A tiny spherical oil drop carrying a net charge q is balanced in still aire with a vertical uniform electric field of strength . 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 ms–2, viscosity of the air = 1.8 × 10–5 Ns m–2 and the density of oil = 900 kg m

Engineering

Physics

Viscosity and Terminal Velocity New

Charge and Coulombs Law

Question

A tiny spherical oil drop carrying a net charge q is balanced in still aire with a vertical uniform electric field of strength $\frac{81 π}{7} \times 10^{5} {Vm}^{- 1}$ . 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 ms^–2, viscosity of the air = 1.8 × 10^–5 Ns m^–2 and the density of oil = 900 kg m^–3, the magnitude of q is

1.6 × 10^–19C

4.8 × 10^–19C

3.2 × 10^–19C

8.0 × 10^–19C

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$2 \times 10^{- 3} = \frac{2}{9} \frac{r^{2} g \times p}{η}$

qE = mg = 6 $πη$ rv

$q = \frac{6 \times π \times 1 .8 \times 10^{- 5} \times r \times 2 \times 10^{- 3} \times 7}{8 π \times 10^{5}}$

$900 \times \frac{4}{3} π rg = 6 πηrv$

$r^{2} = \frac{6 \times 1 .8 \times 10^{- 5} \times 2 \times 10^{- 3}}{1200 \times 9 .8}$

$r = \frac{3}{7} \times 10^{- 5}$

$q = \frac{2 \times 1 .8 \times \frac{3}{7} \times 10^{- 14} \times 2 \times 10^{- 8} \times 7}{27 \times 10^{5}} = 8 \times 10^{- 19}$