From b1fe0875e39ad000d28ac1746b194ce52291f06d Mon Sep 17 00:00:00 2001
From: Nathan Nguyen <159226326+Indigo5684@users.noreply.github.com>
Date: Sun, 17 Nov 2024 18:52:04 -0600
Subject: [PATCH] Electrostatics Section 4-5

---
 docs/physics/electrostatics/4-conductors.md   | 18 +++++++++++++++
 .../electrostatics/5-moving-charges.md        | 23 ++++++++++++++++++-
 2 files changed, 40 insertions(+), 1 deletion(-)

diff --git a/docs/physics/electrostatics/4-conductors.md b/docs/physics/electrostatics/4-conductors.md
index d28c4d4..4e1d614 100644
--- a/docs/physics/electrostatics/4-conductors.md
+++ b/docs/physics/electrostatics/4-conductors.md
@@ -62,3 +62,21 @@ Both $\vb{D}$ and $\vb{P}$ have units of Coulombs/m^2. Additionally, we see that
 $$
 \div \vb{D} = \div (\epsilon_0 \vb{E} + \vb{P}) = \div \epsilon_0 \vb{E} + \div \vb{P} = (\rho_{ef} + \rho_{eb}) - \rho_{eb} = \rho_{ef}
 $$
+
+## Section 4.5 - Induced Electric Charges, their Potentials and Fields
+
+This is an application chapter.
+
+## Section 4.6 - Capacitance
+
+**Definition**. The *capacitance* of an object $C$ is the charge per volt, such that
+
+$$
+C := \frac{Q}{V}
+$$
+
+This unit, $\frac{C}{V}$, is known as a Farad. For a sphere, $C = 4 \pi \epsilon_0 R$. For a parallel plate capacitor, this reduces to $C = \frac{epsilon_0 A}{d}$.
+
+## 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.
\ No newline at end of file
diff --git a/docs/physics/electrostatics/5-moving-charges.md b/docs/physics/electrostatics/5-moving-charges.md
index 9376772..4ce9044 100644
--- a/docs/physics/electrostatics/5-moving-charges.md
+++ b/docs/physics/electrostatics/5-moving-charges.md
@@ -10,4 +10,25 @@ With math, we see that $\div \vb{J}(\vb{r}) = -\frac{\partial \rho(\vb{r})}{\par
 
 For a wire of uniform cross-sectional area, we see that $G = \sigma \frac{A}{L}$, where $A$ is the cross-sectional area, $L$ is the length of the wire, and $\sigma$ is the conductivity of a wire. Inverted, we see that $R$ = $\rho \frac{L}{A}$, where $\rho = \frac{1}{\sigma}$ is the resistivity of the wire.
 
-**Definition**. *Ohm's Law* can be written as $I = G V$, or inverted, $V = IR$. In a wire, we see that current density $\vb{} = \frac{I}{A} = \sigma \frac{V}{L} = \sigma \vb{E}$
\ No newline at end of file
+**Definition**. *Ohm's Law* can be written as $I = G V$, or inverted, $V = IR$. In a wire, we see that current density $\vb{} = \frac{I}{A} = \sigma \frac{V}{L} = \sigma \vb{E}$
+
+## Section 5.2
+
+We know that $\vb{J}_e = \curl{\vb{H}}$ and $\vb{J}_m = -\curl{\vb{E}}$. That is, current densities cause the opposing field to curl.
+
+For a wire with current $I_e$, we see that applying Stoke's theorem to the first equation,
+
+$ \int_S \curl{\vb{H}} \vdot \vu{n} \dd{S} = \int_{\partial S} = \vb{H} \vdot \dd{\vb{l}}$. Apply the identity $\curl{\vb{H}} = \vb{J}_e$ to the left side to see that $\int_S \curl{\vb{H}} \vdot \vu{n} \dd{S} = \int_S \vb{J}_e \vdot \vu{n} \dd{S} = (I_e)_S$, or the current passing through the cross-sectional area. By the original equation, we see that $(I_e)_S = \vb{H} \vdot \dd{\vb{l}}$.
+
+If we assume cylindrical coordinates and that $\vb{H}(vb{r}) = H_\varphi(s) \vu{\varphi}$, then $\vb{H} \vdot \dd{\vb{l}} = \int_0^{2\pi} H_\varphi(S) s \dd{\varphi}$, so then $(I_e)_S = \int_0^{2\pi} H_\varphi(S) s \dd{\varphi}$. Thus, for $s > a$ (where $a$ is the radius of the wire), $2\pi s H_\varphi = I_e$, and for $s < a$, $2\pi s H_\varphi = I_e \frac{s^2}{a^2}$.
+
+---
+
+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
+
+$$
+\vb{H}(\vb{r}) = \int{I_e}{4\pi}\int_{\text{wire}} \frac{-(\vb{r}-\vb{r'}) \cross \dd{\vb{l'}}}{|\vb{r}-\vb{r'}|^3}
+$$
+
+---
+