CBSE Class 12-science Answered
pls solve
Asked by mufeedatvp2000 | 18 Apr, 2020, 02:24: PM
Expert Answer
Forces acting on the ol drop are
(1) downward force due to weight of drop and bouyancy of air V( ρ - ρo ) g
and
(2) electrostatic force qE due to electric field E
where V = (4/3)πr3 is volume of oil drop, ρ is density of oil, ρo is density of air , g is acceleration due to gravity
and q is charge on oil drop
At equilibrium, qE = (4/3)πr3 ( ρ - ρo ) g
Hence charge q = [ 4 π r3 ( ρ - ρo ) g ] / ( 3 E )
To get electrostatic force upward , charge q should be +ve .
If distributed charge q is collection of n fundamental charges e, then q = + ( n e )
we have considered that magnitude of each fundametal charge is magnitude of charge of electron
number of fundamental charges on oil drop , n = [ 4 πr3 ( ρ - ρo ) g ] / ( 3 e E )
Since charge on oil drop is +ve, we write the above equation as ,
deficiency of electrons = [ 4 π r3 ( ρ - ρo ) g ] / ( 3 e E )
Answered by Thiyagarajan K | 19 Apr, 2020, 08:29: AM
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