Question

A charged water drop of radius 0.1 m is under equilibrium in some electric field. The charge on the drop is equivalent to electronic charge. The intensity of electric field is (g = 10 m/s2 )-

Answer 1

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