n Millikan oil drop experiment, a charged dro | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(c) $QE = mg$

$ \Rightarrow Q = \frac{{mg}}{E}$

$ \Rightarrow n = \frac{{mgd}}{{Ve}}$

$ \Rightarrow n = \frac{{1.8 	imes {{10}^{ - 14}} 	im

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(c) $QE = mg$

$ \Rightarrow Q = \frac{{mg}}{E}$

$ \Rightarrow n = \frac{{mgd}}{{Ve}}$

$ \Rightarrow n = \frac{{1.8 	imes {{10}^{ - 14}} 	imes 10 	imes 0.9 	imes {{10}^{ - 2}}}}{{2 	imes {{10}^3} 	imes 1.6 	imes {{10}^{ - 19}}}} = 5$

$n$ Millikan oil drop experiment, a charged drop of mass $1.8 \times {10^{ - 14}}kg $ is stationary between its plates. The distance between its plates is $0.90 cm$ and potential difference is $2.0$ kilo volts. The number of electrons on the drop is

Similar Questions

The fact that electric charges are integral multiples of the fundamental electronic charge was proved experimentally by

An oxide coated filament is useful in vacuum tubes because essentially

The cathode rays have particle nature because of the fact that

The current conduction in a discharged tube is due to

What is the unit of energy in atomic and nuclear physics ? Define it.