ece106: add capacitors and dielectrics

docs/1b/ece106.md

@@ -409,3 +409,111 @@ A non-uniform object, such as a cube, will have larger charge density / stronger
 
 !!! warning
     An off-centre charge in a cavity will require a non-uniform induced charge to cancel out the internal field, but the external surface charge will be uniform (or non-uniform if the surface is odd).
+
+### Nutshell
+
+**Inside** a conductor:
+
+- $\vec E=0$
+- $\Delta V=0$
+- $\rho_v=0$
+
+Inside a cavity, if there exists an external field:
+
+- $\vec E=0$
+- $\rho_s=-Q$
+- $\rho_{s\ outer}=Q$
+
+The inner surface charge distribution matches that of the inner charge, but the outer surface charge distribution is dependent only on the shape of the conductor.
+
+On conductor surfaces, the only $\vec E$ is **normal** to the surface and dependents on the shape of the surface.
+
+$$|\vec E_N|=\frac{|\rho_s|}{\epsilon_0}$$
+
+Grounding a conductor neutralises any free charges.
+
+In slabs, as $A>>d$, assume $Q$ is uniformly distributed.
+
+To solve systems:
+
+- Assigning charge **density** is easier with sheets
+- Assigning **charges** is easier with cylinders/spheres
+
+## Dielectrics
+
+!!! definition
+    - An **insulator** has electrons tightly bound to atoms.
+
+### Polarisation
+
+Polarisation is the act of inducing a dipole to a lesser extent than conductors. The induced field cannot reduce $\vec E$ inside the insulator to zero, but it will reduce its effects. The **polarisation vector** $\vec P$ is an average of the effects of all induced fields on a certain point inside a volume.
+
+$$\vec P=\lim_{\Delta V\to 0}\frac{\sum^{N\Delta v}\vec p_i}{\Delta v}$$
+
+where:
+
+- $\Delta v\approx dv$ is the volume of the insulator
+- $p_i$ is the dipole moment at a point
+- $N$ is the total number of atoms in the volume
+
+Polarisation is proportional to electric field and the **electric susceptibility** $X_e$ of a material to external fields.
+
+$$\boxed{\vec P=\epsilon_0X_e\vec E}$$
+
+The **relative permittivity** $\epsilon_r$ of a material is the ratio of decreasing $\vec E$ inside a medium relative to free space.
+
+$$\epsilon_r=1+X_e$$
+
+The new **flux density** formula includes polarised charges, so now $Q_{enc}$ includes **only free charges** (i.e., not polarised charges).
+
+$$\boxed{\vec D=\epsilon_0\vec E+\vec P=\epsilon_0\epsilon_r\vec E}$$
+
+$$\boxed{\oint\vec D\bullet\vec{dS}=Q_{enc,free}}$$
+
+In uniform charge distributions, the surface charge density is related to its polarisation. Where $\hat n$ is the unit normal of the surface:
+
+$$\rho_s=\vec P\bullet\hat n$$
+
+### Boundary conditions
+
+Regardless of permittivity, the $\vec E$ **tangential to the boundary** between two materials must be equal.
+
+## Capacitors
+
+!!! definition
+    - A **capacitor** is a device that uses the capacitance of materials to store energy in electric fields. It is usually composed of two conductors separated by a dielectric.
+    
+**Capacitance** is a measurement of the charge that can be stored per unit difference in potential.
+
+$$\boxed{Q=C\Delta V}$$
+
+To determine $C$:
+
+1. Place a positive and a negative charge on conductors
+2. Determine charge distribution
+3. Determine $\vec E$ between the conductors
+4. Find a path from the negative to the positive conductor and determine voltage
+
+??? example
+    For two plates separated by distance $d$, with charges of $+Q$ and $-Q$, and a dielectric in between with permittivity $\epsilon_0\epsilon_r$:
+    
+    - Clearly $\rho_0=\frac Q A$ as sheets must have uniform distribution. $-\rho_0$ is on the negative plate.
+    - From Gauss' law, creating a Gaussian surface outside the capacitor to between the plates gives $DA=\rho_0A$.
+    - $D=\epsilon_0\epsilon_rE$ gives $E=\frac{\rho_0}{\epsilon_0\epsilon_r}$
+    - Sheets have uniform fields, thus $\Delta V=Ed$
+    - Finally, $C=\epsilon_0\epsilon_r\frac A d$
+
+!!! warning
+    If three dielectrics with different permittivities are allowed to touch each other, they will create **fringe fields** at their intersection that destroy the boundary condition.
+
+### Capacitors and energy
+
+The stored energy inside capacitors is the same as any other energy.
+
+$$\boxed{U_e=\frac 1 2CV^2}$$
+
+Much like VIR, it's usually easier to work with the form of the equation that has squared constants.
+
+$$U_e=\frac 1 2 \frac {Q^2}{C}=\frac 1 2 QV$$
+
+Adding dielectrics increases capacitance but decrease stored energy.