A tiny spherical oil drop carrying a net charge q is balanced in still air, with a vertical uniform electric field of strèngth 81 7 10^5 ~V / m . When the field is switched OFF, the drop is observed to fall with terminal velocity 2 10⁻³ ~ms ⁻¹ . Here g=9.8 ~m / s ^2 , viscosity of air is 1.8 10⁻⁵ Ns / m ^2 and density of oil is 900 ~kg m ⁻³ . The magnitude of q is

Step-by-step solution

Given, E= 81 7 10^5 ~V / m Terminal velocity , v=2 10⁻³ ~m / s g=9.8 ~m / s ^2 Viscosity, =1.8 10⁻⁵ ~N - s / m ^2 Density, =900 ~kg / m ^3 'Since, q E=m g In the absence of electric field, array rlrl & & m g & =6 r v & & q E & =6 r & & r= q E 6 v array [from Eq. (i)] From Eq. (i), we get gathered m= q E g 4 3 r^3 d= q E g 4 3 ( q E 6 v )^3 d= q E g q= 3 6^3 ^2 ^3 v^3 4 E^2 g gathered aligned & = 3 6^3 (314)^2 (1.8 10⁻⁵ )^3 (2 10⁻³ )^3 4 ( 81 7 10^5 )^2 9.8 & =8 10⁻¹⁹ C aligned