Question 18: Electrostats & Capacitance

Question

An oil drop of radius $ 1 \mu \mathrm{~m} $ is held stationary under a constant electric field of $ 3.65 	imes 10^{4} \mathrm{~N} / \mathrm{C} $ due to some excess electrons presence on it. If the density of oil drop is $ 1.26 \mathrm{~g} / \mathrm{cm}^{3} $, then number of excess electrons on the oil drop approximately are [Take, $ g=10 \mathrm{~ms}^{-2} $ ]

Accepted Answer

An oil drop with a radius of $1 \, \mu \mathrm{m}$ is held stationary in a constant electric field of $3.65 	imes 10^{4} \, \mathrm{N/C}$ due to excess electrons on it. The oil drop's density is $1.26 \, \mathrm{g/cm}^{3}$. To find the number of excess electrons on the oil drop, consider the following details:
Convert the given measurements:
Radius of the drop, $ r = 1 \, \mu \mathrm{m} = 10^{-6} \mathrm{~m} $.
Density of the oil, $ho_{	ext{oil}}$, is converted to:
$   ho_{	ext{oil}} = 1.26 \, \mathrm{g/cm}^{3} = 1.26 	imes 10^{3} \, \mathrm{kg/m}^{3}   $
Since the droplet is stationary, the gravitational force acting downward is balanced by the electric force acting upward:
$ 	ext{Weight of the droplet} = 	ext{Force due to the electric field} $
Mathematically, this can be expressed as:
$ \frac{4}{3} \pi r^{3} ho_{	ext{oil}} \cdot g = qE $
where $ q $ is the charge due to excess electrons. If $ n $ is the number of excess electrons, then:
$ q = ne $
Substitute this into the previous equation:
$ \frac{4}{3} \pi r^{3} ho_{	ext{oil}} \cdot g = neE $
Solving for $ n $:
$ n = \frac{4 \pi r^{3} ho_{	ext{oil}} g}{3eE} $
Substitute the known values into the equation to calculate $ n $:
$ \begin{aligned} n & = \frac{4 	imes 3.14 	imes (10^{-6})^{3} 	imes 1.26 	imes 10^{3} 	imes 10}{3 	imes 1.6 	imes 10^{-19} 	imes 3.65 	imes 10^{4}} \ & = 9.03 \approx 9 \end{aligned} $
Thus, there are approximately 9 excess electrons on the oil drop.

Answer

7

Answer

12

Answer

9

Answer

8

Electrostats & Capacitance

Choose the correct answer: