Path Independence: Difference between revisions
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!Outline of proof | !Outline of proof | ||
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|states that the | |[[Coulomb's law]] states that the electric field due to a stationary [[point charge]] is: | ||
:<math>\mathbf{E}(\mathbf{r}) = \frac{q}{4\pi \varepsilon_0} \frac{\mathbf{e_r}}{r^2}</math> | |||
where | |||
:'''e<sub>r</sub>''' is the radial [[unit vector]], | |||
:''r'' is the radius, |'''r'''|, | |||
:<math>\varepsilon_0</math> is the [[electric constant]], | |||
:''q'' is the charge of the particle, which is assumed to be located at the [[origin (mathematics)|origin]]. | |||
Using the expression from Coulomb's law, we get the total field at '''r''' by using an integral to sum the field at '''r''' due to the infinitesimal charge at each other point '''s''' in space, to give | |||
:<math>\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \int \frac{\rho(\mathbf{s})(\mathbf{r}-\mathbf{s})}{|\mathbf{r}-\mathbf{s}|^3} \, d^3 \mathbf{s}</math> | |||
where <math>\rho</math> is the charge density. If we take the divergence of both sides of this equation with respect to '''r''', and use the known theorem<ref>See, for example, {{cite book | author=Griffiths, David J. | title=Introduction to Electrodynamics (4th ed.) | publisher=Prentice Hall | year=2013 | page=50 }}</ref> | |||
:<math>\nabla \cdot \left(\frac{\mathbf{r}}{|\mathbf{r}|^3}\right) = 4\pi \delta(\mathbf{r})</math> | |||
where ''δ''('''r''') is the [[Dirac delta function]], the result is | |||
:<math>\nabla\cdot\mathbf{E}(\mathbf{r}) = \frac{1}{\varepsilon_0} \int \rho(\mathbf{s})\ \delta(\mathbf{r}-\mathbf{s})\, d^3 \mathbf{s}</math> | |||
Using the "[[Dirac delta function#Translation|sifting<!--Note: This is not a typo, the word is really 'sifting' not 'shifting'--> property]]" of the Dirac delta function, we arrive at | |||
:<math>\nabla\cdot\mathbf{E}(\mathbf{r}) = \frac{\rho(\mathbf{r})}{\varepsilon_0},</math> | |||
which is the differential form of Gauss's law, as desired. | |||
|} | |} | ||
Revision as of 16:51, 13 October 2015
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Outline of proof Coulomb's law states that the electric field due to a stationary point charge is: - [math]\displaystyle{ \mathbf{E}(\mathbf{r}) = \frac{q}{4\pi \varepsilon_0} \frac{\mathbf{e_r}}{r^2} }[/math]
where
- er is the radial unit vector,
- r is the radius, |r|,
- [math]\displaystyle{ \varepsilon_0 }[/math] is the electric constant,
- q is the charge of the particle, which is assumed to be located at the origin.
Using the expression from Coulomb's law, we get the total field at r by using an integral to sum the field at r due to the infinitesimal charge at each other point s in space, to give
- [math]\displaystyle{ \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \int \frac{\rho(\mathbf{s})(\mathbf{r}-\mathbf{s})}{|\mathbf{r}-\mathbf{s}|^3} \, d^3 \mathbf{s} }[/math]
where [math]\displaystyle{ \rho }[/math] is the charge density. If we take the divergence of both sides of this equation with respect to r, and use the known theorem[1]
- [math]\displaystyle{ \nabla \cdot \left(\frac{\mathbf{r}}{|\mathbf{r}|^3}\right) = 4\pi \delta(\mathbf{r}) }[/math]
where δ(r) is the Dirac delta function, the result is
- [math]\displaystyle{ \nabla\cdot\mathbf{E}(\mathbf{r}) = \frac{1}{\varepsilon_0} \int \rho(\mathbf{s})\ \delta(\mathbf{r}-\mathbf{s})\, d^3 \mathbf{s} }[/math]
Using the "sifting property" of the Dirac delta function, we arrive at
- [math]\displaystyle{ \nabla\cdot\mathbf{E}(\mathbf{r}) = \frac{\rho(\mathbf{r})}{\varepsilon_0}, }[/math]
which is the differential form of Gauss's law, as desired.
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