From d5ce9d9dd0bc76b27dd946a9247c55a8073d1b6a Mon Sep 17 00:00:00 2001 From: eggy Date: Thu, 2 Feb 2023 10:42:28 -0500 Subject: [PATCH] ece106: i'm gonna cry --- docs/1b/ece106.md | 32 ++++++++++++++++++++++++++++++++ 1 file changed, 32 insertions(+) diff --git a/docs/1b/ece106.md b/docs/1b/ece106.md index 4ca3b69..7843ef4 100644 --- a/docs/1b/ece106.md +++ b/docs/1b/ece106.md @@ -248,3 +248,35 @@ This implies $\phi_e>0$ is a net positive charge enclosed. !!! warning Gauss's law only applies when $\vec E$ is from all charges in the system + +### Charge distributed over a line/cylinder + +!!! warning "Limitations" + To apply this strategy, the following conditions must hold: + + - $Q$ must not vary with the length of the cylinder or $\phi$ + - The charge must be distributed over either a cylindrical surface or the volume of the cylinder. + - In the real world, $r$ must be significantly smaller than $L$ as an approximation. + - The strategy is more accurate for points closer to the centre of the wire. + +Please see [Maxwell's integral equations#Gauss's law](https://en.wikiversity.org/wiki/MyOpenMath/Solutions/Maxwell%27s_integral_equations) for more information. + +**Outside** the radius $R$ of the cylinder of the Gaussian surface, the enclosed charge is, where $L$ is the length of the cylinder: + +$$Q_{enc}=\pi R^2\rho_0L$ + +such that the field at any radius $r>R$ is equal to: + +$$\vec E(r)=\frac{\rho_0\pi R^2}{2\pi\epsilon_0r}\hat r$$ + +**Inside** the radius $R$ of the cylinder, the enclosed charge depends on $r$. For a uniform charge density: + +$$Q_{enc}=\pi r^2\rho_0L$$ + +such that the field at any radius $r< R$ is equal to: + +$$\vec E(r)=\frac{\rho_0}{2\epsilon_0}r\hat r$$ + +The direction of $\vec E$ should always be equal to that of $\vec r$. Generally, where $lim$ is $r$ if $r$ is *inside* the cylinder or $R$ otherwise, $\rho_v$ is the function for charge density based on radius, and $r_1$ is hell if I know: + +$$\epsilon_0 E2\pi rL=\int^{lim}_0\rho_v(r_1)2\pi r_1L\ dr_1$$