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#+title: electrostatics
#+author: Preston Pan
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* What is Electricity?
Because this is an introduction and not a lesson in quantum mechanics, I will say that electricity is broadly
defined by the charge that an object has that corresponds the force that it both feels and also gives
to other objects. Charge is measured in coulombs and can be negative or positive, which leads us to the man himself:

* Coulomb's Law
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In order to define the phenomena of electric force in the real world, we use
this experimentally verified law known as Coulomb's Law. Let \( \vec{r_{1}} \) be the displacement
of a charge \( Q \), and let \( \vec{r_{2}} \) be the displacement of a charge \( q \), where these two charges are named \( P_{1}\) and \( P_{2} \) respectively.
Then let \( \vec{r} = \vec{r_{1}} - \vec{r_{2}} \) be the distance between the charges. For simplicity, we assume
these charges have no mass or volume; we call these /point charges/. The equation for the force
on \( P_{1} \) is as follows:

\begin{align*}
\vec{F(\vec{r})} = \frac{1}{4\pi\epsilon_{0}}\frac{qQ}{r^{2}} \hat{r}
\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 [[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 [[id:422653e2-daa4-422a-9cb7-3983a5a72554][superposition principle]].
That is, if we have different electrostatic forces acting on one particle:

\begin{align*}
\vec{F_{tot}} = \vec{F_{1}} + \vec{F_{2}} + … = \sum_{i=1}^{n} \vec{F_{i}}.
\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 [[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
think of it like that, you will never need to know what the units actually are, although I'm
sure you can find that online. Just know that it is called the permeability of free space, and
it is defined in terms of the speed of light and a constant relating to magnetism:

\begin{align*}
\epsilon_{0}\mu_{0} = \frac{1}{c^{2}}
\end{align*}

but since this is electrostatics and not electrodynamics, you will not have to worry about
magnetic constants. Again, it is just a shift from speed of light units to our mortal units.

** Electric Field
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Okay, now we can continue to defining the /electric field/ of a particle. Let's call \( P_{1} \) our
/test charge/, and \( P_{2} \) our /source charge/. If we now want to measure the force on \( P_{1} \),
our equation is going to be the same. However, we can define a field \( \vec{E(\vec{r})} \) such that:

\begin{align*}
\vec{F} = Q\vec{E}
\end{align*}

Where:

\begin{align*}
\vec{E} = \frac{\vec{F}}{Q}
\end{align*}

Therefore, the value of \( \vec{E} \) for a point charge must be:

\begin{align*}
\vec{E} := \frac{1}{4\pi\epsilon_{0}}\frac{q}{r^{2}}\hat{r}.
\end{align*}

The result is we find a way to express force in a /test charge independent way/. This is useful
because we often want to find the force if an arbitrary object with an arbitrary charge is next
to the particle in question, instead of focusing specifically on two charges.

Note that it is trivial to prove that \( \vec{E} \) also follows the [[id:422653e2-daa4-422a-9cb7-3983a5a72554][superposition principle]].


** Continuous Charge Distributions
Now that we have a working definition of \( \vec{E} \), we can now find the electric field of an object
that has a continuous charge distribution. Note that there aren't actually infinite charges in real
world objects which is what we are assuming by taking an integral over some space of charge, but
it's close enough because there are so many individual charges in real world objects. Assuming we are in
three dimensions:

\begin{align*}
\vec{E(\vec{r})} = \frac{1}{4\pi\epsilon_{0}} \int_{space} \frac{\sigma(\vec{r_{2}})}{r^{2}}\hat{r}d\tau
\end{align*}

Where \( \tau \) is the patch of volume we are integrating over, and \( \sigma \) is the charge density function,
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 potential
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:ROAM_ALIASES: "potential difference" voltage
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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*}
Note that the difference of two points in the potential scalar field is also known as potential difference
between those two points, or the voltage between those two points.