A oil drop having a mass $4.8 imes 10^{-10} \mathrm{~g}$ and charge $2.4 imes 10^{-18} \mathrm{C}$ stands still between two charged horizontal plates separated by a distance of $1 \mathrm{~cm}$. If now the polarity of the plates is changed, instantaneous acceleration of the drop is :…

Question

A oil drop having a mass $4.8 	imes 10^{-10} \mathrm{~g}$ and charge $2.4 	imes 10^{-18} \mathrm{C}$ stands still between two charged horizontal plates separated by a distance of $1 \mathrm{~cm}$. If now the polarity of the plates is changed, instantaneous acceleration of the drop is : $\left(g=10 \mathrm{~ms}^{-2}ight)$

Aicharya · Accepted Answer

$a=\frac{F}{m}=\frac{q E}{m}=\frac{m g}{m}$ $g=10 \mathrm{~m} / \mathrm{s}^2$

A oil drop having a mass $4.8 \times 10^{-10} \mathrm{~g}$ and charge $2.4 \times 10^{-18} \mathrm{C}$…

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