An oil drop of 12 excess electrons is held st | vedclass

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Question

Excess electrons on an oil drop, $n=12$

Electric field intensity, $E =2.55 	imes 10^{4} \,N \,C ^{-1}$

Density of oil, $ho=1.26 \,gm / cm ^{3

Vedclass · Accepted Answer

$9.82 	imes 10^{-4} \;mm$.

Vedclass · Answer

Excess electrons on an oil drop, $n=12$

Electric field intensity, $E =2.55 	imes 10^{4} \,N \,C ^{-1}$

Density of oil, $ho=1.26 \,gm / cm ^{3}=1.26 	imes 10^{3} \,kg / m ^{3}$

Acceleration due to gravity, $g=9.81 \,m\, s ^{-2}$

Charge on an electron, $e=1.6 	imes 10^{-19} \,C$

Radius of the oil drop $= r$

Force $(F)$ due to electric field $E$ is equal to the weight of the oil drop $(W)$

$F = W \Rightarrow E q=m g \Rightarrow$ $Ene$ $=\frac{4}{3} \pi r^{3} 	imes ho 	imes g$

Where, $q =$ Net charge on the oil drop $= ne$ $m =$ Mass of the oil drop $=$ Volume of the drop $	imes$ Density of oil $=\frac{4}{3} \pi r^{3} 	imes ho$

$r=\left[\frac{3 E n e}{4 \pi ho g}ight]^{\frac{1}{3}}=\left[\frac{3 	imes 2.55 	imes 10^{4} 	imes 12 	imes 1.6 	imes 10^{-19}}{4 	imes 3.14 	imes 1.26 	imes 10^{3} 	imes 9.81}ight]^{\frac{1}{3}}$

$=\left[946.09 	imes 10^{-21}ight]^{\frac{1}{3}}=9.82 	imes 10^{-7} \,mm$

Therefore, the radius of the oil drop is $9.82 	imes 10^{-4} \;mm$.

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