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Grade: 12
        
A charged water drop of radii 0.1 micrometre is under eqillibrium in some electric field . The charge on the drop is equivalent to electronic charge .the intensity of electric field is
5 months ago

Answers : (2)

Arun
23020 Points
							
mass of water drop = volume of water drop × density of water 
radius of water drop is r = 0.1 micro m = 0.1 × 10^-6 m = 10^-7 m
so, volume of water drop = 4/3 πr³ 
= 4/3 × 3.14 × (10^-7)³ 
= 4 × 1.046 × 10^-21 
= 4.184 × 10^-21 m³ 
 and density of water = 1000 kg/m³ 
so, mass of water drop = 4.184 × 10^-21 × 1000 = 4.184 × 10^-18 kg 
at equilibrium, 
weight of water drop = electrostatic force
or, mg = qE 
here, g = 10m/s² and q = 1.6 × 10^-19C 
or, 4.184 × 10^-18 × 10 = 1.6 × 10^-19 × E
or, 4.184 × 10^-17/1.6 × 10^-19 = E 
or, E = 261.5 N/C
 
 
5 months ago
Khimraj
3008 Points
							

Mass of a drop of water = 4/3  x 3.14 x 0.001 x 10^-18 = 4.18 x 10 ^-18 kg

mg = qE

E    = 4.18 x 10^-18 x 10/1.6 x 10^-19

E    = 2.62 x 100

E    = 262 N/C

4 months ago
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