---
title: Crazy Mix of Everything You Could Dread
date: 2025-02-15
---

# Electric Charge and Electric Field

## Laws?

1. *Law of Conservation of Electric Charge*: **the net amount of electric charge produced in any process is zero**.

2. I forgot

## Insulators and Conductors

This image should sum things up nicely.

![](swappy-20250215-165128.png)

## Coulomb's Law 

$$
||\vec{F}||=k\frac{Q_1 Q_2}{r^2}
$$

$k$ is a random constant written in terms of a more definite constant such that:

$$
F=\frac{1}{r\pi\epsilon_0}\frac{Q_1 Q_2}{r_2}
$$
where
$$
\epsilon_0=\frac{1}{4\pi k}=8.85\times 10^{-12}\;\text{ C}^2/\mathrm{N}*\mathrm{m}^2
$$

It's important to consider here, that $r$ is the magnitude of distance, and $F$ always represents the magnitude of the force. Oppositely signed charges attract, and same-signed charges repel with the force computed by the formula.

# The Electric Field 

$$
\vec{E}=\frac{\vec{F}}{q}
$$

- $q$ is the magnitude of the charge 
- The electric field only reflects the *direction* of the magnetic field, not the intensity.
- The electric field $\vec{E}$ represents the force per unit charge, so really: 

$$
\vec{E}=\lim_{q\to 0}{\vec{F}/q}
$$ 

The electric field has the units Newtons per Coulomb: N/C. The magnitude of the electric field vector $\vec{E}$, can be calculated as $E$:
$$
E=\frac{F}{q}=\frac{kqQ/r^2}{q} = k\frac{Q}{r^2}
$$


Notice that the vector field $\vec{E}$ is independent of the test charge $q$, only dependent on the parent charge $Q$ that produces the field. $\therefore$, the force experienced by a particle at a given position $r$ can be expressed by: 

$$
\vec{F}=q\vec{E}
$$

If there are more than one parent charge $Q$, the experienced electric vector field is simply: 

$$
\vec{E}=\vec{E_1}+\vec{E_2}+\vec{E_n}
$$

# Electric Potential Energy and Potential Difference 

A fundamental law of physics is that $\Delta \textbf{PE} =-W$, while the change in kinetic energy is oppositely signed. The potential difference between any two points is $\textbf{PE}_b-\textbf{PE}_a$, while work is expressed like so:

$$
W=\int{F\mathrm{d}r}
$$

> [!NOTE]
> $r$ and $x$ are used interchangeably for displacement. $r$ is more commonly seen in multi-dimensional reference frames.

$\therefore$ the work done by electric field $\vec{E}$ is:

$$
W=Fd=-qEd
$$
$$\Delta \textbf{PE}=-qEd 
$$

This is only true if the electric field is uniform. In reality, it never is.

## Potential Difference 

**Electric Potential** is understood as the electric potential energy per unit charge. This variable is given the symbol $V$. At some displacement $a$ from the parent charge $Q$:

$$
V_a=\frac{\textbf{PE}_a}{q}
$$

Two balls at the same height are not meaningfully different in position, therefore only *potential difference* matters physically. In this case, between two points, $a$ and $b$. When the electric force does positive work on a charge, the kinetic energy increases in tandem with a decrease of that potential difference. Keeping in line with the rules laid out above:

$$
V_{ba}=V_b-V_a=\frac{\Delta \textbf{PE}_{ba}}{q}=\frac{-W_{ba}}{q}
$$

Just as the potential energy of a raised ball does not depend on the gravitational field of the ball, the electric potential of the test charge $q$ doesn't depend on the magnitude of the charge itself. This quantity is given the unit **Volt**, or $1\text{ V}=1\;\text{J}/\mathrm{C}$ 

---
#physics