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

Question

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, 1.6 × 10 - 19 C, 3.2 × 10 - 19 C, 4.8 × 10 - 19 C, 8.0 × 10 - 19 C

MARKS App · Accepted Answer

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 π ρ g 1 / 2 6 π η v 3 / 2 ∴ q = 1 E 3 4 π ρ g 1 2 6 π η v 3 / 2 Substituting the values, we get q = 7 81 π × 10 5 3 4 π × 900 × 9.8 × 216 π 3 × 1.8 × 10 - 5 × 2 × 10 - 3 3 = 8.0 × 10 - 19 C

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