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A tiny spherical oil drop carrying a net charge $q$ is balanced in still air, with a vertical uniform electric field of strèngth $\frac{81}{7} \pi \times 10^5 \mathrm{~V} / \mathrm{m}$. When the field is switched OFF, the drop is observed to fall with terminal velocity $2 \times 10^{-3} \mathrm{~ms}^{-1}$. Here $g=9.8 \mathrm{~m} / \mathrm{s}^2$, viscosity of air is $1.8 \times 10^{-5} \mathrm{Ns} / \mathrm{m}^2$ and density of oil is $900 \mathrm{~kg}$ $\mathrm{m}^{-3}$. The magnitude of $q$ is
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Verified Answer
The correct answer is:
$8 \times 10^{-19} \mathrm{C}$
Given, $E=\frac{81 \pi}{7} \times 10^5 \mathrm{~V} / \mathrm{m}$
Terminal velocity , $v=2 \times 10^{-3} \mathrm{~m} / \mathrm{s}$
$$
g=9.8 \mathrm{~m} / \mathrm{s}^2
$$
Viscosity, $\eta=1.8 \times 10^{-5} \mathrm{~N}-\mathrm{s} / \mathrm{m}^2$
Density, $\rho=900 \mathrm{~kg} / \mathrm{m}^3$
'Since, $q E=m g$
In the absence of electric field,
$$
\begin{array}{rlrl}
& & m g & =6 \pi \eta r v \\
\Rightarrow & & q E & =6 \pi \eta r \nu \\
\Rightarrow & & r=\frac{q E}{6 \pi \eta v}
\end{array}
$$
[from Eq. (i)]
From Eq. (i), we get
$$
\begin{gathered}
m=\frac{q E}{g} \\
\Rightarrow \frac{4}{3} \pi r^3 d=\frac{q E}{g} \Rightarrow \frac{4}{3} \pi\left(\frac{q E}{6 \pi \eta v}\right)^3 d=\frac{q E}{g} \\
\Rightarrow \quad q=\sqrt{\frac{3 \times 6^3 \pi^2 \eta^3 v^3}{4 E^2 g}}
\end{gathered}
$$
$$
\begin{aligned}
& =\sqrt{\frac{3 \times 6^3 \times(314)^2 \times\left(1.8 \times 10^{-5}\right)^3 \times\left(2 \times 10^{-3}\right)^3}{4 \times\left(\frac{81 \pi}{7} \times 10^5\right)^2 \times 9.8}} \\
& =8 \times 10^{-19} \mathrm{C}
\end{aligned}
$$
Terminal velocity , $v=2 \times 10^{-3} \mathrm{~m} / \mathrm{s}$
$$
g=9.8 \mathrm{~m} / \mathrm{s}^2
$$
Viscosity, $\eta=1.8 \times 10^{-5} \mathrm{~N}-\mathrm{s} / \mathrm{m}^2$
Density, $\rho=900 \mathrm{~kg} / \mathrm{m}^3$
'Since, $q E=m g$
In the absence of electric field,
$$
\begin{array}{rlrl}
& & m g & =6 \pi \eta r v \\
\Rightarrow & & q E & =6 \pi \eta r \nu \\
\Rightarrow & & r=\frac{q E}{6 \pi \eta v}
\end{array}
$$
[from Eq. (i)]
From Eq. (i), we get
$$
\begin{gathered}
m=\frac{q E}{g} \\
\Rightarrow \frac{4}{3} \pi r^3 d=\frac{q E}{g} \Rightarrow \frac{4}{3} \pi\left(\frac{q E}{6 \pi \eta v}\right)^3 d=\frac{q E}{g} \\
\Rightarrow \quad q=\sqrt{\frac{3 \times 6^3 \pi^2 \eta^3 v^3}{4 E^2 g}}
\end{gathered}
$$
$$
\begin{aligned}
& =\sqrt{\frac{3 \times 6^3 \times(314)^2 \times\left(1.8 \times 10^{-5}\right)^3 \times\left(2 \times 10^{-3}\right)^3}{4 \times\left(\frac{81 \pi}{7} \times 10^5\right)^2 \times 9.8}} \\
& =8 \times 10^{-19} \mathrm{C}
\end{aligned}
$$
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