Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
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
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 mso, 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 forceor, mg = qE here, g = 10m/s² and q = 1.6 × 10^-19C or, 4.184 × 10^-18 × 10 = 1.6 × 10^-19 × Eor, 4.184 × 10^-17/1.6 × 10^-19 = E or, E = 261.5 N/C
Mass of a drop of water = 4/3 x 3.14 x 0.001 x 10^-18 = 4.18 x 10 ^-18 kgmg = qEE = 4.18 x 10^-18 x 10/1.6 x 10^-19E = 2.62 x 100E = 262 N/C
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
Find solution of question on doubt solving app Instasolv
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -