ece106: add point charge stuffs

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eggy
2023-02-08 17:06:19 -05:00
parent c2cced2ded
commit 9d05a9e67e

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@@ -322,10 +322,29 @@ $$
Because the desired force acts opposite to the force from the electric field, as long as $\vec E$ is known at each point: Because the desired force acts opposite to the force from the electric field, as long as $\vec E$ is known at each point:
$$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
$$
The work done only depends on initial and final positions it is conservative, thus implying Kirchoff's voltage law. The work done only depends on initial and final positions it is conservative, thus implying Kirchoff's voltage law.
Where $\vec dl$ is the path of the test charge and $\vec dr$ is the direct path from infinity through the point to the charge, because $dr=|dl|\cos\theta$: Where $\vec dl$ is the path of the test charge from infinity to the point, and $\vec dr$ is the direct path from the origin through the point to the charge, because $dr=-dl$:
$$\vec E\bullet\vec{dl}=Edr$$ $$\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}$$