Question

A tiny spherical oil drop carrying a net charge $q$ is balanced in still air, with a vertical uniform electric field of strength $\frac{81}{7} \pi \times 10^{5}\, V / m$. When the field is switched off, the drops is observed to fall with terminal velocity $2 \times 10^{-3} \,ms ^{-1}$. Here $g =9.8\, m / s ^{2}$, Viscosity of air is $1.8 \times 10^{-5} \,Ns / m ^{2}$ and the density of oil is $900\, kg\, m ^{-3}$. The magnitude of ' $q$ ' is

• A. $1.6 \times 10^{-19} C$
• B. $3.2 \times 10^{-19} C$
• C. $0.8 \times 10^{-19} C$
• D. $8.0 \times 10^{-19} C$

Answer 1

Answer: (D)

Here,
$
E =\frac{81 \pi}{7} \times 10^{5} V \textrm {m } ^ { - 1 }
$
$
v =2 \times 10^{-3} ms ^{-1}
$
$
\eta=1.8 \times 10^{5} N s m ^{-2}
$
$
\rho=900 kg m ^{-3}
$
When the electric field is switched off, let the drop falls with terminal velocity v , then
$
v =\frac{2 r ^{2}(\rho-\sigma) g}{9 \eta} \text { or } r=\left[\frac{9 v \eta}{2(\rho-\sigma) g}\right]^{\frac{1}{2}}
$
$
\therefore q=\frac{1}{E} \times \frac{4}{3} \pi \rho g\left[\frac{9 v \eta}{2(\rho-\sigma) g}\right]
$
$
=\frac{7}{81 \pi \times 10^{5}} \times \frac{4}{3} \times \pi \times 900 \times 9.8 \times\left[\frac{9 \times 8 \times 10^{-5} \times 2 \times 10^{-3}}{2 \times 900 \times 9.8}\right]^{\frac{3}{2}}
$
On solving we get, $q =8 \times 10^{-19} C$