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\boxaround[Electrodynamics with Jackson - Prof. Jianping Lu - Fall 2008\\
University of North Carolina Chapel Hill
Final Review Guide by Justin Moore]
\noindent Coulomb's Law: $\vec{\nabla}\cdot\vec{D}=\rho(\vec{x})$ $\quad\quad$ Ampere's Law: $\vec{\nabla}\times\vec{H}=\mu_0\vec{J}(\vec{x})+\epsilon_0\mu_0\frac{\partial \vec{D}}{\partial t}$\\
Faraday's Law: $\vec{\nabla}\times\vec{E}=-\frac{\partial \vec{B}}{\partial t}$ $\quad\quad$ No Magnetic Charge: $\vec{\nabla}\cdot\vec{B}=0$\\

\noindent Lorentz Force: $\vec{F}=$q$(\vec{E}+\vec{v}\times\vec{B})$ $\quad$
Continuity Equation(cons. of charge): $\vec{\nabla}\cdot\vec{J}+\frac{\partial \rho(\vec{x})}{\partial t}=0$\\
Energy Density: $W=\frac{1}{2}(\vec{E}\cdot{D}+\vec{H}\cdot\vec{B})$ $\quad$ Note \scshape not \upshape energy -- We must integrate over volume for energy.\\

\noindent Linear Media: $\vec{D}=\epsilon\vec{E}, \vec{B}=\mu\vec{H}$ $\quad$
In vaccuum, $\epsilon=\epsilon_o$, $\mu=\mu_0$, $\frac{1}{c^2}=\epsilon_0\mu_0$, $\frac{1}{v^2}=\epsilon\mu$\\

\noindent \scshape Linear Superposition Theory!\upshape $\quad$ Boundary Conditions:\\
Convention: Region 1 is below the boundary and Region 2 is above the boundary.  The direction $\vec{n}$ points from 1 $\to$ 2.\\
Normal $(\vec{D})$ is discont.: $\vec{n}\cdot(\vec{D}_2-\vec{D}_1)=\sigma$ $\quad$ 
Normal $(\vec{B})$ is cont.: $\vec{n}\cdot(\vec{B}_2-\vec{B}_1)=0$\\
Tangential $(\vec{E})$ is cont.: $\vec{n}\times(\vec{E}_2-\vec{E}_1)=0$ $\quad$
Tangential $(\vec{H})$ is discont.: $\vec{n}\times(\vec{H}_2-\vec{H}_1)=\vec{K}$\\

\noindent Non-Exhaustive Math and Vector Identities:
BAC-CAB Identity: $\vec{A}\times(\vec{B}\times\vec{C})=\vec{B}(\vec{A}\cdot\vec{C})-\vec{C}(\vec{A}\cdot\vec{B})$\\
Cyclic Permutations: $\vec{A}\cdot(\vec{B}\times\vec{C})=\vec{B}\cdot(\vec{C}\times\vec{A})=\vec{C}\cdot(\vec{A}\times\vec{B})$\\
$\vec{\nabla}t=\frac{\partial t}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial t}{\partial \theta}\hat{\theta}+\frac{1}{r S[\theta]}\frac{\partial t}{\partial \phi}\hat{\phi}|\vec{\nabla}^2 t=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial t}{\partial r})+\frac{1}{r^2 S[\theta]}\frac{\partial}{\partial \theta}(S[\theta]\frac{\partial t}{\partial \theta})+\frac{1}{r^2 S[\theta]^2}\frac{\partial^2 t}{\partial \phi^2}$\\
$\vec{\nabla}\cdot v=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 v_r)+\frac{1}{r S[\theta]}\frac{\partial}{\partial \theta}(S[\theta] v_{\theta})+\frac{1}{r S[\theta]}\frac{\partial v_{\phi}}{\partial \phi}$\\
$\vec{\nabla}\times v=\frac{1}{r S\theta}\left[\frac{\partial}{\partial \theta}(S\theta v_{\theta})-\frac{\partial v_{\theta}}{\partial \phi}\right]\hat{r}+\frac{1}{r}\left[\frac{1}{S\theta}\frac{\partial v_r}{\partial \phi}-\frac{\partial}{\partial r}(r v_{\phi})\right]\hat{\theta}+\frac{1}{r}\left[\frac{\partial}{\partial r}(r v_{\theta})-\frac{\partial v_r}{\partial \theta}\right]\hat{\phi}$\\
$\vec{\nabla}t=\frac{\partial t}{\partial s}\hat{s}+\frac{1}{s}\frac{\partial t}{\partial \phi}\hat{\phi}+\frac{\partial t}{\partial z}\hat{z}\quad\vec{\nabla}^2 t=\frac{1}{s}\frac{\partial}{\partial s}\left(s\frac{\partial t}{\partial s}\right)+\frac{1}{s^2}\frac{\partial^2 t}{\partial \phi^2}+\frac{\partial^2 t}{\partial z^2}$\\
$\vec{\nabla}\cdot v=\frac{1}{s}\frac{\partial}{\partial s}(s v_s)+\frac{1}{s}\frac{\partial v_{\phi}}{\partial \phi}+\frac{\partial v_z}{\partial z}$ $\quad$ 
$f(x+a)= \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$\\
$\vec{\nabla}\times v=\left[\frac{1}{s}\frac{\partial v_z}{\partial \phi}-\frac{\partial v_{\phi}}{\partial z}\right]\hat{s}+\left[\frac{\partial v_s}{\partial z}-\frac{\partial v_z}{\partial s}\right]\hat{\phi}+\frac{1}{s}\left[\frac{\partial}{\partial s}(s v_{\phi})-\frac{\partial v_s}{\partial \phi}\right]\hat{z}$\\
Divergence Thm: $\int_{V} (\vec{\nabla} \cdot \vec{A}) d\vec{x} = \int_{S} \vec{A} \cdot \vec{n} da$ $\quad$
Stoke's Thm: $\int_{S} (\vec{\nabla} \times \vec{A})\cdot\vec{n} da = \oint_{c} \vec{A}d\vec{l}$\\
Green Theorem: 
$ \int_{V} (\phi \vec{\nabla}^2 \psi- \psi \vec{\nabla}^2 \phi)d\vec{x} = \oint_{S}\left[ \phi \frac{\partial \psi}{\partial n} - \psi \frac{\partial \phi}{\partial n} \right]\cdot \vec{n} da$\\

\boxaround[Chapter 1]
$\vec{E}$, $\vec{E}=-\vec{\nabla}\Phi$ (Electrostatic case), $\rho(\vec{x})$, Point charge - Laplace's Equation: $\vec{\nabla}^2\Phi=0$\\
$\Phi(\vec{x})=\frac{q}{4\pi\epsilon_0|\vec{x}-\vec{x}'|}$ (Goes as $\frac{1}{r^2}$)\\
$\vec{E}(\vec{x})=\frac{q}{4\pi\epsilon_0}\frac{|\vec{x}-\vec{x}'|}{(|\vec{x}-\vec{x}'|)^3}$(Goes as $\frac{1}{r^3}$)\\
Superposition theory allows us to evaluate these for an arbitrary point charge distribution by integrating over all space.\\
$\Phi(\vec{x})=\frac{1}{4\pi\epsilon_0}\int{}\frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|} d\vec{x}$ (Goes as $\frac{1}{r^2}$)\\
$\vec{E}(\vec{x})=\frac{1}{4\pi\epsilon_0}\int{}\frac{\rho(\vec{x}')|\vec{x}-\vec{x}'|}{(|\vec{x}-\vec{x}'|)^3} d\vec{x}$(Goes as $\frac{1}{r^3}$)\\
\noindent Point dipole is a little more interesting:\\

\noindent$\Phi(\vec{x})=\frac{\vec{p}\cdot(\vec{x}-\vec{x}')}{4\pi\epsilon_0(|\vec{x}-\vec{x}'|)^3}$\\
 $\vec{E}(\vec{x})=\frac{1}{4\pi\epsilon_0}\frac{3 \hat{x}(\vec{p}\cdot\vec{x})-\vec{p}}{(|\vec{x}|)^3}$ Or, more generally, $\vec{E}(\vec{x})=\frac{1}{4\pi\epsilon_0}\frac{3 (\hat{x}-\hat{x}')(\vec{p}\cdot(\hat{x}-\hat{x}'))-\vec{p}}{(|\vec{x}-\vec{x}'|)^3}$\\

\noindent Free space Green Function: $G(\vec{x},\vec{x}')=\frac{1}{|\vec{x}-\vec{x}'|}$ Defined as the solution to a point charge problem.\\
$\vec{\nabla}^2 G(\vec{x},\vec{x}')= -4\pi\delta(\vec{x}-\vec{x}')$\\
\noindent Dirichlet Boundary Conditions:\\ $G_D(\vec{x},\vec{x}')|_{\vec{x}'=on S}=0$, $\Phi(Surface)$ is known.\\
$\Phi(\vec{x})=\frac{1}{4\pi\epsilon_0}\int{}\rho(\vec{x}')G_D(\vec{x},\vec{x}')d\vec{x}-\frac{1}{4\pi}\oint{}\Phi(\vec{x}')\frac{\partial G_d(\vec{x},\vec{x}')}{\partial n'}da'$\\
Two terms: Charge Term and Boundary Term. (Pay attention to the direction of the normal - p.36 in Jackson.  Convention: Away from region of interest. (e.g. $\vec{x}$ inside sphere, $n'=\frac{\partial}{\partial r}$ or $\vec{x}$ outside sphere, $n'=-\frac{\partial}{\partial r}))$\\
\noindent Neumann Boundary Conditions:
Specify Electric Field\\ ($\frac{\partial G(\vec{x},\vec{x}')}{\partial n'}|_{\vec{x}'=on S}=-\frac{4\pi}{S}$ where S is the Surface Area)\\
$\Phi(\vec{x})=\langle\Phi\rangle_s + \frac{1}{4\pi\epsilon_0}\int{}\rho(\vec{x}')G_N(\vec{x},\vec{x}')d\vec{x}-\frac{1}{4\pi}\oint_S\frac{\partial \Phi(\vec{x}')}{\partial n'}G_N(\vec{x},\vec{x}')da'$\\
Capacitance\\ $W=\frac{1}{2}\sum_iQ_iV_i=\frac{1}{2}\sum_{ij}C_{ij}V_iV_j$\\

\boxaround[Chapter 2]
\noindent Image Charge Problem:\\
Charge above a conducting plane, place image charge at same distance but behind plane.  Image charge must be located outside interested region - otherwise Laplace's equation would not be satisfied.  Image charge is \scshape virtual \upshape, and as such, can't be in region of interest!\\
\noindent Spherical image charge problem: point charge $q$ at location $\vec{y}$ will induce image charge $q'$ at location $\vec{y}'$ inside sphere of radius $a$. These two values are as follows (this can be accomplished by unit analysis):\\ $q'=-\frac{q a}{y}$ (as $\vec{y}\to\infty$, $q'\to0$) $\qquad$
$\vec{y}'=\frac{a^2}{y}\hat{y}$ (unit analysis)\\
\noindent Separation of Variables $\quad$
$\vec{\nabla}^2\Phi=-\frac{\rho(\vec{x}')}{\epsilon_0}$\\Assume separable solutions: (For a box, a cube, straight, use Rectangular.)\\
$\Phi(x,y,z)=X(x)Y(y)Z(z)$\\
First weapon: Image Charge -- if that doesn't work, use Separation of Variables.\\
Two Dimensional Cylindrical Problem: a wedge with angle $\beta$.  Qualitative:\\ $\vec{E}_\rho\approx\rho^{\frac{\pi}{\beta}-1}$ If $\beta=\pi$, $\vec{E}_\rho = C_0$. If $\beta=2 \pi$, then $\vec{E}_\rho = \frac{1}{\sqrt{\rho}}$\\
$\sigma=-\epsilon_0 \frac{\partial \Phi}{\partial x}|x=a$ Charge density is calculated by the normal derivative at the surface.\\

\boxaround[Chapter 3]
\noindent $\vec{\nabla}^2\Phi=-\frac{\rho(\vec{x})}{\epsilon_0}$ This is Laplace's Equation($\vec{\nabla}^2\Phi=0$) if $\rho(\vec{x})=0$\\
Boundary Value Problems with $\vec{x}\to r,\theta,\phi$ | Separable Form:
$\Phi(r,\theta,\phi)=\frac{U(r)}{r} P(\theta) Q(\phi)$\\
The General Expansion in terms of Spherical Harmonics:
$\Phi=\sum_{lm}[A_{lm} r^l + B_{lm} r^{-(l+1)}]Y^m_{l}(\theta,\phi)$\\
Then Green Function in Spherical Harmonics:
$\frac{1}{|\vec{x}-\vec{x}'|}=\sum_{lm}\frac{4\pi}{2l+1}\frac{{r_{<}}^l}{{r_{>}}^{l+1}}{Y_{l}^m}^{*}(\theta',\phi')Y^m_{l}(\theta,\phi)$\\
With cylindrical symmetry, we have that $m=0$ and therefore the Green Function Expansion reduces to:\\
$\frac{1}{|\vec{x}-\vec{x}'|}=\sum_{lm}\frac{{r_{<}}^l}{{r_{>}}^{l+1}}P_l(\cos{\gamma})$ with$\quad$
$\Phi(r,\theta,\phi)=\sum_{l}[A_{l0} r^l + B_{l0} r^{-(l+1)}]P_{l}(\cos{\theta})$\\
$Y_{0}^{0}(\theta,\phi)=\frac{1}{\sqrt{4 \pi}}$
$Y_{1}^{0}(\theta,\phi)=\sqrt{\frac{3}{4\pi}} \cos\theta$
$Y_{1}^{\pm1}(\theta,\phi)=\mp\sqrt{\frac{3}{8\pi}} \sin\theta e^{\pm i\phi}$\\
$Y_{2}^{0}(\theta,\phi)=\sqrt{\frac{5}{16\pi}}(3\cos^{2}\theta-1)$
$Y_{2}^{\pm1}(\theta,\phi)=\mp\sqrt{\frac{15}{8\pi}}\sin\theta\cos\theta\, e^{\pm i\phi}$\\
$Y_{2}^{\pm2}(\theta,\phi)=\sqrt{\frac{15}{32\pi}}\sin^{2}\theta e^{\pm 2i\phi}$$\quad$
In Cartesian: ($r=\sqrt{x^2+y^2+z^2})$\\
$Y_{0}^{0}(\theta,\phi)=same$
$Y_{1}^{0}(\theta,\phi)=\sqrt{\frac{3}{4\pi}} \frac{z}{r}$
$Y_{1}^{\pm1}(\theta,\phi)=\mp\sqrt{\frac{3}{8\pi}} \frac{x\pm iy}{r}$\\
$Y_{2}^{0}(\theta,\phi)=\sqrt{\frac{5}{16\pi}}\frac{3z^2-r^2}{r^2}$
$Y_{2}^{\pm1}(\theta,\phi)=\mp\sqrt{\frac{15}{8\pi}}\frac{z(x\pm iy)}{r^2}$
$Y_{2}^{\pm2}(\theta,\phi)=\sqrt{\frac{15}{32\pi}}\frac{(x\pm iy)^2}{r^2}$\\
$P_0^0(x)=1$ $\qquad$ $P_0^0(\cos\theta)=1$\\
$P_1^0(x)=x$ $\qquad$ $P_1^0(\cos\theta)=\cos\theta$\\
$P_1^1(x)=\frac{1}{2}(3x^2-1)$ $\qquad$ $P_1^1(\cos\theta)=-\sin\theta$\\
$P_2^0(x)=\frac{1}{2}(3x^2-1)$ $\qquad$ $P_2^0(\cos\theta)=\frac{1}{2}(3\cos\theta^2-1$\\
$P_2^1(x)=-3\sqrt{1-x^2}$ $\qquad$ $P_2^1(\cos\theta)=-3\sin\theta\cos\theta$\\
$P_2^2(x)=3(1-x^2)$ $\qquad$ $P_2^2(\cos\theta)=3\sin\theta^2$\\
Now for 3D Wedge (Cone) - $\vec{E}_{\rho}\approx\rho^{\nu-1}$ where $\nu$ is the smallest root of $P_{\nu}(\cos{\theta})=0$\\
If $2\beta=\pi, \vec{E}_{\rho}=C_0$;
$2\beta>\pi, \nu<1, \vec{E}_{\rho}$ diverges;
$2\beta\to\pi$, $\frac{1}{\rho}$ (Needle) vs. $\frac{1}{sqrt{\rho}}$ (Wedge)\\
Spherical Shell, inner radius $a$, outer radius $b$.
$G(\vec{x}-\vec{x}')=\sum_{lm}\frac{4\pi}{2l+1}\frac{{r_{<}}^l}{{r_{>}}^{l+1}}{Y_{l}^m}^{*}(\theta',\phi')Y^m_{l}(\theta,\phi)$\\
When using Green function here, make sure to use symmetric case $G(\vec{x},\vec{x}')=G(\vec{x}',\vec{x})$ as well as the matching conditions across the boundary (see above in B.C section) Generally of the form:\\
$(r_{<}^l-\frac{a^{2l+1}}{r_{<}^{l+1}})-(\frac{1}{r_{<}^{l+1}}-\frac{r_{>}^l}{b^{2l+1}})$  *Given $\Phi$ at boundaries, charge should be "calculable". HINT?\\
Finally, separable solutions in cylindrical coordinates:\\
$\Phi(\rho,\phi,z)=R(\rho)Q(\phi)Z(z)=\sum_{mn}J_m(k_{mn}\rho)\sinh(k_{mn}z)(A_{mn}\sin{m\phi}+B_{mn}\cos{m\phi})$\\
$\frac{1}{|\vec{x}-\vec{x}'|}=\frac{2}{\pi}\int{}dk e^{im(\phi-\phi')}\cosh{z-z'}I_m(k\rho_<)K_m(k\rho_>)$\\Emphasis on Spherical Harmonics, but he will give Bessel Function requirements.\\

\boxaround[Chapter 4]
\noindent
Multipole Expansion;Spherical Harmonics.  Given $\rho(\vec{x}')$,
$\Phi(\vec{x})=\frac{1}{4\pi\epsilon_0}\sum_{lm}\frac{4\pi}{2l+1}q_{lm}\frac{Y_{lm}(\theta,\phi)}{r^{l+1}}$\\
$\Phi(\vec{x})=[\frac{Q}{r}+\frac{\vec{p}\cdot\vec{x}}{r^3}+\frac{1}{2}\sum_{ij}Q_{ij}\frac{x_ix_j}{r^5}+...]$ 
Charge,Dipole,Quadropole terms...\\
$\vec{E}(\vec{x})=-\vec{\nabla}\Phi=\frac{1}{4\pi\epsilon_0} [\frac{Q\cdot\vec{r}}{r^2}+\frac{3\hat{x}(\vec{p}\cdot\hat{x}-\vec{p})}{|\vec{x}|^3}+...]$\\
Localized q in slow-varying $\vec{E}$ field.
$W=q\Phi(0)-\vec{p}\cdot\vec{E}(0)-\frac{1}{6}\sum_{ij}Q_{ij}\frac{\partial \vec{E}(0)}{\partial x_i}|_{\vec{x}=0}+...$\\
How to calculate $q=\int \rho(\vec{x})d\vec{x}$,$\quad$ $\vec{p}=\int \vec{x}\rho(\vec{x})d\vec{x}$\\
Dialectric Media:\\
$\vec{D}=\epsilon_0\vec{E}+\vec{P}$ where $\vec{P}=\epsilon_0\chi_{el}\vec{E}$ Polarization Density for linear media where $\chi$ is the polarizability of the material.\\
$\vec{D}=\epsilon_0(1+\chi_{el})\vec{E}=\epsilon\vec{E}$\\

\boxaround[Chapter 5]
\noindent
Ampere's Law in integral form: $\oint_{c} \vec{B}\cdot dl=\mu_0 \oint_{s} \vec{J}(\vec{x})\cdot da$\\
Two current loops,$\qquad$ $\vec{F}_{12} = -\frac{\mu_0}{4\pi}\int \frac{I_1I_2(d\vec{l}_1-d\vec{l}_2)\vec{x}_{12}}{|\vec{x}_{12}|^3}$\\
No Magnetic Charge! $\vec{\nabla}\cdot\vec{B}=0$, implies, $\vec{B}=\vec{\nabla}\times\vec{A}$ then, $\quad$ $\vec{\nabla}^2\vec{A}=\mu_0\vec{J}(\vec{x})$, $\,$
$\vec{A}=\frac{\mu_0}{4\pi}\int\frac{\vec{J}(\vec{x})}{|\vec{x}-\vec{x}'|}d\vec{x}'$\\
Localized current distribution - looking from far field, leading order term of expansion is given by Magnetic Dipole because there is no magnetic charge term!\\
\scshape Far Field Case:\upshape $\qquad$
$\vec{A}=\frac{\mu_0}{4\pi}\frac{\vec{m}\times\vec{r}}{r^3}$,$\quad$ $\vec{B}=\frac{\mu_0}{4\pi}\frac{3\hat{x}(\vec{m}\cdot\hat{x})-\vec{m}}{|\vec{x}|^3}$\\
Simple Loop, $\vec{m}=I \vec{A}$, $\vec{A}$ is area. (Direction given by right hand rule)\\
Notice factor of $\frac{1}{2}$ difference in equation for magnetic dipole vs. electric dipole:\\
$\vec{m}=\frac{1}{2}\int\vec{x}\times\vec{J}(\vec{x})d\vec{x}$ $\quad$ vs. $\quad$
$\vec{p}=\int\vec{x}\cdot\rho(\vec{x})d\vec{x}$\\
Energy stored in system for magnetic moment in slow-varying external field:\\
$W=-\vec{m}\cdot\vec{B}_0$ Force calculated by gradient of the energy just like in Mechanics.\\
Magnetization Density Analogous to Polarization Density\\
Magnetic Field: $\vec{H}=\frac{1}{\mu_0}\vec{B}-\vec{M}$, $\qquad$ $\vec{M}=\chi_m\vec{H}$ 
Linear media: $\vec{H}=\frac{1}{\mu}\vec{B}$\\
$\vec{\nabla}\times\vec{H}=\vec{J}(\vec{x})$ We specifically study the case where there is no current, namely $\vec{J}(\vec{x})=0$.  Then we can write:\\
$\vec{H}=-\vec{\nabla}\Phi_m$ Which says that the magnetic field is the gradient of the magnetic scalar potential. (Only if $\vec{J}(\vec{x})=0$!)\\
$\Phi_m=-\frac{1}{4\pi}\vec{\nabla}\int \frac{\vec{m}(\vec{x}')}{|\vec{x}-\vec{x}'|}d \vec{x}$ Note: If $\vec{M}$ is continuous, then the gradient doesn't contribute.\\
Time Dependence $\to$ Faraday's Law!  $\quad$
$\epsilon_{mf}=-\frac{\partial\vec{F}}{\partial t}$ A change in flux density \scshape GENERATES \upshape $\epsilon_{mf}$\\
$\vec{\nabla}\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}$ $\quad$
$Flux=\int \vec{B}\cdot d\vec{l}$ Know how to calculate\\
Introduce displacement current in Ch. 6 (Now we have full Maxwell Eq's)\\
Energy of EM Field:\\
$W=\frac{1}{2}\int (\vec{E}\cdot\vec{D}+\vec{H}\cdot\vec{B})d\vec{x}$ $\quad$
Electric part + Magnetic part (inside the parenthesis is the energy density, $w$.)\\
$W=\frac{1}{2}\sum_{ij}C_{ij}V_iV_j$ where $C_{ii}=$Capacitance and $C_{ij}=$coefficient of induction\\
Lots of loops:$\quad$
$W=\frac{1}{2}\sum_{ij}L_iL_j$ where $L_{ii}=$Self Inductance and $L_{ij}=$Mutual Inductance\\

\boxaround[Chapter 6]
\noindent
$\frac{\partial\vec{D}}{\partial t}$ added to $\vec{\nabla}\times\vec{B}$ and
Gauge Transformations (Lorentz Conditon) and Coulumb Gauge.\\
All obey "Wave Equations"$\qquad$
$\vec{\nabla}^2\vec{A}-\frac{1}{c^2}\frac{\partial^2\vec{A}}{\partial t^2}=\mu_0 \vec{J}(\vec{x})$\\
\boxaround[PROBLEMS]
An approx. : $\frac{1}{|\vec{x}+R\hat{z}|}=\frac{1}{r^2+R^2+2rR\cos[\theta]^\frac{1}{2}}=\frac{1}{\frac{1}{R}(\frac{r}{R}^2+1+\frac{2r}{R}\cos[\theta])^\frac{1}{2}}$\\ If $R>>r,a \approx \frac{1}{R}[1-\frac{r}{R}\cos[\theta]]\quad
\frac{1}{1+x+x^2}\approx\frac{1}{\sqrt{1+x}}\approx\frac{1}{1+\frac{1}{2}x}\approx1-\frac{1}{2}x$ \\ $\frac{1}{|\vec{x}+\frac{a^2}{R^2}R\hat{z}|}\approx\frac{1}{R}[\frac{R}{r}-\frac{a^2}{r^2}\cos[\theta]]$ $\quad \cos[\gamma]=\frac{\vec{x}\vec{x'}}{|\vec{x}||\vec{x'}|}=\cos[\theta]\cos[\theta']+\sin[\theta]\sin[\theta']\cos[\phi-\phi']$\\
Spherical shell evaluate on z axis: $\Phi(z)=V[1-\frac{z^2-a^2}{z\sqrt{z^2+a^2}}]$ at $z=a$ $\Phi=V$ at $z\rightarrow\infty$ $\Phi\approx\frac{3Va^2}{2z^2}$\\
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%griffiths sep. of vars in rect:
\boxaround[ex. 3.5 griffiths - infinitely long rectangular metal pipe sides a b is grounded but at x=0 has potential v(y,z)]
3d separation of vars:
$\frac{\partial^2 \Phi}{\partial x^2}+\frac{\partial^2 \Phi}{\partial y^2}+\frac{\partial^2 \Phi}{\partial z^2}=0$\\
$\Phi=X(x)Y(y)Z(z) \rightarrow X(x)=A e^{\sqrt{k^2+l^2} x}+B e^{-\sqrt{k^2+l^2}}$\\
$Y(y)=C\sin(ky)+D\cos(ky) \quad
Z(y)=E\sin(ly)+F\cos(ly)$\\
$A=0,D=0,F=0,k=\frac{n\pi}{a},l=\frac{m\pi}{b}$\\
$\Phi(x,y,z)=\sum_{nm}^{\infty}C_{nm}e^{-\pi\sqrt{\frac{n}{a}^2+\frac{n\pi}{a}^2}x}\sin\frac{n\pi y}{a}sin\frac{m\pi z}{b}$\\
Solve for $x=0$ Multiply both sides by $\sin\frac{n'\pi y}{a}sin\frac{m'\pi z}{b}$ and use orthogonality\\
$C_{nm}=\frac{4}{ab}\int_0^a\int_0^b V_0(y,z)\sin\frac{n\pi y}{a}sin\frac{m\pi z}{b}dydz$\\
If V0 is constant: then you have 2 options for $C_{nm}$ -- if either n or m is even, $C_{nm}=0$ otherwise (both n and m are odd) $C_{nm}=\frac{16V_0}{\pi^2nm}$\\
Final result:
$\Phi(x,y,z)=\frac{16V_0}{\pi^2nm}\sum_{nm=odd}^{\infty}\frac{1}{nm}e^-\pi\sqrt{\frac{n}{a}^2+\frac{n\pi}{a}^2}x\sin\frac{n\pi y}{a}sin\frac{m\pi z}{b}$\\With the note that a good approx is only a few terms.\\
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%prob 2.26
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\boxaround[2d wedge w/ circle radius $\beta$ in the middle (problem 2.26)]
2d separation of vars in cylindrical. find potential inside wedge such that all sides are zero potential.
start with laplace eq in cylindrical:\\
$\frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho)\frac{\partial\Phi}{\partial\rho}+\frac{1}{\rho^2}\frac{\partial^2\Phi}{\partial\phi^2}=0$(no charge) 
Break up $\Phi=R(\rho)\Psi(\phi)$
$\frac{\rho}{R}\frac{\partial}{\partial \rho}(\rho)\frac{\partial R}{\partial\rho}=\nu^2$\\
$\frac{1}{\Psi}\frac{\partial^2\Psi}{\partial\phi^2}=-\nu^2$\\
General solutions:
$R(\rho)=C\rho^{\nu}+D\rho^{-\nu}$ and $\Psi(\phi)=A\cos\nu\phi+B\sin\nu\phi$\\
Boundaries: $\Phi=0$ at $\rho=a$, $\phi=0$,$\phi=\beta$ - plug in reduce eq to:\\
$\Phi(\rho,\phi)=\sum_n^{\infty}A_n sin(\nu\phi)(\rho^{\nu}-a^{2\nu}\rho^{-\nu})$ $\nu=\frac{n\pi}{\beta}$\\
Want $E_{\rho}$ and $E_{\phi}$ and $\sigma(\rho,0),\sigma(\rho,\beta),\sigma(a,\phi)$\\
$\vec{E}=-\nabla\Phi \rightarrow E_{\rho}\approx -A_1\sin\frac{\pi\phi}{\beta}(\nu\rho^{\nu-1}+\nu a^{2\nu}\rho^{-\nu-1})$\\
$E_{\phi}=\frac{1}{\rho}\frac{\partial\Phi}{\partial\phi}\approx -\frac{A_1 \phi}{\beta}\cos\frac{\pi\phi}{\beta}(\rho^{\nu-1}-a^{2\nu}\rho^{-\nu-1})$\\
Define region 1 where you're at the top piece of the wedge ($\phi=\beta$) region 2 is on the circle, $\rho=a$ and region 3 is on the bottom piece of the wedge($\phi=0$)\\
Region 1:\\
$\sigma(\rho,\phi=\beta)=\epsilon_0 E_{\rho}=0,\quad\epsilon_0 E_{\phi}=A_1\frac{\epsilon_0\pi}{\beta}(\rho^{\nu-1}-a^{2\nu}\rho^{-\nu-1})$\\
Region 2:\\
$\sigma(\rho=a,\phi)=\epsilon_0 E_{\rho}=-2A_1\epsilon_0\sin(\frac{\pi\phi}{\beta})\nu a^{\nu-1},\quad E_{\phi}=0$\\
Region 3:\\
$\sigma(\rho,\phi=0)=\epsilon_0 E_{\rho}=0,\quad\epsilon_0 E_{\phi}=-A_1\frac{\epsilon_0\pi}{\beta}(\rho^{\nu-1}-a^{2\nu}\rho^{-\nu-1})$\\
If $\beta=\pi$ then you have a flat line with a bump of radius $a$ and by resolving components we get a constnat in the $\hat{z}$ direction.\\
$E=A_1^2\sin(\phi)^2+A_1^2\cos(\phi)^2 \rightarrow \vec{E}=\sqrt{A_1^2}\hat{z}$ To find charge on bulb:\\
$Q=\int_0^{\pi}\sigma\cdot ad\phi$ so $Q_{bulb}=-4A_1\epsilon_0 a$, $Q_{slab}=-2\epsilon_0 a A_1$\\
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%probs 3.7 and 3.9
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\boxaround[Problem 3.7]
Potential with 3 charges: +q at a$\hat{z}$ -2q at origin, and +q at -a$\hat{z}$. Potential is:\\
$\Phi=\frac{1}{4\pi\epsilon_0}\left[\frac{q}{|r-a|}-\frac{2q}{|r-0|}+\frac{q}{|r+a|}\right]$
Simplifying...\\
$\Phi=\frac{q}{4\pi\epsilon_0}\left[\frac{1}{|r-a\hat{z}|}+\frac{1}{|r+a\hat{z}|}-\frac{2}{r}\right]$\\
Now we can expand the Green function in terms of spherical harmonics:
$\frac{1}{|r-r'}\rightarrow \sum_0^{\infty}\frac{r_{\langle}^l}{r_{\rangle}^{l+1}}P_l(\cos\theta)$\\
Then for l is even we get
$\Phi=\frac{q}{2\pi\epsilon_0}(\sum_0^{\infty}\frac{r_{<}^l}{r_{>}^{l+1}}P_l(\cos\theta)-\frac{1}{r})$\\
$a\rightarrow0$ so $r_{\langle}=a$ and $r_{\rangle}=r$ Expand for 2 terms.. l=0 and l=2\\
$\Phi=\frac{q}{2\pi\epsilon_0}\left(\frac{a^0}{r^{0+1}}P_0(\cos\theta)+\frac{a^2}{r^{2+1}}P_2(\cos\theta)-\frac{1}{r}\right)$\\
It's easy to see that the first term is just $\frac{1}{r}$ and this will cancel with the last term and we'll be left with:
$\Phi=\frac{q}{2\pi\epsilon_0}\left(\frac{a^2}{r^3}P_2(\cos\theta)\right)$\\
Which reduces to
$\Phi=\frac{Q}{4\pi\epsilon_0 r^3}(3\cos\theta^2-1)$ after plugging in $P_2(\cos\theta)$\\
Enclose this thing with a spherical shell use image charges...\\
$\Phi=\frac{q}{4\pi\epsilon_0}\left[\frac{1}{|r-a=\hat{z}|}+\frac{1}{|r+a\hat{z}|}-\frac{2}{r}\right]+\frac{q'}{4\pi\epsilon_0}\left[\frac{1}{|r-\hat{z'}|}+\frac{1}{|r+\hat{z'}|}-\frac{2}{r}\right]$\\
Sub in $q'=-\frac{b}{a}q$ and $z'=\frac{b^2}{a}$ -- I've already done some reducing... then we're going to again replace in terms of spherical harmonics...Also for the 2nd term I've put in r lesser as r and r greater as $\frac{b^2}{a}$\\
$\Phi=\frac{q}{2\pi\epsilon_0}\left[-\frac{1}{r}+\frac{1}{b}+(\sum_0^{\infty}\frac{r_{<}^l}{r_{>}^{l+1}}-\frac{1}{b}\left(\frac{ra}{b^2}\right)^l P_l(\cos\theta)\right]$\\
Again only survives for even l, so we take two terms: $l=0$ and $l=2$ to get:\\
$\Phi=\frac{q}{2\pi\epsilon_0}\left(\frac{a^2}{r^3}-\frac{r^2a^2}{b^5}\right)P_2(\cos\theta)$\\
which we can simplify for final answer:
$\Phi=\frac{Q}{2\pi\epsilon_0 r^3}(1-\frac{r^5}{b^5})P_2(\cos\theta)$\\
\boxaround[Problem 3.9 - Tube with $\Phi=0$ on top and bottom and $\Phi=V(\phi,z)$ on side]
Assume separable solutions: $\Phi=R(\rho)Q(\phi)Z(z)$ \\
For Z, $\frac{\partial^2 Z}{\partial z^2}=-k^2 Z \rightarrow Z(z)=A\sin(kz)+B\cos(kz)$\\
But with our BC, $z=0, \Phi=0$ so B must be zero.$Z(z)=A\sin(kz)$ where $k=\frac{n\pi}{L}$ because $z=L, \Phi=0$.\\
For Q, $\frac{\partial^2 Q}{\partial \phi^2}=m^2 Q \rightarrow Q(\phi)=C\sin(m\phi)+D\cos(m\phi)$ But $m=0$ because of azimuthal symmetry so the C term is gone.\\
For R, $\frac{\partial^2 R}{\partial \rho^2}+\frac{1}{\rho}\frac{\partial R}{\partial \rho}-(k^2+\frac{m^2}{\rho^2})R=0$ Substitute $x=k\rho$ And the results for R are modified Bessel functions:$R(x)=E I_m(x)+ F K_m(x)$ but we have a finite potential at $s=0$ and we have an interior problem so F must be zero, because $K_m(x)$ doesn't play nice. Therefore, $R(x)=E I_m(x)$\\
Now we have before using last BC for $\rho=b$:\\
$\Phi(\rho=b,\phi,z)=RQZ=\sum_{nm}^{\infty} E I_m(\frac{n\pi}{\rho})A\sin(\frac{n\pi}{L}z)$\\
$\Phi(\rho,\phi,z)=\sum_{nm}^{\infty} I_m(\frac{n\pi}{\rho})A\sin(\frac{n\pi}{L}z)\frac{2V}{L} I_m\frac{n\pi}{b}\int_0^L\sin(\frac{n\pi}{L}z)dz$\\
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%probs 4.1 6 7 8 10
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\boxaround[Problem 4.1]
calculate multipole moments for all l and 2 sets of nonvanishers
+q at $a\hat{x}$ and $a\hat{y}$ and -q at $-a\hat{x}$ and $-a\hat{y}$\\
Figure out $\rho$ which is just the sum of all the charges at their respective locations in $\theta$ and $\phi$ because all of them have $r=a$\\
$q_{lm}=q a^l \left[ Y_l^{m*}(\frac{\pi}{2},0)+Y_l^{m*}(\frac{\pi}{2},\frac{\pi}{2})-Y_l^{m*}(\frac{\pi}{2},\pi)-Y_l^{m*}(\frac{\pi}{2},\frac{3\pi}{2})\right]$\\
Since all have $\theta=\frac{\pi}{2}$ and all $Y_{lm}$ have cosine dependence, they will be $P_l(0)$ We can use this to simplify... we learn that only m=odd survive.\\
$q_{lm}=q a^l \sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}P_l^m(0)2[1-i^m]$
plug and chug $l=1$,$l=3$ should get $q_{lm}=-q_{lm}^*$ and results specifically as $q_{11},q_{31},q_{33}$\\
If you do the same problem we did in 3.7 with 3 charges all on z, only terms are $q_{20},q_{40}$ because azimuthal symmetry and m=0.\\
To get multipole expansion, take answer $q_{20}$ and plug into multiple expnasion -- reduce $P_l(\cos\theta)$ to $P_l(0)$ again because we're on xy plane and $l\ne0$ because we've got $Q=0$ Answer is:\\
$\Phi(r,\theta,\phi)=\frac{1}{2\pi\epsilon_0}\frac{qa^2}{r^3}(\frac{1}{2}(3(0)^2-1))\rightarrow-\frac{1}{4\pi\epsilon_0}\frac{qa^2}{r^3}$\\
\boxaround[Problem 4.6]
traceless quadrupole tensor - use $Q_{11}=Q_{22}=-\frac{1}{2}Q_{33}$ and $\frac{Q_{33}}{e}=Q$ to get $W=-\frac{e}{4}Q\frac{\partial E_z}{\partial z}$ from $W=\int\rho(\vec{x})\Phi(\vec{x})d\vec{x}$ but for quadrupole term...\\ $W=-\frac{1}{6}\sum_{ij}Q_{ij}\frac{\partial E_i}{\partial x_j}$\\
\boxaround[Problem 4.7]
localized charge distribution - right away we make our lives easier by realizing that there's a $\sin(\theta)^2$ term in the volume charge so we should probably rewrite in terms of spherical harmonics.\\
$\rho(\vec{r})={1}{64\pi}r^2e^{-r}\sin(\theta)^2, r\rightarrow\infty, \rho(\vec{r})=0$\\
$\sin(\theta)^2=1-\cos(\theta)^2=\frac{2}{3}(P_0(\cos(\theta))-P_2(\cos(\theta)))$ - from this we know only $l=0$ and $l=2$ term survive from $q_{lm}$ because of orthogonality. should get $q_{00}=\frac{1}{2\sqrt{\pi}}$ and $q_{20}=-6\sqrt{\frac{5}{4\pi}}$\\
The essence of multipole expansion is only valid in the \scshape far field case \upshape.
$\Phi=\frac{1}{4\pi\epsilon_0}\left[\frac{1}{r}-\frac{6}{r^3}P_2(\cos(\theta))\right]$\\
Now to find $\Phi$ at any point, we have to use the freespace green function -- but wait ! -- this can ALSO be expanded in terms of spherical harmonics, and solved just like part a\\
OR we can rewrite it in terms of no $\phi$ dependence (ie, $m=0$) and use this form:
$G(x,x')=\sum_l\frac{r_{\langle}^l}{r_{\rangle}^{l+1}}P_l(\cos\theta)P_l(\cos\theta')$ Then we have orthogonality again with $l=0$ and $l=2$ but we have to be careful because now our $\int_0^{\infty}dr$ integral must be split to $\int_0^r dr'$ and $\int_r^{\infty} dr$\\
$\Phi(r,\theta,\phi)=\frac{1}{4\pi\epsilon_0}\frac{1}{24}\left[\frac{1}{r}(...+...)\right]$ and for far field, $e^{-r}=0$ and reduces back to part a) -- also for close, you have to taylor expand -- perhaps up to $r^5$\\
If you see a function of $\theta$ it is desirable to express in terms of Spherical Harmonics -- might greatly reduce the problem.  Multipole expansion is only valid for far field case -- not local field.\\
\boxaround[Problem 4.8concentric cylinder filled with dielectric material.]
Solve potential in three regions - inner circle, in between a and b and outside b (inner radius a, outer radius b)\\
$\Phi_I=\sum_{m=0}^{\infty} F_m \rho^m\cos m\phi$ (includes origin - so 2nd term is not allowed)\\
$\Phi_{II}=\sum_{m=0}^{\infty}B_m \rho^m+ D_m \rho^{-m} \quad \Phi_{III}=\sum_{m=0}^{\infty}A_m \rho^m+ C_m \rho^{-m} $\\
Far field case, this should just look like the regular electric field -- $-E_o\rho\cos\phi$\\
All $m>2$ $A_m$ will have to be zero, otherwise first term blows up.Also, all the initial constants are not important in finding $\vec{E}$ because we'll be taking derivative of $\Phi$\\
$\Phi_{I}= F\rho\cos\phi, \Phi_{II}=(B\rho+\frac{D}{\rho})\cos\phi,\Phi{_III}=(\frac{C}{\rho}-E_0\rho)\cos\phi$
Take E as negative gradient of potential:
$E_{\rho}=E_{I}=-F\cos\phi, E_{II}=(D\rho^{-2}-B)\cos\phi, E_{III}=(C\rho^{-2}+E_0)\cos\phi$\\
$E_{\phi}=E_{I}=F\sin\phi, E_{II}=(B+D\rho^{-2})\sin\phi, E_{III}=(C\rho^{-2}-E_0)\sin\phi$\\
D-Boundary and E-Boundary must be continuos across these jumps.\\
$\epsilon_0E_{III}(b)=\epsilon E_{II}(b), E_{III}(b)=E_{II}(b) \quad
\epsilon_0E_{I}(a)=\epsilon E_{II}(b), E_{I}(a)=E_{II}(a)$\\
Four constants, four unknown - solve. this is gross.\\
\boxaround[Problem 4.10concentric cylinder 1/2 dielectric material, 1/2 with empty space]
If E is in the radial dierction, then the normal component is automatically zero and both boundary conditions are continuous.
$\oint\vec{D}\cdot\hat{n}da=Q_{enclosed}$\\ Define the Gaussian surface justoutside of a (inner radius) and get$\vec{E}=\frac{Q}{2\pi r^2(\epsilon+\epsilon_0)}$\\
To get $\sigma$ on inner surface, use boundary condition  $(\vec{D}_2-\vec{D}_1)\cdot n_{12}=\sigma$ with $\vec{D}_1$ zero in both cases because inside of a there is no charge.and we get:\\
$\sigma_{R}(\epsilon) \rightarrow \frac{Q}{2\pi r^2(1+\frac{\epsilon_0)}{\epsilon}}(\epsilon\vec{E}) \quad
\sigma_{L}(\epsilon_0) \rightarrow \frac{Q}{2\pi r^2(1+\frac{\epsilon)}{\epsilon_0}}(\epsilon_0\vec{E})$\\
Calculate prob. density induced on the surface at r=a, 
$\vec{P}=\epsilon_0 \chi_e \vec{E}$\\
Noting that all charge comes from the dielectric side - and that the normal is inward $-\hat{r}$ then we get 
$\vec{P}=-(\epsilon-\epsilon_0)E_r$ or $\vec{P}=-\frac{Q}{2\pi r^2}\frac{\epsilon-\epsilon_0}{\epsilon+\epsilon_0}$ and the charge on this would be + if it were asking at the other side (b) -- lots of little + and - cancel out in the middle, net effect on edges.\\
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%probs 5.3 6 7 22 19 little of 14
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\boxaround[Problem 5.3]
solenoid of N turns - find B in terms of $\theta_1$ and $\theta_2$ For one loop of wire, above z axis all contributions for B field cancel except the z component. Just use ampere's law. $B(z)=\frac{\mu_0}{4\pi}I\int\frac{dl'}{|\vec{x}|^2}\cos\theta$\\
$B(z)=\frac{\mu_0}{2}\frac{a^2}{\sqrt{a^2+z^2}}$ and we can integrate this to figure out solenoid. to get in terms: $B(z)=\frac{\mu_0 I N}{2}(\cos\theta_1+\cos\theta_2)$\\
\boxaround[Problem 5.6]
magnetic flux density, or more familiar,  Find B.\\
Make one of radius a going one direction and current going other direction for radius b and add them together.  when you add in the vector d which separates the centers of the circles, it turns out not to matter where the hole is. $\vec{B}_{hole}(\vec{r})=\frac{1}{2}\mu_0 \vec{J}(\hat{z}\times\vec{d})$\\
\boxaround[Problem 5.7]
so for a part we get the same thing as before on Bz for 1 loop -- when we add a second loop a distance b away, things fer more interesting -- esp. if we put the origin at the halfway point.  Restricting to the center is crying for an expansion. $\sum_0^{\infty}a_n (x-c)^n$ c is usually zero.\\
\boxaround[Problem 5.22]
based on prob 5.19a) -- $B_e=\frac{\mu_0}{2}\left(\frac{z+L}{\sqrt{(z+L)^2+a^2}}-\frac{z}{\sqrt{z^2+a^2}}\right)\hat{z}$
We need to know this at B(0) and B(L) because the force on this cylinder will be nonzero there.  Otherwhere it will be zero because n is perpendicular to M and the M is uniform.\\
In the limit $L>>a$ we add these two up and get that the 1st one cancels with the 2nd term of the 2nd one\\
$F\approx \frac{\mu_0}{2} M^2 A$ relate this to $\Delta W=-\frac{\sigma^2\Delta a\Delta x}{2\epsilon_0}$ for electrostatics.$W=F\Delta x$ and\\
$W=\frac{1}{2}\int_v1\vec{M}\cdot\vec{B}_0d\vec{x}$ analogous to $W=-\frac{1}{2}\int_v1\vec{P}\cdot\vec{E}_0d\vec{x}$\\
$\Phi_M(\vec{x})=-\frac{1}{4\pi}\int_v \frac{\nabla\cdot\vec{M}(\vec{x}')}{|\vec{x}-\vec{x}'|}d\vec{x}'+\frac{1}{4\pi}\oint_S\frac{\vec{n}'\cdot\vec{M}(\vec{x})da'}{|vec{x}-vec{x}'|}$\\
First term goes to zero with uniform M!\\
\boxaround[Problem 5.13]
random tidbit: C for concentric cylinders $C=\frac{2\pi\epsilon_0 L}{ln(\frac{b}{a}}$ and $F=+\frac{\partial W}{\partial t}$ instead of normal -.
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