Q. Two metal plates having a potential difference of $800V$ are $2cm$ apart. It is found that a particle of mass $1.96\times 10^{- 15}kg$ remain suspended in the region between the plates. The charge on the particle must be $\left(\right.e=$ elementary charge $\left.$

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A

$3e$

B

$4e$

C

$6e$

D

$8e$

Solution:

$qE=mg$
$E=\frac{V}{d}=\frac{800}{2 \times 10^{- 2}}=4\times 10^{4}$
$q=\frac{mg}{E}=\frac{1 . 96 \times 10^{- 15} \times 9 . 8}{4 \times 10^{4}}$
$q=4.8\times 10^{- 19}=3e$

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