Formatting Fix

docs/physics/electrostatics/1-math.md

@@ -1,4 +1,5 @@
 # Chapter 1 - Mathematics
+
 ## 1.5 - Dyads and Tensors
 
 **Definition**. A *dyadic* is a representation of two-ish vectors.

docs/physics/electrostatics/2-coulomb.md

@@ -81,7 +81,6 @@ $$
 
 The curl of an electrostatic or magnetostatic is relatively simple.
 
-
 $$
 \begin{align}
     \curl{E(\vb{r})} &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
@@ -132,4 +131,4 @@ $$
 
 **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 symnetry, 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

@@ -25,7 +25,7 @@ The units of magnetostatic potential is Joule/Weber, also known as an Ampere. Th
 With this, we can calculate work. Moving a charge $q$ from $A$ to $B$, we see that
 
 $$
-\delta W = \int_A^B \vb{F} \vdot \dd{\vb{l}} = q_e \int_A^B \vb{E} \vdot \dd{\vb{l}} = -q_e \int_A^B \grad{\vb{V}} \vdot \dd{\vb{l}} = -q_e \delta V_e 
+\delta W = \int_A^B \vb{F} \vdot \dd{\vb{l}} = q_e \int_A^B \vb{E} \vdot \dd{\vb{l}} = -q_e \int_A^B \grad{\vb{V}} \vdot \dd{\vb{l}} = -q_e \delta V_e
 $$
 
 Strictly speaking, this is a potential difference. To find the absolute potential, assume a point charge $Q$ at the origin, and a charge $q$. We take the work as $q$ moves from $\vb{r'} = \vb{\infty}$ to $\vb{r'} = \vb{r}$. Thus,
@@ -62,7 +62,6 @@ $$
 V_e(\vb{r}) = \frac{1}{4 \pi \epsilon_0} \int_{V'} \frac{\rho_e(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'}
 $$
 
-
 $$
 V_m(\vb{r}) = \frac{1}{4 \pi \mu_0} \int_{V'} \frac{\rho_m(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'}
 $$
@@ -75,7 +74,7 @@ $$
 W_2 = W_{21} = \frac{1}{4 \pi \epsilon_0} \frac{Q_{e1} Q_{e2}}{\abs{\vb{r_2} - \vb{r_1}}}
 $$
 
-Superposition applies here. The energy to create $N$ charges is 
+Superposition applies here. The energy to create $N$ charges is
 
 $$
 W_n = \frac{1}{2} \frac{4 \pi \epsilon_0} \sum_{i = 1}^{N} \sum_{j > i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
@@ -177,7 +176,7 @@ $$
 Taking the divergence, we find that
 
 $$
-- \laplacian{V(\vb{r})} = G(\vb{r}, \vb{r'}) \frac{Q_e}{\epsilon_0} = \frac{Q_e}{\epsilon_0} \laplacian({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}}) 
+- \laplacian{V(\vb{r})} = G(\vb{r}, \vb{r'}) \frac{Q_e}{\epsilon_0} = \frac{Q_e}{\epsilon_0} \laplacian({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}})
 =  \frac{Q_e}{\epsilon_0} \div \frac{\vb{r} - \vb{r'}}{\abs{\vb{r} - \vb{r'}}^3} = \frac{Q_e}{\epsilon_0} \delta(\vb{r} - \vb{r'})
 $$
 
@@ -274,4 +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 volate or other equations, however, this becomes very messy.

docs/physics/electrostatics/4-conductors.md

@@ -16,4 +16,4 @@ 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, $\sigma_e = \epsilon_0 E$.
+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, $\sigma_e = \epsilon_0 E$.