Question

1. a chemist from a parallel universe, where electrons have a different charge than they do in our universe, performs the millikan oil drop experiment to measure the electron’s charge. the charges of several drops are recorded here. what is the charge of the electron in this parallel universe?

drop # \t charge
a \t -6.9×10⁻¹⁹ c
b \t -9.2×10⁻¹⁹ c
c \t -11.5×10⁻¹⁹ c
d \t -4.6×10⁻¹⁹ c

Explanation:

Step1: Find differences between charges

Calculate the differences between the charges of the drops.

• Between Drop A ($-6.9\times10^{-19}\,C$) and Drop D ($-4.6\times10^{-19}\,C$): $|-6.9\times10^{-19} - (-4.6\times10^{-19})| = |-2.3\times10^{-19}| = 2.3\times10^{-19}\,C$
• Between Drop B ($-9.2\times10^{-19}\,C$) and Drop A ($-6.9\times10^{-19}\,C$): $|-9.2\times10^{-19} - (-6.9\times10^{-19})| = |-2.3\times10^{-19}| = 2.3\times10^{-19}\,C$
• Between Drop C ($-11.5\times10^{-19}\,C$) and Drop B ($-9.2\times10^{-19}\,C$): $|-11.5\times10^{-19} - (-9.2\times10^{-19})| = |-2.3\times10^{-19}| = 2.3\times10^{-19}\,C$

Step2: Identify the electron charge

The charge of an electron should be the greatest common divisor (GCD) of the charges of the oil drops (since each drop has a multiple of the electron's charge). From the differences, we see that the common value is $2.3\times10^{-19}\,C$ (or we can check that each charge is a multiple of $2.3\times10^{-19}\,C$:

• Drop D: $\frac{-4.6\times10^{-19}}{-2.3\times10^{-19}} = 2$ (2 electrons)
• Drop A: $\frac{-6.9\times10^{-19}}{-2.3\times10^{-19}} = 3$ (3 electrons)
• Drop B: $\frac{-9.2\times10^{-19}}{-2.3\times10^{-19}} = 4$ (4 electrons)
• Drop C: $\frac{-11.5\times10^{-19}}{-2.3\times10^{-19}} = 5$ (5 electrons)

So the charge of the electron is the magnitude of this common divisor (and negative, as electrons are negative).

Answer:

The charge of the electron in this parallel universe is $\boldsymbol{-2.3\times10^{-19}\,C}$ (or $2.3\times10^{-19}\,C$ with a negative sign indicating negative charge).