Capacitance

Formula:

$$\begin{array}{r}C=\frac{Q}{V}=\frac{{ϵ}_{0}A}{D}\end{array}$$

The latter formula represents, in geometry, that the larger the area, the more charge it holds. The further apart the two plates or sides of a capacitor, the less charge it holds.

Surface Charge Density:

$$\begin{array}{rl}\sigma & =\frac{Q}{A}\end{array}$$

Spherical Capacitor

$$\begin{array}{rl}& V_a-V_b\\ & =kQ\left(\frac{1}{r_a}-\frac{1}{r_b}\right)\\ & \\ C& =\frac{Q}{V}=\frac{1}{k}\frac{r_a{r}_b}{r_b-r_a}\end{array}$$

Cylindrical Capacitor

$$\begin{array}{rl}{V}_a-V_b& ={\int }_a^b\vec{E}\cdot d\vec{l}\\ & ={\int }_a^b{E}_r\cdot dr\\ & =\frac{\lambda }{2\pi {ϵ}_0}{\int }_a^b\frac{dr}{r}\\ & =\frac{\lambda }{2\pi {ϵ}_0}\ln \frac{r_b}{r_a}\end{array}$$

$V_p$ - potentiel électrique au point $p$ $r_p$ - Distance radiale au point $p$

$\vec{E}$ - Champ¹ électrique $\lambda$ - Densité linéique de charge ${ϵ}_0$ - Permittivité du vide $r$ - Variable d’intégration représentant la distance radiale $\vec{l}$ - Variable d’intégration représentant le chemin²

Cette équation représente la différence de potentiel électrique entre deux points $a$ et $b$, exprimée en termes de la densité linéique de charge $\lambda$ et de la permittivité du vide ${ϵ}_0$. — Claude

Series

$V_{ab}=V_1+V_2$ $Q_{eq}=Q_1=Q_2$ $\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\dots$ (these guys have the same charge)

Parallel

$V_{ab}=V_1=V_2$ $Q_{eq}=Q_1+Q_2$ $C_{eq}=C_1+C_2+C_3+\dots$ (these guys have same potential difference)

In series charges are the same and potentials add. In parallel, charges add and potentials are the same. — Noah

Electric Potential Energy

During yesterdays recitation, Noah (the recitation instructor that sounds like the good doctor) brought up a fantastic analogy— he said that in parallel, the potential energy does not diminish the same way that your potential energy at the top of a mountain would not diminish were there more than one trail down.

$$U=\frac{1}{2}C{V}^2=\frac{1}{2}QV=\frac{Q^2}{2C}$$

Cette équation représente l’énergie électrostatique $U$ stockée dans un condensateur, exprimée de trois manières équivalentes en termes de capacité $C$, tension³ $V$ et charge $Q$. — Claude

Energy Density

$$\begin{array}{rlr}u=\frac{U}{V}& & u=\frac{1}{2}{ϵ}_0{E}^2\end{array}$$

Où : $u$ - Densité d’énergie $U$ - Énergie totale $V$ - Volume ${ϵ}_0$ - Permittivité du vide $E$ - Champ électrique

GOD WHY DO WE HAVE BOTH $u$ and $U$ I HATE MATHEMATICIANS AND PHYSICISTS 😭

Dielectrics

Not to be confused with Marxist Dialectics or Scientology’s Dianetics. Now, let’s imagine placing a piece of metal in a uniform electric field of strength $E$. This metal is a conductor, and electrons are free to move within it. Subjected to our electric field, the electrons will rush toward one end leaving the other deft of electrons. Surfeit positive charges on one side, and negative the other. This results in an electric dipole.

This dipole generates its own electric field, $\overrightarrow{E^{\prime }}$ that opposes the first electric field $\vec{E}$ until equilibrium is reached and the charges do not move. Because the charges are free to move, this only happens when the induced field $E^{\prime }=E$, so the resulting field in the material is zero. The applied field is kind of cancelled. This is how a Faraday cage works. The conducting field acts as a barrier, the electric field cannot pass through.

Now, lets introduce a dielectric field to the material. Since it is an insulator, the charges are not free to move.. okay they’re free to move a little bit (this is crucial!) Materials, depending on their ability to get electrically polarized have an Electric Susceptibility, typically represented with $\chi$

Constante diélectrique : $K=\frac{E_0}{E}=\frac{C}{C_0}$

$\overrightarrow{E^{\prime }}=\frac{\vec{E}}{K}$

$\vec{P}={ϵ}_0\chi \vec{E}$

$\chi =\frac{P}{{ϵ}_{0}E}$

$K=1+\chi$

The dielectric increases the capacitance by a factor $k$

Permittivity

Permittivité : $ϵ=K{ϵ}_0$

Capacité : $C=K{C}_0=K{ϵ}_0\frac{A}{d}=ϵ\frac{A}{d}$

Densité d’énergie : $u=\frac{1}{2}K{ϵ}_0{E}^2=\frac{1}{2}ϵE^2$

Où : $K$ - Constante diélectrique relative $E_0$ - Champ électrique sans diélectrique $E$ - Champ électrique avec diélectrique $C$ - Capacité avec diélectrique $C_0$ - Capacité sans diélectrique $ϵ$ - Permittivité du matériau diélectrique ${ϵ}_0$ - Permittivité du vide $A$ - Surface des plaques $d$ - Distance entre les plaques $u$ - Densité d’énergie

2/1/25

Footnotes

1. field ↩

2. path ↩

3. voltage, “potential difference” ↩