Quantum Mechanics demystified, a try

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First bit in electromagnetism.

If an object has more electrons than protons, object’s charge is said to be negative. Positive if vice versa. Neutral otherwise.

An electron’s charge is called elementary charge and is marked with $e$

Unit of measure for electric charge is Coulomb (C).

Electric charge is denoted by letter Q.

$1C = \frac{e}{1.60217653\cdot10^{-19}} \hspace{1em} \hspace{1em} i.e. \hspace{1em} 1C \sim 6.242\cdot 10^{18}e$

When you have two electric charges , Q and q, nearby each other, they will have a force affecting between them. Opposite charges have attractive force between them, like charges repel each other. Let’s say charges are at distance r, in which case the force between them is characterized by formula

$\text{Coulomb’s law:} \hspace{1em} \vec{F_{e}}= \frac{1}{4\pi\epsilon_{0}} \frac{Qq}{r^2} \sim \frac{kg\cdot m^3}{s^2\cdot C^2}\cdot \frac{C^2}{m^2}=\frac{kg m}{s^2}=J$

Note: Vector force is along line connecting charges. Positive force is repulsive, negative is attractive.

$\text{Where $\epsilon_{0}$ is vacuum permittivity:} \hspace{1em}\epsilon_{0}=8.8541878128(13)×10^{-12} \frac{F}{m}$

$F=farad \sim \frac{s^{4}A^{2}}{kgm^{2}} $

An electric charge (Q) creates an electric field around it. If another electric charge (q) is put in this field, at any point of it, there is force affecting this charge. This force is described by Coulomb’s law.

Strength of the electric field at each point is defined to be the force effecting a charge Q at that point divided by Q itself:

$\vec{E}=\frac{\vec{F}}{q}=\frac{1}{4\pi\epsilon_{0}} \frac{Q}{r^{2}}$

*Q* is the source charge constituting electric field $\vec{E}$, which induces electric force $\vec{F}$ affecting charge *q* put at the distance of *r* from the source charge *Q*.

$\text{Unit of vacuum permittivity}=\frac{s^{4}A^{2}}{kgm^{3}}$

$\implies \text{Unit of } \vec{E} \text{ :} \hspace{1em} \frac{kgm^{3}}{s^{4}A^{2}} \frac{As}{m^2}= \frac{kgm}{s^{3}A}= \frac{kgm}{s^{3}\frac{C}{s}} = \frac{kgm}{s^{2}\cdot C} = \frac{N}{C}$

Electric field makes electric charges to move in it. Moving electric electric charges create an electric current in turn.

Electric current is measured by amperes (amps, A): One Coulomb of elementary charges in a second.

$1A=1\frac{C}{s}, \implies C=As$

On a metallic wire, an electric field is created within it when it’s ends are connected to a power supply. This makes loosely connected electrons on outer shell of the metal to start moving in the field.

Power supply is a device with two poles, with the other (negative) having more electrons than the other (positive). Power supplies can be created e.g. with chemical reactions and electromagnetic radiation.

Moving electric charges can also be bigger entities, like ions. If table salt, $Na^+ Cl^-$ is dissolved in water, $Na^+$ and $Cl-$ ions get separated due to water molecules dipole effect. Once an electric field is generated in this liquid, ions start to travel along the field.

When and electron moves in a homogeneous electric field a distance of $\Delta x$, electric force makes a work of

$W=\vec{F}\Delta x$, by which amount potential energy is also changed.

As $\vec{F}=\vec{E}q$, potential energy of the electron is $W=E_p=\vec{E}e\Delta x$ when compared to electron’s starting point.

It is interesting how this correlates with gravitational potential energy $E_g=mgh$.

Objects mass $m$ relates to electric charge $e$, $h$ relates to $\Delta x$ and acceleration of gravity $g$ relates to $\vec{E}$. Only that $g$ is a simplified constant derived empirically to fit exactly gravitational field $\vec{g}$ on earth’s surface.

The further charge is from another, the more is potential energy, the less is electric force.

The same goes with gravity on earth. The higher the object, the bigger potential energy and the less gravity.

To get potential for electric field itself, we devide field $\vec{E}$ by the charge $Q$. Thus we get a general feature describing the field itself, independently from the charge in question.

We mark electric field potential by $V$.

It’s unit is Volt, also marked $V=\frac{J}{C}=\frac{N\cdot m}{As}=\frac{\frac{kg\cdot m^2}{s^2}}{As}
=\frac{kg\cdot m^2}{As^3}$

$V=\frac{\vec{E_p}}{Q}$

In a homogeneous electric field potential energy for an electron was $E_p=\vec{E}e\Delta x$.

As potential in general electric field for an electron is $V=\frac{\vec{E_p}}{e}$,

we get $V=\frac{\vec{E}e\Delta x}{e}=\vec{E}\Delta x$

Thus unit of electric field’s strength can also be expressed as

$[E]=\frac{V}{m}=\frac{\frac{J}{C}}{m}=\frac{N\cdot m}{As\cdot m}=\frac{N}{C}$

Note: Electric field in metallic conductor is homogeneous.