83 lines
5.4 KiB
Markdown
83 lines
5.4 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 throughout 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 symmetric 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 $\int_A^B \mathbf{E} \cdot d{\mathbf{l}}$. Since $\mathbf{E} = 0$ inside the conductor, the voltage 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 $d{\mathbf{l}}$. Now, $\mathbf{E} \cdot d{\mathbf{l}} = \mathbf{E}_s \cdot d{\mathbf{l}}_s + \mathbf{E}_p \cdot d{\mathbf{l}}_p = d{V_s} + d{V_p}$, in terms of parallel and perpendicular components. The parallel 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 cylinder 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, $\int_V \nabla \cdot \mathbf{E} d{V} = \frac{Q_e}{\varepsilon_0} = \frac{A\sigma_e}{\varepsilon_0}$. Thus, $\sigma_e = \varepsilon_0 E$.
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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 symmetry, $V(r, \theta) = \sum_{l=0}^\infty a_l r^l P_l(cos\theta)$ 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 $\int_{d{\Omega}}[\Phi(\mathbf{r})\nabla{\Phi{\mathbf{r}}} \cdot \hat{\mathbf{n}} d{S} = 0]$. With this, consider a surface $d{\Omega}$ 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 $\Phi(\mathbf{r})$ as the difference between any two potential solutions to the Laplace equation at point $\mathbf{r}$. Since potential must be a constant,
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$$
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\int_{d{\Omega}}[\Phi(\mathbf{r})\nabla{\Phi{\mathbf{r}}}] \cdot \hat{\mathbf{n}} d{S}
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= \sum_{i=1}^N \Phi_i \int_{d{\Omega_i}} \nabla{\Phi{\mathbf{r}}} \cdot \hat{\mathbf{n}} d{S}
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$$
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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 $\mathbf{P}$. Additionally, we note the charge density of bound charges is $\rho_{eb}$. Thus,
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$$
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\nabla \cdot \mathbf{P} = - \rho_{eb}
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$$
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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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$$
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\varepsilon_0 \nabla \cdot \mathbf{E} = \rho_{ef} + \rho_{eb}
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$$
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**Definition**. The electric flux density field $\mathbf{D}$ is defined as
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$$
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\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P}
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$$
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Both $\mathbf{D}$ and $\mathbf{P}$ have units of Coulombs/m^2. Additionally, we see that
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$$
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\nabla \cdot \mathbf{D} = \nabla \cdot (\varepsilon_0 \mathbf{E} + \mathbf{P}) = \nabla \cdot \varepsilon_0 \mathbf{E} + \nabla \cdot \mathbf{P} = (\rho_{ef} + \rho_{eb}) - \rho_{eb} = \rho_{ef}
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$$
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## Section 4.5 - Induced Electric Charges, their Potentials and Fields
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This is an application chapter.
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## Section 4.6 - Capacitance
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**Definition**. The *capacitance* of an object $C$ is the charge per volt, such that
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$$
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C := \frac{Q}{V}
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$$
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This unit, $\frac{C}{V}$, is known as a Farad. For a sphere, $C = 4 \pi \varepsilon_0 R$. For a parallel plate capacitor, this reduces to $C = \frac{epsilon_0 A}{d}$.
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## Section 4.7 - Forces on Charged Conductors in Electric Fields
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We know that $\mathbf{F} = \int \mathbf{E}_{ext}(\mathbf{r}) \rho_e(\mathbf{r}) dV$, where $\mathbf{E}_{ext}(\mathbf{r})$ is the external electric field and $\rho_e(\mathbf{r})$ is the charge density of the object.
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