Revision as of 02:42, 29 November 2023

Claimed by Spencer Boebel (Fall 2023)

Coulomb's Law

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}_{on1} }[/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. Coloumb'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. Coloumb'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 frac{farads}{meter} }[/math] 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 Coloumb'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 Coloumb's Law, then we can write Coloumb'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. Therefore if we have many charges [math]\displaystyle{ q_{1},...,q_{j} }[/math], we can write Coloumb'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. Moreover, a unit of charge is called a Coloumb, and all charges are measured in Coloumbs. An electron has [math]\displaystyle{ 1.602\times10^-19 }[/math] C of charge.

A Computational Model

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>\hat{r}_12 = \frac{\vec{r}_{12}}{|\vec{r}_{12}|} = <\frac{4}{5},\frac{3}{5},0></math> m. Plug into Coloumb's Law [math]\displaystyle{ \vec{F}_{2on1} = {\frac{q_{1}q_{2}}{4\pi\epsilon_{0}|\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.

Connectedness

History

Coulomb's Law was first formulated in

References

1. https://www.feynmanlectures.caltech.edu/II_04.html 2. https://physics.nist.gov/cgi-bin/cuu/Value?ep0

@@ Line 20: / Line 20: @@
 Given charges <math>q_{1} = 5</math> C located at <math>\vec{r}_{1} = <1,-1,0></math> m and <math>q_{2} = -10</math> C at location <math>\vec{r}_{2} = <5,2,0></math> m, find the force on charge 2 by charge 1.
-''Solution.'' Find the vector distance between the two charges <math>\vec{r}_{12} = \vec{r}_{2} - \vec{r}_{1} = <4,3,0></math> m. Find the magnitude <math>|\vec{r}_{12}| = 5</math> and unit vector </math>\hat{r}_12 = \frac{\vec{r}_{12}}{|\vec{r}_{12}|} = <\frac{4}{5},\frac{3}{5},0></math> m. Plug into Coloumb's Law, we get <math>\vec{F}_{2on1} = {\frac{(9\times10^9)\times(-10)\times(5)}{25}}<\frac{4}{5},\frac{3}{5},0> =  <-1.44\times10^10,-1.08\times10^10,0></math> N.
+''Solution.'' Find the vector distance between the two charges <math>\vec{r}_{12} = \vec{r}_{2} - \vec{r}_{1} = <4,3,0></math> m. Find the magnitude <math>|\vec{r}_{12}| = 5</math> and unit vector </math>\hat{r}_12 = \frac{\vec{r}_{12}}{|\vec{r}_{12}|} = <\frac{4}{5},\frac{3}{5},0></math> m. Plug into Coloumb's Law <math>\vec{F}_{2on1} = {\frac{q_{1}q_{2}}{4\pi\epsilon_{0}|\vec{r}_{12}|^2}}\hat{r}_{12}</math>, we get <math>\vec{F}_{2on1} = {\frac{(9\times10^9)\times(-10)\times(5)}{25}}<\frac{4}{5},\frac{3}{5},0> =  <-1.44\times10^10,-1.08\times10^10,0></math> N.
 == Connectedness ==

Coulomb's law: Difference between revisions