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docs/physics/electrostatics/4-conductors.md

@@ -12,11 +12,11 @@ We note that there is no electric fiend inside a conductor, as charges internal
 
 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.
 
-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 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.
+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.
 
-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 parallel voltage difference is zero, so the electric field must be zero.
+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.
 
-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, $\int_V \div{\vb{E}} \dd{V} = \frac{Q_e}{\varepsilon_0} = \frac{A\sigma_e}{\varepsilon_0}$. Thus, $\sigma_e = \varepsilon_0 E$.
+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, $\sigma_e = \varepsilon_0 E$.
 
 ## Section 4.3 - Exercises involving conductors at fixed potentials
 
@@ -24,23 +24,23 @@ Consider a square with left and right potentials $V(0, y) = V(l, y) = V_1$ and $
 
 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.
 
-**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.
+**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.
 
-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,
+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,
 
 $$
-\int_{\dd{\Omega}}[\Phi(\vb{r})\grad{\Phi{\vb{r}}}] \vdot \vu{n} \dd{S}
-= \sum_{i=1}^N \Phi_i \int_{\dd{\Omega_i}} \grad{\Phi{\vb{r}}} \vdot \vu{n} \dd{S}
+\int_{d{\Omega}}[\Phi(\mathbf{r})\nabla{\Phi{\mathbf{r}}}] \cdot \hat{\mathbf{n}} d{S}
+= \sum_{i=1}^N \Phi_i \int_{d{\Omega_i}} \nabla{\Phi{\mathbf{r}}} \cdot \hat{\mathbf{n}} d{S}
 $$
 
 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,
+**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,
 
 $$
-\div \vb{P} = - \rho_{eb}
+\nabla \cdot \mathbf{P} = - \rho_{eb}
 $$
 
 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.
@@ -48,19 +48,19 @@ This field is zero outside of a material, and if non-zero inside a material, wil
 **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,
 
 $$
-\varepsilon_0 \div \vb{E} = \rho_{ef} + \rho_{eb}
+\varepsilon_0 \nabla \cdot \mathbf{E} = \rho_{ef} + \rho_{eb}
 $$
 
-**Definition**. The electric flux density field $\vb{D}$ is defined as
+**Definition**. The electric flux density field $\mathbf{D}$ is defined as
 
 $$
-\vb{D} = \varepsilon_0 \vb{E} + \vb{P}
+\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P}
 $$
 
-Both $\vb{D}$ and $\vb{P}$ have units of Coulombs/m^2. Additionally, we see that
+Both $\mathbf{D}$ and $\mathbf{P}$ have units of Coulombs/m^2. Additionally, we see that
 
 $$
-\div \vb{D} = \div (\varepsilon_0 \vb{E} + \vb{P}) = \div \varepsilon_0 \vb{E} + \div \vb{P} = (\rho_{ef} + \rho_{eb}) - \rho_{eb} = \rho_{ef}
+\nabla \cdot \mathbf{D} = \nabla \cdot (\varepsilon_0 \mathbf{E} + \mathbf{P}) = \nabla \cdot \varepsilon_0 \mathbf{E} + \nabla \cdot \mathbf{P} = (\rho_{ef} + \rho_{eb}) - \rho_{eb} = \rho_{ef}
 $$
 
 ## Section 4.5 - Induced Electric Charges, their Potentials and Fields
@@ -79,4 +79,4 @@ This unit, $\frac{C}{V}$, is known as a Farad. For a sphere, $C = 4 \pi \varepsi
 
 ## Section 4.7 - Forces on Charged Conductors in Electric Fields
 
-We know that $\vb{F} = \int \vb{E}_{ext}(\vb{r}) \rho_e(\vb{r}) dV$, where $\vb{E}_{ext}(\vb{r})$ is the external electric field and $\rho_e(\vb{r})$ is the charge density of the object.
+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.