A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength `(81pi)/(7)xx10^5Vm^-1`. When the field is switched off, the drop is observed to fall with terminal velocity `2xx10^-3ms^-1`. Given `g=9.8ms^-2`, viscoisty of the air `=1.8xx10^-5Nsm^-2` and the denisty of oil `=900kg m^-3`, the magnitude of q is

Options

A. `1.6xx10^(-19)C`
B. `3.2xx10^(-19)C`
C. `4.8xx10^(-19)C`
D. `8.0xx10^(-19)C`

Answer

D

Explanation

When electric field is switched ON, mg = qE
`m to` mass of oil drop `=(4)/(3)pi r^(3)xx rho`, where r is radius of drop and `rho` is density of oil. When electric field is switched OFF `mg = 6 pi eta rv`
where v is terminal velocity of oil drop. Solving above equation, we get `q=8xx10^(-19)C`