A tiny spherical oil drop carrying a net charge q is balanced in still air with vertical uniform electric field of strength 81 π 7 × 10 5 V m - 1 . When the field is switched off, the drop is observed to fall with terminal velocity 2 × 10 - 3 m s - 1 . Given g = 9.8 m s - 2 , viscosity of the air = 1.8 × 10 - 5 N s m - 2 and the density of oil = 900 kg m - 3 , the magnitude of q is

Step-by-step solution

q E   = m g ...(i) 6 π η   r v = m g       4 3 π r 3   ρ g = m g ...(ii) ∴ r = 3 m g 4 π ρ g 1 / 3       ...(iii) Substituting the value of r in Eq. (ii), we get 6 π η v 3 m g 4 π ρ g 1 / 3 = m g or 6 π η v 3   3 m g 4 π ρ g         = m g 3 Again substituting m g = q E ,   we get         q E 2 = 3 4 π ρ g 6 π η v 3 Or q E = 3 4