A tiny spherical oil drop carrying a net charge q is balanced in still air when a vertical uniform electric field of strength `(81pi)/(7)xx 10^(5) V m^(-1)`. When the field is switched off, the drop is observed to fall with terminal velocity `2 xx 10^(-3) m s^(-1)` . The magnitude of q is (given g = `9.8 m s^(-1), eta_("air") = 1.8 xx 10^(-5)N s m^(-1), rho_("oil") = 900 kg m s^(-3)`, Electric force = charge `xx`electric field )

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

`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`