Revision as of 03:22, 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. 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 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 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. Therefore 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. Moreover, 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

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 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 multiply by [math]\displaystyle{ 1.602\times10^-19 }[/math] C, the charge of an electron, to find that there were approximately 2.645\times10^13 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]C/m.

Solution.

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 19: / Line 19: @@
 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 Coulomb's Law <math>\vec{F}_{2on1} = {\frac{kq_{1}q_{2}}{|\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.
 '''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>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>q_{2}</math>) is equal to the aggregate positive charges on your hair (if the positive charges are <math>q_{2}</math>, then <math>q_{1} = q_{2}</math>). Thus plugging into Coulomb's law we get <math>{\frac{(9\times10^9)q_{1}^2}{(0.1)^2}}<0,1,0> = 2.45<0,1,0></math>. This implies <math>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 multiply by <math>1.602\times10^-19</math> C, the charge of an electron, to find that there were approximately 2.645\times10^13 electrons that were transferred from your hair to the balloon.
 '''Difficult'''
 Derive a formula for the magnitude of the electric field a distance <math>r</math> away from an infinite line of charges, spaced out with charge density <math>\rho</math>C/m.
 ''Solution.''

Coulomb's law: Difference between revisions