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 $= 1 .8 \times 10^{- 5} {Nsm}^{- 2}$ and the density of oil $= 900 {kg m}^{- 1}$ , the magnitude of q is $x x 10^{- 19} C .$ The value of $x$ is___

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Detailed Solution

When the electric field is on

In equilibrium, force due to electric field = weight qE=mg

$\Rightarrow qE = \frac{4}{3} {πR}^{3} ρg ∴ q = \frac{4 {πR}^{3} ρg}{3 E} . . . (i)$ When the electric field is switched off

Weight of the drop= viscous & force on the drop $mg = 6 πη {Rv}_{1}$

$\frac{4}{3} {πR}^{3} ρg = 6 {πηR}_{1} ∴ R = \sqrt{\frac{9 {ηv}_{1}}{2 ρg}} . . (ii)$

from eq. (i) & (ii) $q = \frac{4}{3} π {[\frac{9 {ηv}_{1}}{2 ρg}]}^{\frac{3}{2}} \times \frac{ρg}{E}$

$\begin{array}{l} = \frac{4}{3} \times π {[\frac{9 \times 1 .8 \times 10^{- 5} \times 2 \times 10^{- 3}}{2 \times 900 \times 9 .8}]}^{\frac{3}{2}} \times \frac{900 \times 9 .8 \times 7}{81 π \times 10^{5}} \\ or, q = 7 .8 \times 10^{- 19} C ≃ 8 x 10^{- 19} C \end{array}$