Electrons

1. Introduction

The electron is a fundamental subatomic particle carrying the smallest negative electric charge. Three classic experiments shaped our understanding of it: J.J. Thomson's measurement of its specific charge (the charge-to-mass ratio $e/m$), Millikan's oil drop experiment which measured the charge $e$ itself and proved that charge is quantized, and the study of how electron beams move in electric and magnetic fields, which underlies devices such as the cathode ray oscilloscope.

The accepted modern values are:

$$ \begin{aligned} e &= 1.6 \times 10^{-19}\ \text{C} \ \qquad \frac{e}{m} &= 1.76 \times 10^{11}\ \text{C kg}^{-1} \ \qquad m &= 9.1 \times 10^{-31}\ \text{kg} \end{aligned} $$

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2. Millikan's Oil Drop Experiment

Aim: to determine the charge of an electron and to show that electric charge is quantized (always an integer multiple of a basic unit $e$).

Principle: a charged oil drop is balanced or made to move between two horizontal plates. By comparing its motion with and without an electric field, and applying Stokes' law of viscous drag, the charge on the drop is found.

Apparatus: two horizontal metal plates $A$ and $B$, about $20\ \text{cm}$ in diameter and separated by about $1.5\ \text{cm}$, form a parallel plate capacitor. The upper plate $A$ has a small hole $H$ through which fine oil droplets (charged by friction during atomization) enter. The region is illuminated and viewed through a microscope.

2.1 Motion with no electric field (finding the radius)

When no field is applied, a drop falls under gravity and quickly reaches a terminal velocity $v_1$, at which the viscous force balances the effective weight (weight minus upthrust of air). Let $r$ be the radius of the drop, $\rho$ the density of oil and $\sigma$ the density of air.

Effective weight of the drop:

$$W = \frac{4}{3}\pi r^3 (\rho - \sigma) g$$

By Stokes' law the viscous drag at terminal velocity $v_1$ is $6\pi \eta r v_1$, where $\eta$ is the viscosity of air. At terminal velocity:

$$6\pi \eta r v_1 = \frac{4}{3}\pi r^3 (\rho - \sigma) g$$

Solving for the radius:

$$r = \sqrt{\frac{9\eta v_1}{2(\rho - \sigma) g}} \qquad \text{(1)}$$

2.2 Motion with the electric field (finding the charge)

A strong electric field $E$ is now applied so that the electric force lifts the negatively charged drop upward. The drop then attains a new terminal velocity $v_2$ in the upward direction. The upward electric force $qE$ is balanced by the effective weight (downward) plus the viscous drag (now downward, opposing the upward motion):

$$qE = \frac{4}{3}\pi r^3 (\rho - \sigma) g + 6\pi \eta r v_2$$

Using the field-free balance, $\tfrac{4}{3}\pi r^3(\rho-\sigma)g = 6\pi\eta r v_1$, so:

$$ \begin{aligned} qE &= 6\pi \eta r v_1 + 6\pi \eta r v_2 \ &= 6\pi \eta r (v_1 + v_2) \ \boxed{q = \frac{6\pi \eta r (v_1 + v_2)}{E}} \end{aligned} $$

Substituting $r$ from equation (1) gives $q$ in terms of measurable quantities.