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