From 9d05a9e67e051519d5d343cf4581e12964a66bf6 Mon Sep 17 00:00:00 2001 From: eggy Date: Wed, 8 Feb 2023 17:06:19 -0500 Subject: [PATCH] ece106: add point charge stuffs --- docs/1b/ece106.md | 23 +++++++++++++++++++++-- 1 file changed, 21 insertions(+), 2 deletions(-) diff --git a/docs/1b/ece106.md b/docs/1b/ece106.md index 3c5adc3..35ee429 100644 --- a/docs/1b/ece106.md +++ b/docs/1b/ece106.md @@ -322,10 +322,29 @@ $$ Because the desired force acts opposite to the force from the electric field, as long as $\vec E$ is known at each point: -$$V_p=-\int^p_\infty\vec E\bullet\vec{dl}$$ +$$ +V_p=-\int^p_\infty\vec E\bullet\vec{dl} +V_p=-\int^p_\infty E\ dr +$$ The work done only depends on initial and final positions — it is conservative, thus implying Kirchoff's voltage law. -Where $\vec dl$ is the path of the test charge and $\vec dr$ is the direct path from infinity through the point to the charge, because $dr=|dl|\cos\theta$: +Where $\vec dl$ is the path of the test charge from infinity to the point, and $\vec dr$ is the direct path from the origin through the point to the charge, because $dr=-dl$: $$\vec E\bullet\vec{dl}=Edr$$ + +Therefore, the potential due to a point charge is equal to: + +$$V_p=-\int^p_\infty\frac{kQ}{r^2}dr=\frac{kQ}{r}$$ + +**Positive** charges naturally move to **lower** potentials ($V$ decreases) while negative charges do the opposite. + +In order to calculate the voltage for charge distributions: + +- If $\vec E$ is easy to find via Gauss law: + +$$V_p=-\int^p_\infty\vec E\bullet\vec{dl}$$ + +- If the charge is asymmetric: + +$$V_p=\int_\text{charge dist}\frac{kdQ}{r}$$