The intensity of the electric field required | vedclass

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

Question

(B) For the water drop to remain suspended in air,the electric force must balance the gravitational force.$F_e = mg$$eE = mg$$E = \frac{mg}{e}$Here,$m

Vedclass · Accepted Answer

$260 \, N/C$

Vedclass · Answer

(B) For the water drop to remain suspended in air,the electric force must balance the gravitational force.$F_e = mg$$eE = mg$$E = \frac{mg}{e}$Here,$m = 	ext{volume} 	imes 	ext{density} = \frac{4}{3} \pi r^3 ho$.Given: $r = 10^{-5} \, cm = 10^{-7} \, m$,$ho = 1000 \, kg/m^3$,$e = 1.6 	imes 10^{-19} \, C$,$g = 10 \, m/s^2$.$m = \frac{4}{3} 	imes 3.14 	imes (10^{-7})^3 	imes 1000 = \frac{4}{3} 	imes 3.14 	imes 10^{-18} \, kg$.$E = \frac{\frac{4}{3} 	imes 3.14 	imes 10^{-18} 	imes 10}{1.6 	imes 10^{-19}} \approx 261.6 \, N/C$.Rounding to the nearest provided option,the correct value is $260 \, N/C$.

The intensity of the electric field required to keep a water drop of radius $10^{-5} \, cm$ and carrying a charge of $1$ electron suspended in air is:

Similar Questions

$A$ proton and an $\alpha$-particle having equal kinetic energy are projected into a uniform transverse electric field as shown in the figure.

$A$ proton and an $\alpha$-particle are both accelerated from rest in a uniform electric field. The ratio of work done by the electric field on the proton and the $\alpha$-particle in a given time is

$A$ drop of $10^{-6} \ kg$ water carries $10^{-6} \ C$ charge. What electric field should be applied to balance its weight? (Assume $g = 10 \ m/s^2$)

Select the correct statement: (Only force on a particle is due to electric field)

An electron,initially at rest,is accelerated through a potential difference of $200 \ V$,so that it acquires a velocity of $8.4 \times 10^6 \ m/s$. The value of $e/m$ for the electron will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module