Electrostatics Ques 191
- A tiny spherical oil drop carrying a net charge $q$ is balanced in still air with a vertical uniform electric field of strength $\frac{81 \pi}{7} \times 10^{5} \mathrm{Vm}^{-1}$. When the field is switched off, the drop is observed to fall with terminal velocity $2 \times 10^{-3} \mathrm{~ms}^{-1}$. Given $g=9.8 \mathrm{~ms}^{-2}$, viscosity of the air $=1.8 \times 10^{-5} \mathrm{Ns} \mathrm{m}^{-2}$ and the density of oil $=900 \mathrm{~kg} \mathrm{~m}^{-3}$, the magnitude of $q$ is
(a) $1.6 \times 10^{-19} \mathrm{C}$
(b) $3.2 \times 10^{-19} \mathrm{C}$
(c) $4.8 \times 10^{-19} \mathrm{C}$
(d) $8.0 \times 10^{-19} \mathrm{C}$
(2010)
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Answer:
Correct Answer: 191.(d)
Solution:
Formula:
Coulomb Force Between Two Point Charges:
$$ \begin{aligned} q E & =m g \cdots(i)\\ 6 \pi \eta r v & =m g \\ \frac{4}{3} \pi r^{3} \rho g & =m g \cdots(ii) \end{aligned} $$
$$ \therefore \quad r=\frac{3 m g}{4 \pi \rho g}^{1 / 3} \cdots(iii) $$
Substituting the value of $r$ in Eq. (ii) we get,
$$ \begin{aligned} & 6 \pi \eta \nu \frac{3 m g}{4 \pi \rho g}^{1 / 3}=m g \\ & (6 \pi \eta v)^{3} \frac{3 m g}{4 \pi \rho g}=(m g)^{3} \end{aligned} $$
Again substituting $m g=q E$ we get,
$$ \begin{aligned} & (q E)^{2}=\frac{3}{4 \pi \rho g}(6 \pi \eta v)^{3} \\ & \text { or } \quad q E=\frac{3}{4 \pi \rho g}^{1 / 2}(6 \pi \eta v)^{3 / 2} \\ & \therefore \quad q=\frac{1}{E} \frac{3}{4 \pi \rho g}^{1 / 2}(6 \pi \eta v)^{3 / 2} \end{aligned} $$
Substituting the values we get,
$$ \begin{aligned} q & =\frac{7}{81 \pi \times 10^{5}} \sqrt{\frac{3}{4 \pi \times 900 \times 9.8} \times 216 \pi^{3}} \\ & \times \sqrt{\left(1.8 \times 10^{-5} \times 2 \times 10^{-3}\right)^{3}} \\ & =8.0 \times 10^{-19} \mathrm{C} \end{aligned} $$
BETA
