| Wiki Markup |
|---|
h1. Question 1
*Divergence theorem:* For a cube of arbitrary size, demonstrate that the flux of a vector field that varies linearly as a function of <math>x</math>, <math>y</math>, and <math>z</math> is equal to the integral of the divergence of the vector field over the volume of the cube.
Consider a cube with one point at the origin, a face along the x-axis where x=0, and a face along the x-axis where x=dy. There is a y component of A on the surface of the face at x=0, and it is denoted by <math>A_y</math>. From a Taylor extrapolation to first order, the flux on the face at x=dy is <math>A_Y + \frac{\partial A_y}{\partial y} dy</math>. Find the flux of a vector field across surface and repeat for other surfaces.
<center>
<br>
<math>\left ( A_y + \frac{\partial A_y}{\partial y} dy \right ) dS - A_y dS = \frac{\partial A_y}{\partial y} dV</math>
<br>
<math>\frac{\partial A_x}{\partial x} dV + \frac{\partial A_y}{\partial y} dV + \frac{\partial A_z}{\partial z} dV</math>
<br>
<math>\left ( \frac{\partial A_y}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z} \right ) dV</math>
<br>
<math>\int_V \vec \nabla \cdot \vec A d^3 x = \int_S \vec A \cdot \hat n dS</math>
<br>
</center>
h1. Question 2
*Continuity conditions:* Use Stokes theorem to show what are the conditions that the tangential components of <math>E</math> and <math>H</math> need to satisfy as we cross the boundary between two different dielectric layers.
<p>
</p>
Resources
* [Stoke's Theorem|http://en.wikipedia.org/wiki/Stoke%27s_theorem]
* [Dielectric|http://en.wikipedia.org/wiki/Dielectric]
* [Electromagnetic Field|http://en.wikipedia.org/wiki/Electromagnetic_field]
* [Maxwell's Equations|http://en.wikipedia.org/wiki/Maxwell's_equations]
||! Name! Differential form! Integral form
| Gauss's law: How electric charges produce electric fields |
| <math>\vec{\nabla} \cdot \vec{D} = \rho</math> |
| <math>\oint_S \vec{D} \cdot \mathrm{d}\vec{A} = \int_V \rho\, \mathrm{d}V</math> |
| Gauss' law for magnetism <br /> (absence of magnetic monopoles): |
| <math>\vec{\nabla} \cdot \vec{B} = 0</math> |
| <math>\oint_S \vec{B} \cdot \mathrm{d}\vec{A} = 0</math> |
| Faraday's law of induction: How currents and changing electric fields produce magnetic fields |
| <math>\vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}} {\partial t}</math> |
| <math>\oint_C \vec{E} \cdot \mathrm{d}\vec{l} - \oint_C \vec{B} \times \vec{v} \cdot \mathrm{d}\vec{l} = - \ { \mathrm{d} \over \mathrm{d}t } \int_S \vec{B} \cdot \mathrm{d}\vec{A}</math> |
| Amp��re's law<br /> (with Maxwell's extension): How changing magnetic fields produce electric fields |
| <math>\vec{\nabla} \times \vec{H} = \vec{J} + \frac{\partial \vec{D}} {\partial t}</math> |
| <math>\oint_C \vec{H} \cdot \mathrm{d}\vec{l} = \int_S \vec{J} \cdot \mathrm{d} \vec{A} + \int_S \frac{\partial\vec{D}}{\partial t} \cdot \mathrm{d} \vec{A}</math> |
||! Name! Differential form! Integral form (using Kelvin-Stokes theorem)
| Faraday's law of induction: |
| <math>\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}</math> |
| <math>\oint_C \mathbf{E} \cdot d\mathbf{l} = \int_S \nabla \times \mathbf{E} \cdot d\mathbf{A} = - \ { d \over dt } \int_S \mathbf{B} \cdot d\mathbf{A}</math> |
| Amp��re's law<br /> (with Maxwell's extension): |
| <math>\nabla \times \mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}} {\partial t}</math> |
| <math>\oint_C \mathbf{H} \cdot d\mathbf{l} = \int_S \nabla \times \mathbf{H} \cdot d \mathbf{A} = \int_S \mathbf{J} \cdot d \mathbf{A} + |
<center>
<br>
<math>\mathbf{D} = \varepsilon_{0} \mathbf{E} + \mathbf{P} = \varepsilon_{0} \mathbf{E} + \varepsilon_{0}\chi\mathbf{E} = \varepsilon_{0} \mathbf{E} \left( 1 + \chi \right)</math>
<br>
</center>
h2. Boundary Conditions
To investigate the tangential components of <math>E</math> and <math>H</math> consider a closed loop. The loop consists of the elements <math>\delta_{s1}</math>, <math>\delta_{so}</math>, and two short connectors, each of length <math>\delta h</math>. The surface normal of the loop dA is in the direction of unit vector b. Apply Stokes' theorem to the loop, one finds the relation below.
<center>
<br>
<math>\int_{loop area} \left ( \nabla \times \vec E \right ) \cdot dA = \int_{contour} E \cdot ds</math>
<br>
</center>
The integration path of the contour integral is along <math>\delta_{s1}</math>, <math>\delta_{h}</math>, <math>\delta_{s2}</math>, and <math>\delta_{h}</math>. Use an equation below and replace the contour integral by its elements.
<center>
<br>
<math> B = - \nabla \times E </math>
<br>
<math>- \int B_b \delta_s \delta h = E_t^{1} \delta_{s1} - E_n \delta h + E_t^{0} \delta_{s0} + E_n \delta h</math>
<br>
</center>
Let <math>\delta h</math> approach zero. The integral over the area of the loop vanishes, and, considering that <math>\delta s_1</math> and <math>\delta s_0</math> are opposite in sign find the relation below.
<center>
<br>
<math>E_t^{(1)} - E_t^{(0)} = 0</math>
<br>
</center>
The tangential component of the electric field strength is continuous across the boundary. Follow a similar procedure for the tangential component of <math>H</math> one finds the relation below.
<center>
<br>
<math>H_t^{(1)} - H_t^{(1)} = j_{surf}</math>
<br>
</center>
The tangential component of the magnetic field strength changes abruptly in the presence of a surface current, but it is continuous in the absense of such a current.
<p>
</p>
*Source:* [Exploration of the Solar System by Infrared Remote Sensing|http://www.amazon.com/gp/reader/0521818974/ref=sib_books_pg/102-5385840-4382552?ie=UTF8&keywords=tangential%20components%20E%20and%20H&p=S00Y&checkSum=PwQeaWo8oiDAMXMv5fHIpcp4bx5MiSfolojH2DBYlPM%253D]
h2. Boundary Conditions for Electric and Magnetic Fields
Light waves are electromagnetic waves. Consider an interface between medium 1 and medium 2. Start with Gauss's law for the electric field and Gauss's law for the magnetic field and use vector analysis. Obtain the following boundary conditions.
<center>
<br>
<math>\hat n \cdot \left (\vec B_2 - \vec B_1 \right ) = 0</math>
<br>
<math>\hat n \cdot \left (\vec D_2 - \vec D_1 \right ) = \rho_s</math>
<br>
</center>
The term <math>\hat n</math> is the unit vector normal to the interface and <math>\rho_s</math> is the density of surface charge. For insulators, there is no surface charge, so <math>\rho_s</math> is zero. Below are resulting equations.
<center>
<br>
<br>
</center>
The terms <math>B_{2n}</math>, <math>B_{1n}</math>, <math>D_{2n}</math>, and <math>D_{1n}</math> are projections of fields along the direction normal to the boundary. The normal projections of <math>B</math> and <math>D</math> field need to be continuous even at the interface of two media if medium 1 and medium 2 are both insulators.
<p>
</p>
At the interface, the tangential components <math>E</math> field and <math>H</math> field also have to be continuous. Start with Faraday's law and Ampere's law and use vector analysis to obtain the following boundary conditions.
<center>
<br>
<math>\hat n \times \left (\vec E_2 - \vec E_1 \right ) = 0</math>
<br>
<math>\hat n \times \left (\vec H_2 - \vec H_1 \right ) = \vec K</math>
<br>
</center>
The term <math>\vec K</math> is the surface current density at the interface. If medium 1 and medium 2 are insulators, there is no surface current density, so <math>\vec K = 0</math>. The cross product of <math>\hat n</math> and <math>\vec E_2</math> is the tangential direction along the interface. For insulators, the equations can be written as below.
<center>
<br>
<math>E_{2t} - E_{1t} = 0</math>
<br>
<math>H_{2t} - H_{1t} = 0</math>
<br>
<math>E_{2t} = E_{1t}</math>
<br>
<math>H_{2t} = H_{1t}</math>
<br>
</center>
The terms in the equations above are the tangential components of <math>E</math> and <math>H</math> along the interface boundary.
<p>
</p>
*Source:* [Optical Components for Communications|http://www.amazon.com/gp/reader/1402076363/ref=sib_books_pg/102-5385840-4382552?ie=UTF8&keywords=tangential%20components%20E%20and%20H&p=S00R&checkSum=PwQeaWo8oiC5lu8IBKFmOXXYiFJPXBXGgCoDfU%252BGKUk%253D]
h1. Question 3
*Metals:* Show that in a metallic conductor any free charge can only be found at the surface.
Start with the fact that there is no electric field inside a conductor. Use a Maxwell equation below.
<center>
<br>
<math>\nabla \cdot E = \frac{4 \pi \rho}{\epsilon}</math>
<br>
<math>\int \int \int \nabla \cdot E = \int \int \int \frac{4 \pi \rho}{\epsilon}</math>
<br>
</center>
The left side of the equation is equal to zero from the fact that the electric field inside the conductor is equal to zero.
<center>
<br>
<math>0 = \int \int \int \frac{4 \pi \rho}{\epsilon}</math>
<br>
</center>
This equation is only true if the charge density inside the conductor is equal to zero. Since there is charge in the material and the total charge is equal to the amount in the material plus the amount the surface, charge must be on the surface.
|
Page History
Overview
Content Tools
Activity