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A small oil drop of mass $10^{-6} \mathrm{~kg}$ is hanging in at rest between two plates separated by $1 \mathrm{~mm}$ having a potential difference of $500 \mathrm{~V}$. The charge on the drop is $\left(g=10 \mathrm{~ms}^{-2}\right.$ )
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Verified Answer
The correct answer is:
$2 \times 10^{-11} \mathrm{C}$
Given, the drop rests between the two plates
$\begin{aligned}
&\therefore \quad q E=m g \\
&\text { or } \quad q \frac{V}{r}=m g \quad\left(\because E=\frac{V}{r}\right) \\
&\Rightarrow \quad q=\frac{m g r}{V} \\
&\text { here, } m=10^{-6} \mathrm{~kg}, g=10 \mathrm{~m} / \mathrm{s}^{2}, r=1 \mathrm{~mm}=10^{-3} \mathrm{~m} \\
&\text { and } \\
&\text { Substituting all the values, we get } \\
&q=\frac{10^{-6} \times 10 \times 10^{-3}}{500}, q=2 \times 10^{-11} \mathrm{C}
\end{aligned}$
$\begin{aligned}
&\therefore \quad q E=m g \\
&\text { or } \quad q \frac{V}{r}=m g \quad\left(\because E=\frac{V}{r}\right) \\
&\Rightarrow \quad q=\frac{m g r}{V} \\
&\text { here, } m=10^{-6} \mathrm{~kg}, g=10 \mathrm{~m} / \mathrm{s}^{2}, r=1 \mathrm{~mm}=10^{-3} \mathrm{~m} \\
&\text { and } \\
&\text { Substituting all the values, we get } \\
&q=\frac{10^{-6} \times 10 \times 10^{-3}}{500}, q=2 \times 10^{-11} \mathrm{C}
\end{aligned}$
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