A tiny spherical oil drop carrying a net charge is balanced in still air with a vertical uniform electric field of strength When the field is switched off, the drop is observed to fall with terminal velocity Given viscosity of the air and the density

Question

A tiny spherical oil drop carrying a net charge $q$ is balanced in still air 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 $2 \times 10^{- 3} {ms}^{- 1} .$ Given $g = 9.8 {ms}^{- 2},$ viscosity of the air $= 1.8 \times 10^{- 5} N s m^{- 2}$ and the density of oil $= 900 kg m^{- 3},$ the magnitude of $q$ is [2010]

Smart Achievers · Accepted Answer

(4)

$When the electric field is on$

$In equilibrium, force due to electric field = weight$

$q E = m g$

$\Rightarrow q E = \frac{4}{3} π R^{3} ρ g$

$∴$ $q = \frac{4 π R^{3} ρ g}{3 E}$ ...(i)

$When the electric field is switched off$

$Weight of the drop = viscous force on the drop$

$m g = 6 π η R v_{t}$

$\frac{4}{3} π R^{3} ρ g = 6 π η R v_{t}$

$∴$ $R = \sqrt{\frac{9 η v_{t}}{2 ρ g}}$ ...(ii)

$From Eqs. (i) and (ii)$

$q = \frac{4}{3} π {(\frac{9 η v_{t}}{2 ρ g})}^{\frac{3}{2}} \times \frac{ρ g}{E}$

$= \frac{4}{3} \times π {[\frac{9 \times 1.8 \times 10^{- 5} \times 2 \times 10^{- 3}}{2 \times 900 \times 9.8}]}^{\frac{3}{2}} \times \frac{900 \times 9.8 \times 7}{81 π \times 10^{5}}$

$∴$ $q = 7.8 \times 10^{- 19} C$

Solution

1 $1.6 \times 10^{- 19} C$

2 $3.2 \times 10^{- 19} C$

3 $4.8 \times 10^{- 19} C$

4 $8.0 \times 10^{- 19} C$

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Solution

1 1.6×10-19C

2 3.2×10-19C

3 4.8×10-19C

4 8.0×10-19C

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1 $1.6 \times 10^{- 19} C$

2 $3.2 \times 10^{- 19} C$

3 $4.8 \times 10^{- 19} C$

4 $8.0 \times 10^{- 19} C$