From 80da24887ac760a9d18936634d8d46c0643521ee Mon Sep 17 00:00:00 2001 From: Preston Pan Date: Sun, 23 Jul 2023 09:12:03 -0700 Subject: add a lot of mindmap articles --- mindmap/electrostatics.org | 23 +++++++++++++++++++++-- 1 file changed, 21 insertions(+), 2 deletions(-) (limited to 'mindmap/electrostatics.org') diff --git a/mindmap/electrostatics.org b/mindmap/electrostatics.org index e71a05e..55310de 100644 --- a/mindmap/electrostatics.org +++ b/mindmap/electrostatics.org @@ -26,7 +26,7 @@ on \( P_{1} \) is as follows: \end{align*} Where \( \hat{r} \) is the unit vector pointing in the direction of \( P_{2} \). Note that there are a -couple of interesting things about this force. First, it is an inverse square law, and the formula looks a lot like the one for gravitation, +couple of interesting things about this force. First, it is an [[id:2a543b79-33a0-4bc8-bd1c-e4d693666aba][inverse square]] law, and the formula looks a lot like the one for gravitation, only charge can be negative and mass cannot. Second, it is symmetrical, in the sense that the force felt by \( P_{2} \) is going to be the same, only \( \hat{r} \) is pointing in the other direction. Also, note that due to linearity, this force calculation follows the /superposition principle/. @@ -37,7 +37,7 @@ That is, if we have different electrostatic forces acting on one particle: \end{align*} Wait, where does the \( \frac{1}{4\pi\epsilon_{0}} \) term come from? Well, the surface area of a sphere -is \( 4\pi r^{2}\) , which explains both the inverse square law and this \( 4\pi \) term in the denominator, +is \( 4\pi r^{2}\) , which explains both the [[id:2a543b79-33a0-4bc8-bd1c-e4d693666aba][inverse square]] law and this \( 4\pi \) term in the denominator, but what about \( \epsilon_{0} \), what does it even mean? Well, it is simply a conversion of units from /speed of light/ terms to /SI unit terms/. If you @@ -95,3 +95,22 @@ Where \( \tau \) is the patch of volume we are integrating over, and \( \sigma \ which takes a position vector and returns the charge at that vector. Of course, surface and line integrals have their own analogues -- simply replace \( d\tau \) with \( da \) or \( dl \), and make sure your charge distribution is in the correct amount of dimensions. + +** [[id:12a2d5b3-f98c-45e5-9107-5560288b5aa8][Divergence]] and [[id:b25e0e44-c764-4f0a-a5ad-7f9d79c7660d][Curl]] of Electric Field +The divergence and curl of the electric field are essential to solving electrostatic configurations with more +ease, as well as proving some qualities about the electric field. Because the electric field is an [[id:2a543b79-33a0-4bc8-bd1c-e4d693666aba][inverse square]] +field: +\begin{align*} +\vec{\nabla} \cdot \vec{E} = \frac{\sigma(\vec{r_{1}})}{\epsilon_{0}} \\ +\oint\vec{E} \cdot d\vec{a} = \frac{q_{enc.}}{\epsilon_{0}} \\ +\vec{\nabla} \times \vec{E} = \vec{0} \\ +\oint\vec{E} \cdot d\vec{l} = \vec{0} +\end{align*} + +* Electrostatic Potentials +Because \(\vec{E}\) is a [[id:6f2aba40-5c9f-406b-a1fa-13018de55648][conservative field]]: +\begin{align*} +\vec{E} = -\vec{\nabla}V \\ +\nabla^{2}V = -\frac{\sigma(\vec{r_{1}})}{\epsilon_{0}} \\ +V(\vec{r}) = \frac{1}{4\pi \epsilon_{0}}\int\frac{\sigma(\vec{r_{2}})}{r}dr +\end{align*} -- cgit