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docs/physics/electrostatics/2-coulomb.md

@@ -98,11 +98,11 @@ $$
 
 We can verify that $\curl{(\vb{r}-\vb{r'})} = 0$, cancelling the first term. Additionally, $\curl{\frac{1}{\abs{\vb{r}-\vb{r'}}^3}} = -3 \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^5}$, which when crossed with $\vb{r}-\vb{r'}$, will cancel. Thus, all terms in the curl cancel, so for a static field, the curl is zero.
 
-## Section 2.4 - Eletric and Magnetic Flux Densities
+## Section 2.4 - Electric and Magnetic Flux Densities
 
 The electric and magnetic flux density vectors are given by $\epsilon_0 \vb{E}$ and $\mu_0 \vb{H}$.
 
-Now, given $S$ is a surfance enclosing $Q_e$ or $Q_m$ total charge, we denotate flux as following:
+Now, given $S$ is a surface enclosing $Q_e$ or $Q_m$ total charge, we denote flux as following:
 
 $$
 \Phi_e = \epsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd S = Q_m
@@ -131,4 +131,4 @@ $$
 
 **Definition**. This is known as *Gauss' Law*.
 
-With applicable symnetry, the integral factor becomes simply $E(r)*A$, where $A$ is the area of the surface at $r$.
+With applicable symmetry, the integral factor becomes simply $E(r)*A$, where $A$ is the area of the surface at $r$.

docs/physics/electrostatics/3-electro-magnetic-potentials.md

@@ -14,7 +14,7 @@ $$
 \vb{E}(\vb{r}) = \frac{1}{q_e} \vb{F_e}(\vb{r}) = - \frac{1}{q_e} \grad{U_e(\vb{r})} = -\grad{V_e(\vb{r})}
 $$
 
-The units of electrostatic potential is Joule/Coublomb, also known as a Volt. Thus, the units of the electric field should be expressed in Volts/meter. Similarly,
+The units of electrostatic potential is Joule/Coulomb, also known as a Volt. Thus, the units of the electric field should be expressed in Volts/meter. Similarly,
 
 $$
 \vb{H}(\vb{r}) = \frac{1}{q_m} \vb{F_m}(\vb{r}) = - \frac{1}{q_m} \grad{U_m(\vb{r})} = -\grad{V_m(\vb{r})}
@@ -80,7 +80,7 @@ $$
 W_n = \frac{1}{2} \frac{4 \pi \epsilon_0} \sum_{i = 1}^{N} \sum_{j > i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
 $$
 
-For the sake of symnetry, sum overall charges and divide by 2.
+For the sake of symmetry, sum overall charges and divide by 2.
 
 $$
 W_n = \frac{1}{2} \frac{1}{4 \pi \epsilon_0} \sum_{i = 1}^{N} \sum_{j \neq i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
@@ -93,7 +93,7 @@ W_n = \frac{1}{2} \sum_{i = 1}^{N}Q_{ei} \sum_{i \neq j}^{N} \frac{1}{4 \pi \eps
 = \frac{1}{2}\sum_{i = 1}^{N} Q_{ei} V(\vb{r_i})
 $$
 
-We can rewrite this as a Reimann sum and convert to an integral.
+We can rewrite this as a Riemann sum and convert to an integral.
 
 $$
 W_e = \frac{1}{2} \int_V p_e(\vb{r}) V_e(\vb{r}) \dd V ; \quad
@@ -129,7 +129,7 @@ $$
 
 We know that $\vb{E}(\vb{r}) = -\div{V_e(\vb{r})}$ and $\vb{H}(\vb{r}) = -\div{V_m(\vb{r})}$
 
-Combinind this, as well as the first of the Maxwell equations, we see that
+Combined this, as well as the first of the Maxwell equations, we see that
 
 $$
 \div{\vb{E}} = -\div{\grad{V_e}} = - \laplacian{V_e} = \frac{\rho_e}{\epsilon_0}
@@ -139,7 +139,7 @@ $$
 \div{\vb{H}} = -\div{\grad{V_m}} = - \laplacian{V_m} = \frac{\rho_m}{\mu_0}
 $$
 
-The last inequatlity is called the Poisson Equation, or the inhomogenous Laplace equation.
+The last inequality is called the Poisson Equation, or the inhomogeneous Laplace equation.
 
 To solve this equation, we define a Green function as follows:
 
@@ -147,7 +147,7 @@ $$
 \laplacian G(\vb{r}, \vb{r'}) = \delta(\vb{r} - \vb{r'})
 $$
 
-Now, we can construct a potential function in terms of said green function that satisfies the lapalce equation.
+Now, we can construct a potential function in terms of said green function that satisfies the Laplace equation.
 
 $$
 V_e(\vb{r}) = - \int_V G(\vb{r}, \vb{r'}) \frac{\rho_e(\vb{r'})}{\epsilon_0} \dd{V'}
@@ -273,4 +273,4 @@ $$
 
 Note that as a quirk of the function, $P_n(1) = 1$ for all $n$.
 
-We can apply these quadrupole and beyond terms to the volate or other equations, however, this becomes very messy.
+We can apply these quadrupole and beyond terms to the violate or other equations, however, this becomes very messy.

docs/physics/electrostatics/4-conductors.md

@@ -6,23 +6,23 @@ We will focus primarily on electric fields and charges. For the purposes for thi
 
 ## Section 4.2 - Electrostatic Properties of a Conductor
 
-In a metal or conductor, there are plentiful charges not bound to a particular atom and are thus free to move throughought the material.
+In a metal or conductor, there are plentiful charges not bound to a particular atom and are thus free to move throughout the material.
 
 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.
 
-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.
+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 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.
+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.
 
-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.
+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.
 
-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, $\sigma_e = \epsilon_0 E$.
+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, $\int_V \div{\vb{E}} \dd{V} = \frac{Q_e}{\epsilon_0} = \frac{A\sigma_e}{\epsilon_0}$. 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, $V(r, \theta) = \sum_{l=0}^\infty a_l r^l P_l(cos\theta)$ where $P_l(x)$ are Legendre polynomials.
+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.
 
@@ -79,4 +79,4 @@ This unit, $\frac{C}{V}$, is known as a Farad. For a sphere, $C = 4 \pi \epsilon
 
 ## 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 $\vb{F} = \int \vb{E}_{ext}(\vb{r}) \rho_e(\vb{r}) dV$, where $\vb{E}_{ext}(\vb{r})$ is the external electric field and $\rho_e(\vb{r})$ is the charge density of the object.

docs/physics/electrostatics/5-moving-charges.md

@@ -1,6 +1,6 @@
 # Chapter 5 - Electrodynamics with Moving Charges
 
-## Section 5.1 - Currents in Steady-State Regine
+## Section 5.1 - Currents in Steady-State Regime
 
 We want to work in a steady-state system. Thus, we restrict ourselves to currents that do not change in time.
 
@@ -24,7 +24,7 @@ If we assume cylindrical coordinates and that $\vb{H}(vb{r}) = H_\varphi(s) \vu{
 
 ---
 
-By Helmholtz Theorem, we know that $\vb{H}(\vb{r}) = \curl{\vb{A}(\vb{r})}$. For a current-carying wire, $\vb{A}(\vb{r}) = \frac{I_e}{4\pi} \int_{\text{wire}} \frac{\dd{\vb{l'}}}{|\vb{r}-\vb{r'}|}$. Applying identities, we see the *Law of Biot and Savart$, where
+By Helmholtz Theorem, we know that $\vb{H}(\vb{r}) = \curl{\vb{A}(\vb{r})}$. For a current-carrying wire, $\vb{A}(\vb{r}) = \frac{I_e}{4\pi} \int_{\text{wire}} \frac{\dd{\vb{l'}}}{|\vb{r}-\vb{r'}|}$. Applying identities, we see the *Law of Biot and Savart$, where
 
 $$
 \vb{H}(\vb{r}) = \int{I_e}{4\pi}\int_{\text{wire}} \frac{-(\vb{r}-\vb{r'}) \cross \dd{\vb{l'}}}{|\vb{r}-\vb{r'}|^3}
@@ -60,7 +60,7 @@ $$
 
 Consider an electric charge moving with velocity $\vb{v}$ in a magnetic parallel plate capacitor with charge densities $\plusminus \sigma_m$. That is, $\mu_0 \vb{H} = \sigma_m \vu{z}$. Then, we can apply theorems to see the resulting force.
 
-**Theorem**. *Lorentz Force Law* states that $\vb{F} = q_e \vb{v} \cross \u_0 \vb{H}$ in the presence of a magnetic field. In the presence of both an electic andmagnetic field, $\vb{F} = q_e (\vb{E} + \vb{v} \cross \u_0 \vb{H})$.
+**Theorem**. *Lorentz Force Law* states that $\vb{F} = q_e \vb{v} \cross \u_0 \vb{H}$ in the presence of a magnetic field. In the presence of both an electric and magnetic field, $\vb{F} = q_e (\vb{E} + \vb{v} \cross \u_0 \vb{H})$.
 
 **Theorem**. *Ampere's Force Law* states that generalizing the previous theorem, we can see that
 
@@ -70,4 +70,4 @@ $$
 
 ## Section 5.4 - Multipole Expansion of a Vector Potential
 
-This is messy. Skipped.
+This is messy. Skipped.