From 38a8ee0814e39ebfddbc6e5f33133a27636a063c Mon Sep 17 00:00:00 2001
From: Indigo5684 <159226326+Indigo5684@users.noreply.github.com>
Date: Tue, 30 Sep 2025 13:19:42 -0500
Subject: [PATCH] Electrodynamicx Fix

---
 docs/physics/electrodynamics/10-electromagnetic-waves.md | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/docs/physics/electrodynamics/10-electromagnetic-waves.md b/docs/physics/electrodynamics/10-electromagnetic-waves.md
index cc8a1d2..398b0ef 100644
--- a/docs/physics/electrodynamics/10-electromagnetic-waves.md
+++ b/docs/physics/electrodynamics/10-electromagnetic-waves.md
@@ -524,25 +524,25 @@ $$V(z + \Delta z, t) - V(z, t) = -L' \Delta z \frac{dI(z, t)}{dt}$$
 
 We can take the limit to see that
 
-$$\frac{\partial V(z, t)}{\partial z} = -L' \frac{\partial I{z, t}}{\partial t}$$
+$$\frac{\partial V(z, t)}{\partial z} = -L' \frac{\partial I(z, t)}{\partial t}$$
 
 This gives is the two telegrapher's equations:
 
 $$\begin{align}
 \frac{\partial I(z, t)}{\partial z} &= -C' \frac{\partial V(z, t)}{\partial t} \\
-\frac{\partial V(z, t)}{\partial z} &= -L' \frac{\partial I{z, t}}{\partial t}
+\frac{\partial V(z, t)}{\partial z} &= -L' \frac{\partial I(z, t)}{\partial t}
 \end{align}$$
 
 We can then take the derivative of the first with respect to $t$ and the derivative of the second with respect to $z$, we see that
 
 $$\begin{align}
 \frac{\partial^2 I(z, t)}{\partial z^2} &= -C' \frac{\partial}{\partial z}(\frac{\partial V(z, t)}{\partial t}) \\
-\frac{\partial}{\partial t}(\frac{\partial V(z, t)}{\partial z}) &= -L' \frac{\partial^2 I{z, t}}{\partial t^2}
+\frac{\partial}{\partial t}(\frac{\partial V(z, t)}{\partial z}) &= -L' \frac{\partial^2 I(z, t)}{\partial t^2}
 \end{align}$$
 
 This tells us that
 
-$$\frac{\partial^2 I(z, t)}{\partial z^2} = -C'(-L' \frac{\partial^2 I{z, t}}{\partial t^2}) = C'L' \frac{\partial^2 I{z, t}}{\partial t^2}$$
+$$\frac{\partial^2 I(z, t)}{\partial z^2} = -C'(-L' \frac{\partial^2 I(z, t)}{\partial t^2}) = C'L' \frac{\partial^2 I(z, t)}{\partial t^2}$$
 
 We can instead take the derivative of the first equation with respect to $z$ and the derivative of the second with respect to $t$ to see that