1. Forces when the drop falls with terminal velocity

When the electric field is switched off, the oil drop falls with terminal velocity $v_t = 2\times 10^{-3}\,\text{m s}^{-1}$.

At terminal velocity, viscous drag balances the effective weight:

$6\pi \eta r v_t = \frac{4}{3}\pi r^3 (\rho-\rho_{air})g$

Since density of air is negligible compared to oil, we take:

$6\pi \eta r v_t = \frac{4}{3}\pi r^3 \rho g$

Cancel $\pi r$:

$6\eta v_t = \frac{4}{3}r^2 \rho g$

So,

$r^2 = \frac{9\eta v_t}{2\rho g}$

Substitute values:

$r^2 = \frac{9(1.8\times10^{-5})(2\times10^{-3})}{2(900)(9.8)}$

$r^2 = \frac{3.24\times10^{-7}}{17640}$

$r^2 \approx 1.837\times10^{-11}$

$r \approx 4.29\times10^{-6}\,\text{m}$

2. Forces when the drop is balanced in electric field

In the electric field, the drop is balanced, so electric force equals weight:

$qE = mg$

where

$m = \frac{4}{3}\pi r^3 \rho$

Hence,

$q = \frac{\frac{4}{3}\pi r^3 \rho g}{E}$

Given

$E = \frac{81\pi}{7}\times10^5\,\text{V m}^{-1}$

Now compute $r^3$:

$r^3 = (4.29\times10^{-6})^3 \approx 7.89\times10^{-17}$

Then

$mg = \frac{4}{3}\pi (7.89\times10^{-17})(900)(9.8)$

$mg \approx 2.91\times10^{-12}\,\text{N}$

Thus,

$q = \frac{2.91\times10^{-12}}{\left(\frac{81\pi}{7}\times10^5\right)}$

Using

$\frac{81\pi}{7}\times10^5 \approx 3.63\times10^6$

we get

$q \approx \frac{2.91\times10^{-12}}{3.63\times10^6}$

$q \approx 8.0\times10^{-19}\,\text{C}$

3. Match with the options

$q = 8.0\times10^{-19}\,\text{C}$

So the correct option is:

D

4. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.