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An oil drop of 10 excess electron is hel...

An oil drop of 10 excess electron is held stationary under a consatnt electric field of `3.6xx10^(4)NC^(-1)` in Millikan's oil drop experiment. The density of oil is `1.26gcm^(-3)`. Radius of the oil drop is
(Take, `g=9.8ms^(-2), e=1.6xx10^(-19)C`)

A

`1.04xx10^(-6)m`

B

`4.8xx10^(-5)m`

C

`4.8xx10^(-18)m`

D

`1.13xx10^(-18)m`

Text Solution

Verified by Experts

The correct Answer is:
A
Here, n = 10, `E=3.65xx10^(4)NC^(-1)`
`rho_(oil)=1.26gcm^(-3)=1.26xx10^(3)kgm^(-3)`
As the droplet is stationary,
weight of droplet = force due to electric field
`or" "4/3pir^(3)rhog= "ne"E" "therefore" "r^(3)=(3"ne"E)/(4pirhog)`
`therefore" "r^(3)=(3xx10xx1.6xx10^(-19)xx3.65xx10^(4))/(4xx3.14xx1.26xx10^(3)xx9.8)=1.13xx10^(-18)`
`or" "r=(1.13xx10^(-18))^(1//3)=1.04xx10^(-6)m`
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Knowledge Check

  • An oil drop of radius r carrying a charge q remain stationary in the presence of electric field of intensity E. If the density of oil rho , then

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    D
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