Claimed by Spencer Boebel (Fall 2023)

## Contents

- 1 Coulomb's Law
- 1.1 Overview
- 1.2 Main Idea
- 1.3 Example
- 1.4 Connectedness
- 1.5 History
- 1.6 See Also
- 1.7 References

## Overview

Coulomb's Law states that [math]\displaystyle{ \vec{F}_{2on1} = {\frac{q_{1}q_{2}}{4\pi\epsilon_{0}|\vec{r}_{12}|^2}}\hat{r}_{12} }[/math] , where [math]\displaystyle{ \vec{F}_{2on1} }[/math] is the force on charge [math]\displaystyle{ q_{1} }[/math] by charge [math]\displaystyle{ q_{2} }[/math], [math]\displaystyle{ |\vec{r}_12| }[/math] is the distance between the charges, and [math]\displaystyle{ \hat{r}_{12} }[/math] is the unit vector pointing from [math]\displaystyle{ q_{1} }[/math] to [math]\displaystyle{ q_{2} }[/math]. [math]\displaystyle{ \epsilon_{0} }[/math] is the electric permittivity of free space. In words, it says that the force on one charged particle ([math]\displaystyle{ A }[/math]) by another charged particle ([math]\displaystyle{ B }[/math]) is directed at [math]\displaystyle{ A }[/math], is jointly proportional to the charges themselves (signs included), and is inversely proportional to the distance between [math]\displaystyle{ A }[/math] and [math]\displaystyle{ B }[/math] squared. Coulomb's Law holds only when the two particles are at rest, and it can be derived from Gauss's law, which is one of Maxwell's Equations. Coulomb's Law can be used to analyze electrostatic situations such as the movement of a charge suspended between two plates as well as determining the electric field from a given charge distribution.

## Main Idea

**A Mathematical Model**

According to [1], [math]\displaystyle{ \vec{F}_{on1} = {\frac{q_{1}q_{2}}{4\pi\epsilon_{0}|\vec{r}_{12}|^2}}\hat{r}_{12} }[/math]. [math]\displaystyle{ \epsilon_{0} }[/math], the electric permittivity of free space, has a value of [math]\displaystyle{ \epsilon_{0} = 8.8541878128\times10^-12 }[/math] F/m according to [2]. [math]\displaystyle{ \frac{1}{4\pi\epsilon_{0}} }[/math] can be conveniently approximated as [math]\displaystyle{ 9\times10^9 }[/math], as is done in [1]. Thus Coulomb's Law is sometimes written as [math]\displaystyle{ \vec{F}_{on1} = {\frac{kq_{1}q_{2}}{|\vec{r}_{12}|^2}}\hat{r}_{12} }[/math], where [math]\displaystyle{ k = 9\times10^9 }[/math]. Moreover, if we define the *electric field* [math]\displaystyle{ \vec{E} }[/math] as the force felt per unit charge as a result of Coulomb's Law, then we can write Coulomb's Law as [math]\displaystyle{ \vec{E} = \frac{q}{4\pi\epsilon_{0}|\vec{r}|^2}\hat{r} }[/math], where [math]\displaystyle{ \vec{r} }[/math] is the vector distance between the charge and the point in space at which the electric field is being measured. Also, if we have many charges [math]\displaystyle{ q_{1},...,q_{j} }[/math], we can write Coulomb's Law as [math]\displaystyle{ \vec{E} = \sum_{n=0}^{j} {\frac{q_{n}}{4\pi\epsilon_{0}|\vec{r}_{n}|^2}}\hat{r}_{n} }[/math], where [math]\displaystyle{ \vec{r}_{n} }[/math] is the distance to the [math]\displaystyle{ n }[/math]th charge. A unit of charge is called a Coulomb, and all charges are measured in Coulombs. An electron has [math]\displaystyle{ 1.602\times10^-19 }[/math] C of charge.

**A Computational Model**

Here is a model of an electron orbiting a proton using Coulomb's law. To see the electron orbit the proton, click the "Run Program" button at the top of the linked page.

## Example

**Simple**

Given charges [math]\displaystyle{ q_{1} = 5 }[/math] C located at [math]\displaystyle{ \vec{r}_{1} = \lt 1,-1,0\gt }[/math] m and [math]\displaystyle{ q_{2} = -10 }[/math] C at location [math]\displaystyle{ \vec{r}_{2} = \lt 5,2,0\gt }[/math] m, find the force on charge 2 by charge 1.

*Solution.* Find the vector distance between the two charges [math]\displaystyle{ \vec{r}_{12} = \vec{r}_{2} - \vec{r}_{1} = \lt 4,3,0\gt }[/math] m. Find the magnitude [math]\displaystyle{ |\vec{r}_{12}| = 5 }[/math] and unit vector [math]\displaystyle{ \hat{r}_{12} = \frac{\vec{r}_{12}}{|\vec{r}_{12}|} = \lt \frac{4}{5},\frac{3}{5},0\gt }[/math] m. Plug into Coulomb's Law [math]\displaystyle{ \vec{F}_{2on1} = {\frac{kq_{1}q_{2}}{|\vec{r}_{12}|^2}}\hat{r}_{12} }[/math], we get [math]\displaystyle{ \vec{F}_{2on1} = {\frac{(9\times10^9)\times(-10)\times(5)}{25}}\lt \frac{4}{5},\frac{3}{5},0\gt = \lt -1.44\times10^{10},-1.08\times10^{10},0\gt }[/math] N.

**Middling**

You shake an inflatable party balloon on your head in order to increase the static electricity on the balloon and impress your friends when your hair sticks up while the balloon is near it. This happens because some electrons in your hair get transferred to the balloon during the shaking. You have 0.25 kg of hair. If the balloon had no charge before the shaking, you had no charge before the shaking, and the distance between the balloon and your head is 10 cm, and all of your hair is stuck up, approximately how many electrons did your hair lose? You may model the balloon as a single charge and the hair as a point mass.

*Solution.* Since your hair is suspended in midair, as opposed to sitting on your head, the balloon must be exerting a force on your hair equivalent to the weight of your hair. The magnitude of this force is [math]\displaystyle{ m_{hair}g = 0.25\times9.81 = 2.45 }[/math]N. It is exerting this force on the positive charges that are left in your hair, and since you had no charge before the shaking, the balloon electrons' aggregate charge (let's call it [math]\displaystyle{ q_{2} }[/math]) is equal to the aggregate positive charges on your hair (if the positive charges are [math]\displaystyle{ q_{2} }[/math], then [math]\displaystyle{ q_{1} = q_{2} }[/math]). Thus plugging into Coulomb's law we get [math]\displaystyle{ {\frac{(9\times10^9)q_{1}^2}{(0.1)^2}}\lt 0,1,0\gt = 2.45\lt 0,1,0\gt }[/math]. This implies [math]\displaystyle{ q_{1} = \sqrt{\frac{(2.45)(0.01)}{9\times10^9}} = 16.5\times10^-7 }[/math] C. To get to the number of electrons, simply divide [math]\displaystyle{ q_{1} }[/math] by [math]\displaystyle{ 1.602 \times10^-19 }[/math] C, the charge of an electron, to find that there are approximately [math]\displaystyle{ 2.645\times10^{13} }[/math] electrons that were transferred from your hair to the balloon.

**Difficult**

Derive a formula for the magnitude of the electric field a distance [math]\displaystyle{ r }[/math] away from an infinite line of charges, spaced out with charge density [math]\displaystyle{ \rho }[/math].

*Solution.* Let [math]\displaystyle{ x }[/math] be a point on the line of charges. Without loss of generality, we can assume that the point [math]\displaystyle{ x }[/math] is [math]\displaystyle{ x }[/math] meters away from the point on the line where the distance to the point being measured is at its minimum. Again without loss of generality, for the arbitrary point [math]\displaystyle{ x }[/math] we can set up a right triangle with [math]\displaystyle{ r }[/math] on the [math]\displaystyle{ y }[/math] axis and [math]\displaystyle{ x }[/math] on the [math]\displaystyle{ x }[/math] axis. Thus the distance between the point being measured and the point [math]\displaystyle{ x }[/math] is equal to [math]\displaystyle{ \sqrt{x^2 + r^2} }[/math]. Let [math]\displaystyle{ (x, x + {\Delta}x) }[/math] be a segment on the line. It contains [math]\displaystyle{ \rho{\Delta}x }[/math] C of charge. If [math]\displaystyle{ {\Delta}x }[/math] is small enough, then we can approximate the force of this line segment on the point being measured by Coulomb's Law. Plugging in, we get that [math]\displaystyle{ \vec{E}_{x} = \frac{k\rho{\Delta}x}{x^2+r^2} }[/math]. Using the sum formula for Coulomb's Law [math]\displaystyle{ \vec{E} = \sum_{n=0}^{j} {\frac{q_{n}}{4\pi\epsilon_{0}|\vec{r}_{n}|^2}}\hat{r}_{n} }[/math], and noting that if we let [math]\displaystyle{ {\Delta}x }[/math] become small, we get for the total force per unit charge at a distance [math]\displaystyle{ r }[/math] is equal to [math]\displaystyle{ |\vec{E}| = \frac{1}{4\pi\epsilon_{0}}\rho\int_{-\infty}^{\infty}\frac{dx}{x^2+r^2} = \frac{1}{4\pi\epsilon_{0}}\rho(\frac{\pi}{r}) }[/math]. Thus [math]\displaystyle{ |\vec{E}| = \frac{\rho}{4\epsilon_{0}r} }[/math].

## Connectedness

This topic is connected to electrostatics. Read more about where Coulombs law came from at https://www.physicsbook.gatech.edu/Gauss%27s_Law . Electrostatics has been interest of mine ever since I started reading Feynman's lectures in high school, and I found this page uncompleted and decided to complete it. Coulomb's law is also important for applications in my major, Aerospace Engineering, since it tells me things about how electrical components behave on satellites in space. Moreover, Coulomb's law has huge industrial applications mostly stemming from the fact that you can use it to derive all sorts of chemical equations, including the constants for Van Der Waal's modified ideal gas law which are useful in high-pressure applications such as rocket engines.

## History

Coulomb's Law was first formulated in 1785 in France by Charles Augustin de Coulomb. It was formulated using experimental evidence from a torsion balance. See [3]. Later, when Maxwell formulated his laws of electromagnetism, it was found to be a special case of the more general Maxwell's equations, which apply even when the charges are moving. Specifically, as mentioned above, Coulomb's Law is an easy consequence of Gauss's law of electric flux.

## See Also

https://www.physicsbook.gatech.edu/Maxwell%27s_Electromagnetic_Theory

https://www.physicsbook.gatech.edu/Ampere-Maxwell_Law

https://www.physicsbook.gatech.edu/Biot-Savart_Law

## References

1. https://www.feynmanlectures.caltech.edu/II_04.html