Physics 4.3-4

docs/physics/electrostatics/4-conductors.md

@@ -17,3 +17,48 @@ Consider any two points internal to the conductor. The voltage between said poin
 The electric field at the surface of a conductor is perpendicular to its surface. Consider some displacement $\dd{\vb{l}}$. Now, $\vb{E} \vdot \dd{\vb{l}} = \vb{E}_s \vdot \dd{\vb{l}}_s + \vb{E}_p \vdot \dd{\vb{l}}_p = \dd{V_s} + \dd{V_p}$, in terms of parallel and perpendicular components. The paralell voltage difference is zero, so the electric field must be zero.
 
 Consider the surface of a conductor with surface charge density $\sigma_e$. A cyliner with one end inside and one end outside said surface, with its axis normal to said surface, will be a Gaussian "pillbox", which will show that with V being the volume of the pillbox, $\int_V \div{\vb{E}} \dd{V} = \frac{Q_e}{\epsilon_0} = \frac{A\sigma_e}{\epsilon_0}$. Thus, $\sigma_e = \epsilon_0 E$.
+
+## Section 4.3 - Exercises involving conductors at fixed potentials
+
+Consider a square with left and right potentials $V(0, y) = V(l, y) = V_1$ and $V(x, 0) = V(x, l) = V_2$. Since we are uniform in $z$, we can say that $V(x, y) = X(x)Y(y)$ and apply separation of variables.
+
+In spherical polar coordinates, we see that with azimuthal symnetry, $V(r, \theta) = \sun_{l=0}^\infty a_l r^l P_l(cos\theta)$ where $P_l(x)$ are Legendre polynomials.
+
+**Theorem**. 4.3.3: A Laplace equation's solution must be unique inside a volume $\Omega$ if $\int_{\dd{\Omega}}[\Phi(\vb{r})\grad{\Phi{\vb{r}}} \vdot \vu{n} \dd{S} = 0]$. With this, consider a surface $\dd{\Omega}$ that surrounds conductors. The integral vanishes if a) the potential is specified on each conductor or b) the total charge on each conductor is specified.
+
+Now, define $\Phi(\vb{r})$ as the difference between any two potential solutions to the Laplace equation at point $\vb{r}$. Since potential must be a constant,
+
+$$
+\int_{\dd{\Omega}}[\Phi(\vb{r})\grad{\Phi{\vb{r}}}] \vdot \vu{n} \dd{S}
+= \sum_{i=1}^N \Phi_i \int_{\dd{\Omega_i}} \grad{\Phi{\vb{r}}} \vdot \vu{n} \dd{S}
+$$
+
+Thus, if potential is specified, $\Phi_i$ vanishes for that conductor. If the total charge is instead specified, the gradient vanishes because there is no difference in charge between any two points.
+
+## Section 4.4 - Electric Field, Polarization Field, and Flux Density in the Presence of Conductors
+
+**Definition**. A *bound charge* is any charge in a conductor that is bound to an atom and not free to be redistributed at the surface. We say that bound charges are the source of the polarization field $\vb{P}$. Additionally, we note the charge density of bound charges is $\rho_{eb}$. Thus,
+
+$$
+\div \vb{P} = - \rho_{eb}
+$$
+
+This field is zero outside of a material, and if non-zero inside a material, will drop to zero at the surface discontinuously. If there is a component perpendicular to the surface, the discontinuity will generate curl. If there is a component parallel to the surface, it will generate divergence.
+
+**Definition**. Charges not bound are called *free*, with density denoted as $\rho_{ef}$. Combined with $\rho_{eb}$, they form the basis of the electric field. THat is,
+
+$$
+\epsilon_0 \div \vb{E} = \rho_{ef} + \rho_{eb}
+$$
+
+**Definition**. The electric flux density field $\vb{D}$ is defined as
+
+$$
+\vb{D} = \epsilon_0 \vb{E} + \vb{P}
+$$
+
+Both $\vb{D}$ and $\vb{P}$ have units of Coulombs/m^2. Additionally, we see that
+
+$$
+\div \vb{D} = \div (\epsilon_0 \vb{E} + \vb{P}) = \div \epsilon_0 \vb{E} + \div \vb{P} = (\rho_{ef} + \rho_{eb}) - \rho_{eb} = \rho_{ef}
+$$