Question

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

• A. $1.6\times 10^{- 19}C$
• B. $3.2\times 10^{- 19}C$
• C. $4.8\times 10^{- 19}C$
• D. $8.0\times 10^{- 19}C$

Answer 1

Answer: (D)

$qE \, =mg$ ...(i)
$6\pi \eta \, rv=mg \, \, \, $
$\frac{4}{3}\pi r^{3} \, \rho g=mg$ ...(ii)
$\therefore r=\left(\frac{3 m g}{4 \pi \rho g}\right)^{1 / 3} \, \, \, $ ...(iii)
Substituting the value of r in Eq. (ii), we get
$6\pi \eta v\left(\frac{3 m g}{4 \pi \rho g}\right)^{1 / 3}=mg$
or $\left(6 \pi \eta v\right)^{3} \, \left(\frac{3 m g}{4 \pi \rho g}\right) \, =\left(m g\right)^{3}$
Again substituting $mg=qE, \, $ we get
$ \, \left(q E\right)^{2}=\left(\frac{3}{4 \pi \rho g}\right)\left(6 \pi \eta v\right)^{3}$
Or $qE=\left(\frac{3}{4 \pi \rho g}\right)^{1 / 2} \, \, \left(6 \pi \eta v\right)^{3 / 2}$
$\therefore \, q=\frac{1}{E}\left(\frac{3}{4 \pi \rho g}\right)^{\frac{1}{2}}\left(6 \pi \eta v\right)^{3 / 2}$
Substituting the values, we get
$q=\frac{7}{81 \pi \times 10^{5}}\sqrt{\frac{3}{4 \pi \times 900 \times 9.8} \times 216 \pi ^{3}} \, \, \, $
$ \, \times \sqrt{\left(1.8 \times \left(10\right)^{- 5} \times 2 \times \left(10\right)^{- 3}\right)^{3} \, }=8.0\times \left(10\right)^{- 19}C$