Question

An oil drop of 10 excess electrons is held stationary under a constant electric field of 3.14 × 104 N/C. The density of oil is 0.80 g/cm3. If g = 10 m/s2 and e = 1.6 × 10­–19 C, then radius of the drop will be nearly

Answer 1

Given:

An oil drop of 10 excess electrons is held stationary under a constant electric field of 3.14 × 104 N/C. The density of oil is 0.80 g/cm3.  

To find:

If g = 10 m/s2 and e = 1.6 × 10­–19 C, then radius of the drop will be nearly

Solution:

As the droplet is stationary, the weight of droplet will be equal to the force due to the electric field

so, we have,

(4/3)πr³ρg = neE

r³ = 3neE/4πρg  

From given, we have,

n = 10, e = 1.6 × 10­–19, E = 3.14 × 10^4, ρ = 0.80 = 0.80 × 10^3 kg/m^3   and g = 10

substituting all these values in the equation, we get,  

r³ = [3 × 10 × 1.6 × 10­–19 × 3.14 × 10^4] / [4 × 3.14 × 0.80 × 10^3 × 10]

r³ = 1.5072 × 10­–13 / 100480

r³ = 1.5 × 10­–18

r = 1.14 × 10­–6

∴ The radius of the drop will be nearly 1.14 × 10­–6 m