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

Here, `E = (81 pi)/(7) xx10^(5) Vm^(-1), v = 2xx10^(-3) ms^(-1)`
`eta = 1.8xx10^(5) N sm^(-2) , rho = 900 kg m^(-3)`, When drop is balanced in still air under the effect of electric field, then
`q E = (4)/(3) pi r^(3) rho g` or `q = (4)/(3 E) pi r^(3) rho g`.....(i)
When the electric field is swiched off, let the drop fall with terminal velocity `v`, then
`v = (2 r^(2) (rho - sigma) g)/(9 eta)` or `r =[(9 eta v)/(2 (rho - sigma) g)]^(1/2)`
`:. q = (1)?(E) xx (4)/(3) pi rho g xx [(9 eta v)/(2 (rho - sigma) g)]^(1/2)`
`= (7)/(81 pi xx 10^(5)) xx (4)/(3) xx pi xx900xx9.8`
`xx [(9xx1.8xx10^(-5)xx2xx10^(-3))/(2xx900xx9.8)]^(3/2)`
On solving, `q = 8.0xx10^(-19) C`