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CBSE Class 12-science Answered

pls solve
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Asked by mufeedatvp2000 | 18 Apr, 2020, 02:24: PM
answered-by-expert 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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