Question

In an experiment replicating Millikan’s oil drop experiment, a pair of parallel plates are placed 0.0200 m apart and the top plate is positive. When the potential difference across the plates is 240.0 V, an oil drop of mass 2.0 × 10-11 kg gets suspended between the plates. (e = 1.6 × 10-19 C)
a) Draw a free-body diagram for the charge.

b) What is the charge on the oil drop?

c) Is there an excess or deficit of electrons on the oil drop? How many electrons are in excess or deficit?

Answer 1

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,

$E=\frac{V}{d}$

On applying force balance, the force on oil drop is equal to the weight of the oil,

$\begin{aligned}F_{E} &=m g \\q E &=m g \\q \frac{V}{d} &=m g \\q &=\frac{m g d}{V}\end{aligned}$

Substituting the given values in the above equation,

$\begin{aligned}&q=\frac{2 \times 10^{-11} \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2} \times \frac{1 \mathrm{~N}}{1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}} \times 0.02 \mathrm{~m}}{240 \mathrm{~V} \times \frac{1 \mathrm{~N} \cdot \mathrm{m} / \mathrm{C}}{1 \mathrm{~V}}} \\&q=1.63 \times 10^{-14} \mathrm{C}\end{aligned}$

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,

$\begin{aligned}q &=n e \\1.63 \times 10^{-14} \mathrm{C} &=n \times 1.6 \times 10^{-19} \mathrm{C} \\n &=1.01 \times 10^{5}\end{aligned}$

Therefore, the number of excess electrons is 1.01x10^5