A liquid drop having 6 excess electrons is ke | vedclass

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Question

$\mathrm{F}=\mathrm{qE}=\mathrm{mg}\left(\mathrm{q}=6 \mathrm{e}=6 	imes 1.6 	imes 10^{-19}ight)$

Density $(\mathrm{d})=\frac{	ext { mass }}{\

Vedclass · Accepted Answer

$7.8	imes10^{-7}\, m$

Vedclass · Answer

$\mathrm{F}=\mathrm{qE}=\mathrm{mg}\left(\mathrm{q}=6 \mathrm{e}=6 	imes 1.6 	imes 10^{-19}ight)$

Density $(\mathrm{d})=\frac{	ext { mass }}{	ext { volume }}=\frac{\mathrm{m}}{\frac{4}{3} \pi r^{3}}$

or $r^{3}=\frac{m}{\frac{4}{3} \pi d}$

Putting the value of $\mathrm{d}$ and $\mathrm{m}\left(=\frac{\mathrm{qE}}{\mathrm{g}}ight)$ and

solving we get $r=7.8 	imes 10^{-7} \,\mathrm{m}$

A liquid drop having $6$ excess electrons is kept stationary under a uniform electric field of $25.5\, k\,Vm^{-1}$ . The density of liquid is $1.26\times10^3\, kg\, m^{-3}$ . The radius of the drop is (neglect buoyancy)

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