubble is balanced by electrostatic force, thus;
$W = F_{E}$
$mg= qE$
$\implies \rho Vg = qE$
$\implies V = \dfrac{ qE}{\rho g}$
Also $V = \dfrac{4}{3}\pi r^3$
Thus $r = \left(\dfrac{3V}{4\pi}\right)^{1/3}$
Now, as the electric field is turned off, at equilibrium, bubble achieves the terminal velocity = $2\times 10^{-3} m/s$:
Thus, for $\sum F_{net} = 0$, we can write:
$mg = 6\pi \eta rv$
$\implies \rho Vg = 6\pi\eta rv$
$\implies V = \dfrac{6\pi\eta rv}{\rho g}$
Also, $r = \left(\dfrac{3V}{4\pi}\right)^{1/3}$
Thus $V = \dfrac{6\pi\eta (\dfrac{3V}{4\pi})^{1/3}v}{\rho g}$
$\implies V = \left(\dfrac{6\pi \eta v}{\rho g}\right)^{3/2} \left(\dfrac{3}{4\pi} \right)^{1/2}$
But $V = \dfrac{ qE}{\rho g}$, thus;
$\left(\dfrac{6\pi \eta v}{\rho g}\right)^{3/2} \left(\dfrac{3}{4\pi} \right)^{1/2} = \dfrac{qE}{\rho g}$
$\implies q = \left(\dfrac{6\pi \eta v}{\rho g}\right)^{3/2} \left(\dfrac{3}{4\pi} \right)^{1/2} \times \dfrac{\rho g}{E}$
Hence, putting values:
$\implies q = \left(\dfrac{6\pi (1.8 \times 10^{-5}) (2\times 10^{-3})}{(900) (9.8)}\right)^{3/2} \left(\dfrac{3}{4\pi} \right)^{1/2} \times \dfrac{(900\times 9.8)}{\dfrac{81\pi\times 10^5}{7}}$
$q = 8 \times 10^{-19}C$.

So, the correct answer is “Option D”.

Note:
Let’s suppose we are also given the value of density of air or fluid in which the body is moving, in that case we have to take the buoyant force also along with the other forces. Hence a very important advice to all the students is that we have to act according to the statement of question. First of all, write all the data that is given in the question, and then think what type of forces must be taken to solve the question.