| Both sides previous revision Previous revision Next revision | Previous revision |
| 184_notes:symmetry [2017/12/01 00:17] – dmcpadden | 184_notes:symmetry [2021/07/06 17:51] (current) – bartonmo |
|---|
| Chapter 21 in Matter and Interactions (4th edition) | Chapter 21 in Matter and Interactions (4th edition) |
| | |
| | /*[[184_notes:maxwells_eq|Next Page: Maxwell's Equations]] |
| | |
| | [[184_notes:conservation_theorems|Previous Page: Conservation Theorems]]*/ |
| |
| ===== Symmetry and Mathematical Tools ===== | ===== Symmetry and Mathematical Tools ===== |
| {{youtube>I4Utm80K8hM?large}} | {{youtube>I4Utm80K8hM?large}} |
| |
| ==== Gauss' Law ==== | ===== Gauss' Law ===== |
| |
| [[184_notes:gauss_motive|Gauss' Law]] helps us to calculate the electric field when there is sufficient symmetry to use it. That is, Gauss' Law, as mathematical statement, is always true, but it's only useful in limited cases (namely, for planes, cylinders and spheres of charge). The total electric flux is always proportional to the enclosed charge, but that doesn't mean we can always calculate the electric field from Gauss' Law. | [[184_notes:gauss_motive|Gauss' Law]] helps us to calculate the electric field when there is sufficient symmetry to use it. That is, Gauss' Law, as mathematical statement, is always true, but it's only useful in limited cases (namely, for planes, cylinders and spheres of charge). The total electric flux is always proportional to the enclosed charge, but that doesn't mean we can always calculate the electric field from Gauss' Law. |
| |
| $$\Phi_E = \oint \vec{E} \cdot d\vec{A} = \dfrac{q_{enc}}{\varepsilon_0}$$ | $$\Phi_E = \oint \vec{E} \bullet d\vec{A} = \dfrac{q_{enc}}{\varepsilon_0}$$ |
| |
| [[184_notes:eflux_curved|For a chosen Gaussian surface]], if the electric field points the same direction as the area vector everywhere on the surface and we can argue that the electric field is constant in magnitude over the surface, then we can simplify Gauss' Law sufficiently where it becomes useful, | [[184_notes:eflux_curved|For a chosen Gaussian surface]], if the electric field points the same direction as the area vector everywhere on the surface and we can argue that the electric field is constant in magnitude over the surface, then we can simplify Gauss' Law sufficiently where it becomes useful, |
| |
| $$\oint \vec{E} \cdot d\vec{A} = E \oint dA = \dfrac{q_{enc}}{\varepsilon_0}$$ | $$\oint \vec{E} \bullet d\vec{A} = E \oint dA = \dfrac{q_{enc}}{\varepsilon_0}$$ |
| |
| {{ 184_notes:smallsphere.jpg?300}} | [{{ 184_notes:smallsphere.jpg?300|Flux through a spherical gaussian surface}}] |
| |
| The example we have seen a number of times is the point charge, $q$. If we encapsulate the point charge with an imaginary spherical surface of radius $r$, such that the point charge is at the center, we can easily find the electric field of the charge, | The example we have seen a number of times is the point charge, $q$. If we encapsulate the point charge with an imaginary spherical surface of radius $r$, such that the point charge is at the center, we can easily find the electric field of the charge, |
| where we understand the direction to point radially outward as usual for a positive point charge. This was necessary to argue the simplification of Gauss' Law from the first to the second line. | where we understand the direction to point radially outward as usual for a positive point charge. This was necessary to argue the simplification of Gauss' Law from the first to the second line. |
| |
| ==== Ampere's Law ==== | ===== Ampere's Law ===== |
| |
| [[184_notes:motiv_amp_law|Ampere's Law]] helps us calculate the magnetic field when there is sufficient symmetry to use it. Like Gauss' Law, Ampere's Law, as a mathematical statement, is always true, but it's only useful in limited contexts (like long wires or solenoids). The total integral around any closed loop is always proportional to the total current enclosed by the loop, but that doesn't mean we can compute the magnetic field using Ampere's Law for every case, | [[184_notes:motiv_amp_law|Ampere's Law]] helps us calculate the magnetic field when there is sufficient symmetry to use it. Like Gauss' Law, Ampere's Law, as a mathematical statement, is always true, but it's only useful in limited contexts (like long wires or solenoids). The total integral around any closed loop is always proportional to the total current enclosed by the loop, but that doesn't mean we can compute the magnetic field using Ampere's Law for every case, |
| |
| $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}.$$ | $$\oint \vec{B} \bullet d\vec{l} = \mu_0 I_{enc}.$$ |
| |
| [[184_notes:loop|For a chosen Amperian loop]], if the magnetic field aligns with the direction of the loop and we can argue that the magnetic field is constant over the loop (or part of it and zero elsewhere), then we can simplify Gauss' Law sufficiently where it becomes useful, | [[184_notes:loop|For a chosen Amperian loop]], if the magnetic field aligns with the direction of the loop and we can argue that the magnetic field is constant over the loop (or part of it and zero elsewhere), then we can simplify Gauss' Law sufficiently where it becomes useful, |
| |
| $$\oint \vec{B} \cdot d\vec{l} = B \oint dl = \mu_0 I_{enc}.$$ | $$\oint \vec{B} \bullet d\vec{l} = B \oint dl = \mu_0 I_{enc}.$$ |
| |
| {{ 184_notes:week10_3.png?400}} | [{{ 184_notes:week10_3.png?400|Current through an amperian loop}}] |
| |
| The example that we have seen a number of times is the very long thin wire with current $I$. If we encircle the wire with a loop of radius $r$ with the wire centered inside the loop, we can easily find the magnetic field, | The example that we have seen a number of times is the very long thin wire with current $I$. If we encircle the wire with a loop of radius $r$ with the wire centered inside the loop, we can easily find the magnetic field, |
| |
| $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$$ | $$\oint \vec{B} \bullet d\vec{l} = \mu_0 I_{enc}$$ |
| $$B \oint dl \mu_0 I$$ | $$B \oint dl \mu_0 I$$ |
| $$B 2 \pi r = \mu_0 I$$ | $$B 2 \pi r = \mu_0 I$$ |