warehouse/content/physics/CrazyStuff.md

title, date

title	date
Crazy Mix of Everything You Could Dread	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.

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 $1\text{ V}=1\;\text{J}/\mathrm{C}$ 

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