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An oil drop of 12 excess electrons is held stationary under a constant electric field of $2.55 \times 10^4 \mathrm{NC}^{-1}$ in Millikan's oil drop experiment. The density of the oil is $1.26 \mathrm{~g} \mathrm{~cm}^{-3}$. Estimate the radius of the drop $\left(g=9.81 \mathrm{~ms}^{-2} ; e=1.60 \times 10^{-19} C\right)$
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Given, $\mathrm{n}=12, \mathrm{e}=1.6 \times 10^{-9} \mathrm{C}$
$\mathrm{E}=2.55 \times 10^4 \mathrm{Vm}^{-1}$
$\sigma=1.26 \mathrm{~g} \mathrm{~cm}^{-3}=1.26 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$, $\mathrm{g}=9.81 \mathrm{~m} \mathrm{~s}^{-2}$,
Radius of drop, $r=$ ?
Now, downward force of gravity on drop $=$ upward force due to electric field
i.e. $\mathrm{mg}=\mathrm{neE} \quad$ or,$(4 / 3) \pi \mathrm{r}^3 \sigma \mathrm{g}=\mathrm{ne} E$
$$
\begin{aligned}
\text { or, } & r=\left(\frac{3 n \mathrm{neE}}{4 \pi \sigma \mathrm{g}}\right)^{1 / 3} \\
=&\left(\frac{3 \times 12 \times 1.6 \times 10^{-19} \times 2.55 \times 10^4}{4 \times 3.14 \times 1.26 \times 10^3 \times 9.81}\right)^{1 / 3} \\
=9.8175 \times 10^{-7} \mathrm{~m}=9.8175 \times 10^{-4} \mathrm{~mm}
\end{aligned}
$$
$\mathrm{E}=2.55 \times 10^4 \mathrm{Vm}^{-1}$
$\sigma=1.26 \mathrm{~g} \mathrm{~cm}^{-3}=1.26 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$, $\mathrm{g}=9.81 \mathrm{~m} \mathrm{~s}^{-2}$,
Radius of drop, $r=$ ?
Now, downward force of gravity on drop $=$ upward force due to electric field
i.e. $\mathrm{mg}=\mathrm{neE} \quad$ or,$(4 / 3) \pi \mathrm{r}^3 \sigma \mathrm{g}=\mathrm{ne} E$
$$
\begin{aligned}
\text { or, } & r=\left(\frac{3 n \mathrm{neE}}{4 \pi \sigma \mathrm{g}}\right)^{1 / 3} \\
=&\left(\frac{3 \times 12 \times 1.6 \times 10^{-19} \times 2.55 \times 10^4}{4 \times 3.14 \times 1.26 \times 10^3 \times 9.81}\right)^{1 / 3} \\
=9.8175 \times 10^{-7} \mathrm{~m}=9.8175 \times 10^{-4} \mathrm{~mm}
\end{aligned}
$$
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