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. `1.6 xx 10^(-19)` C
B. `3.2 xx 10^(-19)` C
C. `4.8 xx 10^(-19)` C
D. `8.0 xx 10^(-19)` C

Answer

D

Explanation

Here,
`E=(81pi)/(7)xx10^(5) V m^(-1),v=2xx10^(-3) m s^(-1)`,
`eta=1.8xx10^(-5) N s m^(-2), rho=900 kg m^(-3)`
When drop is balanced in still air under the effect of electric field, then
`qE=(4)/(3)pir^(3)rhog` or `q=(4)/(3E)pir^(3)rhog`
When the electric field is switched off, let the drop falls with terminal velocity `v`, then
`v=(2r(rho-sigma)g)/(9eta)` or `r=[(9veta)/(2(rho-sigma)g)]^(1//2)`
`therefore q=(1)/(E)xx(4)/(3)pirhog[(9etav)/(2(rho-sigma)g)]^(3//2)`
`=(7)/(81pixx10^(5))xx(4)/(3)xxpixx900xx9.8xx[(9xx1.8xx10^(-5)xx2xx10^(-3))/(2xx900xx9.8)]^(3//2)`
On solving we get, `q=8xx10^(-19)C`