Mars Ultor
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109 lines
3.3 KiB
Markdown
109 lines
3.3 KiB
Markdown
---
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title: Crazy Mix of Everything You Could Dread
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date: 2025-02-15
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---
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# Electric Charge and Electric Field
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## Laws?
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1. *Law of Conservation of Electric Charge*: **the net amount of electric charge produced in any process is zero**.
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2. I forgot
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## Insulators and Conductors
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This image should sum things up nicely.
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## Coulomb's Law
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$$
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||\vec{F}||=k\frac{Q_1 Q_2}{r^2}
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$$
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$k$ is a random constant written in terms of a more definite constant such that:
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$$
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F=\frac{1}{r\pi\epsilon_0}\frac{Q_1 Q_2}{r_2}
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$$
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where
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$$
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\epsilon_0=\frac{1}{4\pi k}=8.85\times 10^{-12}\;\text{ C}^2/\mathrm{N}*\mathrm{m}^2
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$$
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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.
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# The Electric Field
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$$
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\vec{E}=\frac{\vec{F}}{q}
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$$
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- $q$ is the magnitude of the charge
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- The electric field only reflects the *direction* of the magnetic field, not the intensity.
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- The electric field $\vec{E}$ represents the force per unit charge, so really:
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$$
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\vec{E}=\lim_{q\to 0}{\vec{F}/q}
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$$
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The electric field has the units Newtons per Coulomb: N/C. The magnitude of the electric field vector $\vec{E}$, can be calculated as $E$:
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$$
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E=\frac{F}{q}=\frac{kqQ/r^2}{q} = k\frac{Q}{r^2}
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$$
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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:
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$$
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\vec{F}=q\vec{E}
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$$
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If there are more than one parent charge $Q$, the experienced electric vector field is simply:
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$$
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\vec{E}=\vec{E_1}+\vec{E_2}+\vec{E_n}
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$$
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# Electric Potential Energy and Potential Difference
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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:
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$$
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W=\int{F\mathrm{d}r}
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$$
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> [!NOTE]
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> $r$ and $x$ are used interchangeably for displacement. $r$ is more commonly seen in multi-dimensional reference frames.
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$\therefore$ the work done by electric field $\vec{E}$ is:
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$$
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W=Fd=-qEd
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$$
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$$\Delta \textbf{PE}=-qEd
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$$
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This is only true if the electric field is uniform. In reality, it never is.
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## Potential Difference
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**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 $Q$:
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$$
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V_a=\frac{\textbf{PE}_a}{q}
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$$
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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:
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$$
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V_{ba}=V_b-V_a=\frac{\Delta \textbf{PE}_{ba}}{q}=\frac{-W_{ba}}{q}
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$$
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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 $1\text{ V}=1\;\text{J}/\mathrm{C}$
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---
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#physics
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