Capacitance of one surface: $C=q/\phi$
Capacitance of two objects: $C= q/\Delta V$. For convenience, if we define $I=dQ/dt$, there is a voltage drop in the direction of the current as the capacitor gets charged
A capacitor always has two surfaces of opposite charges, as they have to be quasistatic, exerting no electric force on other charges within each conductor on both sides. An easy way to look at this is to consider a gaussian surface with two infinitely large surfaces, one in each conductor. Flux must be 0, so total charge inside must be 0.
Thin wires have negligible capacitance, meaning you can ignore any charge “stored” in the wire. This holds for all thin objects.
A conducting plate sandwiched in between a capacitor can be treated as splitting the capacitor into two capacitors in series, as one capacitor exerts no electric force on other charges.
Total energy: $E=\frac{1}{2}C\,\Delta V^2$. This only works when capacitors are far away from one another though.
Combining capacitance is opposite of resistance
$$ \quad L=\frac{\Phi'}{I} \textrm{ by definition} $$
$$ \frac{d \Phi}{dt}=L\frac{dI}{dt} \implies \varepsilon=-V=-L\frac{dI}{dt} $$
Remember that emfs can be converted to voltage rises.
In a circuit
Mutual inductance: $M \leq \sqrt{L_1 L_2}$
Combining inductors is same as resistance
Faraday’s law is often given as $\varepsilon = - \mathrm{d} \Phi_B/\mathrm{d}t$ (use right-hand rule for direction). This is sound for a changing magnetic field (see Maxwell’s equations). But it is often applied to a static magnetic field with a moving wire. However, both are two different phenomena. One is due to the unconservative electric field, while the other (motional emf) arises from the Lorentz force, which is a result of relativity.