Two large circular discs separated by a dista | vedclass

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

Question

(A) The electric field $E$ between the plates is given by $E = \frac{V}{d} = \frac{200}{0.01} = 2 	imes 10^4 \ V/m$.For the oil drop to float,the ele

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) The electric field $E$ between the plates is given by $E = \frac{V}{d} = \frac{200}{0.01} = 2 	imes 10^4 \ V/m$.For the oil drop to float,the electric force must balance the gravitational force: $qE = mg$.Here,$q = ne$,where $n$ is the number of electrons and $e = 1.6 	imes 10^{-19} \ C$.The mass of the spherical oil drop is $m = ho V_{drop} = ho \left( \frac{4}{3} \pi r^3 ight)$.Substituting the values: $n 	imes (1.6 	imes 10^{-19}) 	imes (2 	imes 10^4) = 900 	imes \frac{4}{3} 	imes 3.14 	imes (8 	imes 10^{-7})^3 	imes 10$.$n 	imes 3.2 	imes 10^{-15} = 1200 	imes 3.14 	imes 512 	imes 10^{-21} 	imes 10$.$n 	imes 3.2 	imes 10^{-15} = 1.93 	imes 10^{-14}$.$n = \frac{1.93 	imes 10^{-14}}{3.2 	imes 10^{-15}} \approx 6.03$.Thus,the number of electrons is $6$.

Similar Questions

Electric field in a certain region is given by $\overrightarrow{E} = (\frac{A}{x^2} \hat{i} + \frac{B}{y^3} \hat{j})$. The $SI$ units of $A$ and $B$ are:

Two equal negative charges $-q$ are fixed at points $(0, -a)$ and $(0, a)$ in the $x-y$ plane. $A$ positive charge $Q$ is released from rest at a point $(2a, 0)$. The charge $Q$ will

Two charges $+5\,\mu C$ and $+10\,\mu C$ are placed $20\,cm$ apart. The net electric field at the midpoint between the two charges is

Vedclass Test Series

Exam Paper Generator

Online Exam Module