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A tiny spherical oil drop carrying a net...

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

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`

Text Solution

Verified by Experts

The correct Answer is:
D
`qE = m g " "….(i)`
`6 pi eta rv = ng`
` (4)/(3) pi r^(3) rho q " "…..(ii)`
`therefore " " r = ((3 m g)/(4 pi rho g))^(1//3) " "….(iii)`
Substituting the value of r in equation (ii) we get (ii) we get `6 pi eta v(( 3m g)/( 4 pi rhp g))^(1//3) = mg or (6 pi eta v)^(3) (( 3m g)/( 4 pi rho g)) = (mg)^(3)`
Again substituting , `(qE)^(3) = ((3)/(4 pi rho g)) ( t pi n eta v)^(3) or qE ((3)/( 4 pi rho g))^(1//2) ( 6 pi eta n v)^(3//2)`
`therefore q = (1)/(E) ((3)/(4 pi rho g))^(1//3) ( 6 pi eta n)^(3//2)`
Substituting the values we get,
` q = (7)/(81 xx 10^(5)) sqrt((3)/(4 pi xx 9.00 xx 9.8))xx 216 pi^(2)) xx sqrt(( 1.8xx 10^(-5) xx 2 xx 10^(-3))^(3)) = 8.0 xx 10^(-19) C`
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