Question 24: Electrostats & Capacitance

Question

A dust particle of mass $ 4 	imes 10^{-12} \mathrm{mg} $ in suspended in air under the influence of an electric field of $ 50 \mathrm{~N} / \mathrm{C} $ directed vertically upwards. How many electrons were removed from the neutral dust particle? [Take, $ g=10 \mathrm{~m} / \mathrm{s} $ $ { }^{2} $ ]

Accepted Answer

To determine how many electrons have been removed from a neutral dust particle suspended in an electric field, let's go through the calculations step by step.
Given:
Mass of the dust particle, $ m = 4 	imes 10^{-12} 	ext{ mg} $
Conversion to kilograms:
$  m = 4 	imes 10^{-12} 	imes 10^{-3} 	ext{ g} = 4 	imes 10^{-12} 	imes 10^{-3} 	imes 10^{-3} 	ext{ kg} = 4 	imes 10^{-18} 	ext{ kg}  $
Electric field strength, $ E = 50 	ext{ N/C} $
Gravitational acceleration, $ g = 10 	ext{ m/s}^2 $
Calculate the Weight:
The weight $ W $ of the dust particle is given by:
$  W = m 	imes g = 4 	imes 10^{-18} 	imes 10 = 4 	imes 10^{-17} 	ext{ N}  $
Equating Electric Force to Weight:
The electric force $ F_e $ experienced by the dust particle is:
$  F_e = q 	imes E  $
Where $ q = n 	imes e $, with $ e = 1.6 	imes 10^{-19} 	ext{ C} $ as the charge of one electron, and $ n $ being the number of electrons removed.
At equilibrium, the electric force equals the weight of the particle:
$  n 	imes e 	imes E = W  $
$  n 	imes 1.6 	imes 10^{-19} 	imes 50 = 4 	imes 10^{-17}  $
Solving for $ n $:
$  n = \frac{4 	imes 10^{-17}}{1.6 	imes 10^{-19} 	imes 50}  $
$  n = \frac{400}{80} = 5  $
Therefore, 5 electrons were removed from the neutral dust particle.

Answer

15

Answer

8

Answer

5

Answer

4

Electrostats & Capacitance

A dust particle of mass 4 \times 10^{-12} \mathrm{mg} in suspended in air under the influence of an electric field of 50 \mathrm{~N} / \mathrm{C} directed vertically upwards. How many electrons were removed from the neutral dust particle? [Take, g=10 \mathrm{~m} / \mathrm{s} { }^{2} ]

Choose the correct answer: