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An oil drop of 12 excess electrons is fi...

An oil drop of 12 excess electrons is field stationary under constant clectric field of `2.55 xx 10^(4) NC^(-1)` in Millikan's oil drop experiment. Then density of the oil is 1.26g `cm^(-3)`. Estimate the radius of the drop. `(g=9.81 ms^(-2), e=1.60 xx 10^(-19)C)`.

Text Solution

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Number of electrons = 12
Electric field = `2.55 xx 10^4Vm^-1`
Density `rho= 1.26 xx 10^3kgm^-3`
As the drop is stationary,
Weight of droplet = Force due to electric field
`4/3pir^3rhog= Eq`
= E r e
`r^3 = (3Ere)/(4pirhog)`
`= (3 xx 2.55 xx10^4xx12xx1.6xx10^-19)/(4xx3.14xx1.26xx10^3xx9.8)`
`r^3 = 0.94xx10^-18`
`:.r= (0.94 xx10^-18)^(1//3)`
`= 9.81 xx 10^-7m`
`:.` Radius of th frop `= 9.81 xx 10^-7m`
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