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 \pi }{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}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 {3mg}{4 \pi\rho g}\right )^{1/3} ...(iii) $
Substituting the value of rin Eq. (ii) we get,
$6 \pi \eta v \left(\frac {3mg}{4 \pi \rho g}\right)^{1/3}=mg $
or $(6 \pi \eta v)^3 \left (\frac {3mg}{4 \pi \rho g}\right)=(mg)^3 $
Again substituting mg = qE we get,
$(qE)^2= \left(\frac {3}{4 \pi \rho g} \right)(6 \pi \eta v)^3 $
or $qE= \left(\frac {3}{4 \pi \rho g}\right)^{1/2}(6 \pi \eta v)^{3/2}$
$\therefore q=\frac {1}{E}\left(\frac {3}{4 \pi \rho g}\right)^{1/2}(6 \pi \eta v)^{3/2} $
Substituting the values we get,
$q= \frac {7}{81 \pi \times 10^5}\sqrt {\frac {3}{4 \pi \times900 \times 9.8}\times 216 \pi ^3 }$
$\times \sqrt {(1.8 \times 10^{-5}\times 2 \times 10^{-3})^3} $
$=8.0 \times 10^{-19}C$