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`        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

```							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
```
5 months ago
```							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
```
4 months ago
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