After reaching the terminal velocity the speed of the oil drop is given as
[tex]v = \frac{d}{t}[/tex]
it drop by 3 mm in 7.33 s
[tex]v = \frac{0.003}{7.33}[/tex]
[tex]v = 4.1 * 10^-4 m/s[/tex]
now by the formula of terminal speed
[tex]6\pi \eta r v = \rho* \frac{4}{3}\pi r^3g[/tex]
[tex]v = \frac{2}{9/eta}r^2\rho g[/tex]
[tex]4.1*10^{-4} = \frac{2}{9*1.83 * 10^{-5}}*r^2*860*9.8[/tex]
[tex]r = 0.16 mm[/tex]
so the radius of drop is 0.16 mm
Now the electric field between the plates is given as
[tex]E = \frac{\Delta V}{\Delta x}[/tex]
[tex]E = \frac{1177}{0.01} = 1.177* 10^5 N/C[/tex]
now we will use the condition of equilibrium
[tex]qE = mg[/tex]
[tex]q*1.177 * 10^5 = \rho* \frac{4}{3}\pi r^3 *9.8[/tex]
[tex]q*1.177 * 10^5 = 860* \frac{4}{3}\pi *(0.16*10^{-3})^3*9.8[/tex]
[tex]q = 1.2 * 10^{-12} C[/tex]
so above is the charge on the drop
