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

Balancing forces we have, `qE=mg` . . . (i)
`6pietarv=mg`
`(4)/(3)pir^(3)rhog=mg` . . . (ii)
`therefore r=((3mg)/(4pirhog))^(1//3)` . . . (iii)
Substituting the value of r in Eq. (ii), we get
`(qE)^(2)=((3)/(4pirhog))(6pietav)^(3)`
`therefore q=(1)/(E)((3)/(4pirhog))^(1//2)(6pietav)^(3//2)`
Substituting the values, we get
`q=(7)/(81pixx10^(5))sqrt((3)/(4pixx900xx9.8)xx216pi^(3))`
`xxsqrt((1.8xx10^(-5)xx2xx10^(-3))^(3))=8.0xx10^(-19)C`