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A charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20000 \mathrm{~V} / \mathrm{m}$. If mass of the particle is $9.6 \times 10^{-16} \mathrm{~kg}$, the charge on it and excess number of electrons on the-particle respectively are $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$
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Verified Answer
The correct answer is:
$4.8 \times 10^{-19} \mathrm{C}, 3$
Charge on the particle is given by
$\begin{aligned}q E & =m g \text { or } q=\frac{m g}{E} \\& =\frac{9.6 \times 10^{-16} \times 10}{20000}=4.8 \times 10^{-19}
\end{aligned}$
When there is an excess of $n$ electrons on the particle, then
$q=n e$
So,
$n=\frac{q}{e}=\frac{4.8 \times 10^{-19}}{1.6 \times 10^{-19}}=3$
$\begin{aligned}q E & =m g \text { or } q=\frac{m g}{E} \\& =\frac{9.6 \times 10^{-16} \times 10}{20000}=4.8 \times 10^{-19}
\end{aligned}$
When there is an excess of $n$ electrons on the particle, then
$q=n e$
So,
$n=\frac{q}{e}=\frac{4.8 \times 10^{-19}}{1.6 \times 10^{-19}}=3$
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