Question

In an oil drop experiment, the terminal velocity of oil droplet was 1 m/sec in absence of electric field and 2 m/s upward in presence of electric field of strength 6.75 x 10 v/m. If d = 1 g/mL and viscosity of air is 2 x 10-5 N-S/m². Calculate: ((i)) radius of oil drop (ii) the charge on each drop ​

Answer 1

Answer:

Explanation:

Answer:

Answer 2

The oil drop experiment is the terminal velocity is oil drop $1 m/sec$ in radius of oil drop.

Explanation:

• Small drops quickly achieve terminal velocity as the downward gravitational force is balanced by the air drag force.
• For small drops, the drag force can be described by Stokes's law, F $drag=6\pi$rⁿv Here $n=1.81*10^{-5} [kg/s m]$ is the viscosity of air, r is the radius of the oil drop, and v is the downwards velocity.
• When the charged drops fell at a constant rate, the gravitational and electric forces on it were equal.