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

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

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

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

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

answer is D.

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Detailed Solution

Terminal velocity $v_{T}$ is given by

$6 π η r v_{T} = m g = (\frac{4 π r^{3}}{3}) ρ g 6 π (1.8 \times 10^{- 5}) r \times 2 \times 10^{- 3} = (\frac{4 π r^{3}}{3}) 900 \times 9.8$

we get r = $\frac{3}{7} \times 10^{- 5} m$ .

The oil drop will be balanced in air if

$q E = m g = (\frac{4 π r^{3}}{3}) ρ g q (\frac{81 π}{7} \times 10^{5}) = (\frac{4 π {(\frac{3}{7} \times 10^{- 5})}^{3}}{3}) 900 \times 9.8$

we get q = $8.0 \times 10^{- 19} C$ .

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