65 lines
4.5 KiB
Markdown
65 lines
4.5 KiB
Markdown
# Chapter 4 - Conductors and Static Electric Fields
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## Section 4.1 - Introduction
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We will focus primarily on electric fields and charges. For the purposes for this section, we will assume insulators are perfect.
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## Section 4.2 - Electrostatic Properties of a Conductor
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In a metal or conductor, there are plentiful charges not bound to a particular atom and are thus free to move throughought the material.
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We note that there is no electric fiend inside a conductor, as charges internal to the material would move under the force it generates until they find a configuration that eliminates the field. This may happen, but not in electrostatics.
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Additionally, as the field is zero, it follows from Maxwell's equations that there is no charge inside a conductor. However, charge may be present at the surface. For sufficiently symnetric charges, this charge may be calculated.
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Consider any two points internal to the conductor. The voltage between said points is defined as $\int_A^B \vb{E} \vdot \dd{\vb{l}}$. Since $\vb{E} = 0$ inside the conductor, the volage difference must be zero. Thus, any two points in or on the surface (TODO: Why on the surface?) of a conductor must be at the same potential.
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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.
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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, $\sigma_e = \epsilon_0 E$.
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## Section 4.3 - Exercises involving conductors at fixed potentials
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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.
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In spherical polar coordinates, we see that with azimuthal symnetry, $V(r, \theta) = \sum_{l=0}^\infty a_l r^l P_l(cos\theta)$ where $P_l(x)$ are Legendre polynomials.
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**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.
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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,
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$$
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\int_{\dd{\Omega}}[\Phi(\vb{r})\grad{\Phi{\vb{r}}}] \vdot \vu{n} \dd{S}
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= \sum_{i=1}^N \Phi_i \int_{\dd{\Omega_i}} \grad{\Phi{\vb{r}}} \vdot \vu{n} \dd{S}
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$$
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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.
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## Section 4.4 - Electric Field, Polarization Field, and Flux Density in the Presence of Conductors
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**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,
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$$
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\div \vb{P} = - \rho_{eb}
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$$
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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.
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**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,
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$$
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\epsilon_0 \div \vb{E} = \rho_{ef} + \rho_{eb}
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$$
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**Definition**. The electric flux density field $\vb{D}$ is defined as
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$$
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\vb{D} = \epsilon_0 \vb{E} + \vb{P}
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$$
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Both $\vb{D}$ and $\vb{P}$ have units of Coulombs/m^2. Additionally, we see that
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$$
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\div \vb{D} = \div (\epsilon_0 \vb{E} + \vb{P}) = \div \epsilon_0 \vb{E} + \div \vb{P} = (\rho_{ef} + \rho_{eb}) - \rho_{eb} = \rho_{ef}
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$$
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