Answer: See below
Explanation:
Given:
The potential between plates, V = 240 V
Distance between plates, d = 0.02 m
The mass of drop, m = 2x10^-11
Charge on electron, e = 1.6x10^-19
Part (a)
The free-body diagram is attached below
Part (b)
The electric field is given by,
[tex]E=\frac{V}{d}[/tex]
On applying force balance, the force on oil drop is equal to the weight of the oil,
[tex]$\begin{aligned}F_{E} &=m g \\q E &=m g \\q \frac{V}{d} &=m g \\q &=\frac{m g d}{V}\end{aligned}$[/tex]
Substituting the given values in the above equation,
[tex]\begin{aligned}&q=\frac{2 \times 10^{-11} \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2} \times \frac{1 \mathrm{~N}}{1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}} \times 0.02 \mathrm{~m}}{240 \mathrm{~V} \times \frac{1 \mathrm{~N} \cdot \mathrm{m} / \mathrm{C}}{1 \mathrm{~V}}} \\&q=1.63 \times 10^{-14} \mathrm{C}\end{aligned}[/tex]
Therefore, the charge on the oil drop is 1.63x10^-14 C
Part (c)
There will be an excess of electrons on the oil drop.
The number of electrons on oil drop can be calculated as,
[tex]\begin{aligned}q &=n e \\1.63 \times 10^{-14} \mathrm{C} &=n \times 1.6 \times 10^{-19} \mathrm{C} \\n &=1.01 \times 10^{5}\end{aligned}[/tex]
Therefore, the number of excess electrons is 1.01x10^5
