Capacitance is geometry. The wiring decides the rest

Charge two conductors to $+Q$ and $-Q$ and measure the potential difference $V$ between them. The ratio

$$C = \frac{Q}{V}$$

is the same for every $Q$ you try. Double the charge and the field doubles everywhere, so $V = \int \vec{E}\cdot d\vec{\ell}$ doubles too, and the ratio does not move. That is why $C$ is a property of the arrangement rather than of what you did to it.

For parallel plates of area $A$ separated by $d$, with $d$ small enough to ignore fringing: the field between the plates is $E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)$, uniform, so $V = Ed = Qd/(\varepsilon_0 A)$ and

$$C = \frac{\varepsilon_0 A}{d}.$$

Read that formula for what it does not contain: no $Q$, no $V$, no battery. Two other geometries are worth deriving once, because both come from the same $V = -\int \vec{E}\cdot d\vec{\ell}$ with a Gauss's law field:

$$C_{\text{cyl}} = \frac{2\pi\varepsilon_0 L}{\ln(b/a)}, \qquad C_{\text{sph}} = \frac{4\pi\varepsilon_0 ab}{b - a}, \qquad C_{\text{isolated sphere}} = 4\pi\varepsilon_0 R.$$

An uncharged capacitor does not have $C = 0$. It has the same $C$ it will have when charged, waiting.

Two facts about where the charge is, both of which get drawn wrong.

1. The charge sits on the INNER faces. $+Q$ and $-Q$ attract, so they pull each other onto the two surfaces that face across the gap. In the ideal parallel-plate capacitor the outer faces carry nothing. That is why the field is confined to the gap and why $C$ uses the facing area once rather than twice.
2. Nothing crosses the gap. The gap is an insulator. Current flows in the WIRES, delivering electrons to one plate and removing them from the other, at equal rates, so the circuit behaves as though current passed through. No charge makes the jump.

The field arithmetic follows from the first fact and is worth writing out, because two different answers are correct for two different questions. Each plate alone is a sheet with density $\sigma = Q/A$, producing $\sigma/2\varepsilon_0$ on both sides. Between the plates the two contributions point the same way and add; outside they point oppositely and cancel:

$$E_{\text{between}} = \frac{\sigma}{\varepsilon_0}, \qquad E_{\text{outside}} = 0.$$

But a plate cannot exert a force on itself. The force on one plate is its charge times the field of the other plate only:

$$F = Q\left(\frac{\sigma}{2\varepsilon_0}\right) = \frac{Q^2}{2\varepsilon_0 A},$$

always attractive. Using the full gap field here overcounts by a factor of two. Match the field to whose force you are computing: anything else placed in the gap feels $\sigma/\varepsilon_0$; a plate feels $\sigma/2\varepsilon_0$.

In a DC circuit, once the capacitor's voltage has risen to the battery emf there is nothing left to drive current, and the wire current is zero. A fully charged capacitor in a steady DC circuit is an open switch.

Anchor on the quantity that is physically shared and the addition rule is forced.

Series. The plates between the two capacitors form an isolated island of conductor. Whatever charge is pushed onto one side of it is pulled off the other, so every capacitor in a series string carries the same $Q$. Voltages add:

$$V = V_1 + V_2 = \frac{Q}{C_1} + \frac{Q}{C_2} \quad\Longrightarrow\quad \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}.$$

The equivalent capacitance is smaller than either one, which makes physical sense: stacking capacitors in series is like increasing the gap.

Parallel. Both capacitors are wired across the same two nodes, so they share the same $V$. Charges add:

$$Q = Q_1 + Q_2 = C_1 V + C_2 V \quad\Longrightarrow\quad C_{\text{eq}} = C_1 + C_2.$$

This is exactly the reverse of the resistor rules, and that reversal is the commonest source of lost points in the unit. Two consequences to keep on hand:

1. In series, voltage divides inversely with capacitance: the SMALL capacitor takes the LARGE share of the voltage, since $V_i = Q/C_i$ with $Q$ common.
2. In parallel, charge divides in proportion to capacitance, since $Q_i = C_i V$ with $V$ common.

A quick check for a series pair: $C_{\text{eq}}$ must come out below the smaller of the two. If your number sits between them or above them, you used the wrong rule.

Charging a capacitor means moving charge across a growing potential difference. Integrating $dW = v\,dq = (q/C)\,dq$ from $0$ to $Q$ gives

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

The three forms are algebraically identical at a single instant. They stop being interchangeable the moment the capacitor changes, because then one of $Q$ and $V$ is held fixed by the circuit and the other one moves. This is the single most important habit in the unit:

1. Battery still attached: $V$ is clamped. Use $U = \tfrac12 CV^2$. Then $Q = CV$ follows $C$.
2. Battery disconnected: $Q$ is clamped. Use $U = Q^2/2C$. Then $V = Q/C$ follows $C$ inversely.

Run the standard problem both ways. Pull the plates apart so that $d$ doubles, halving $C$:

$$\text{attached: } U = \tfrac12 CV^2 \ \text{halves.} \qquad \text{isolated: } U = \frac{Q^2}{2C} \ \text{doubles.}$$

Opposite directions, same physical action, and both are right. In the isolated case you did work pulling against the plates' attraction and it went into the field. In the attached case the plates' attraction still opposes you and you still do positive work, but the battery simultaneously reclaims charge, and the accounting comes out with less energy stored than before. Never quote an energy change without first writing down which variable the wiring froze.

The energy can also be located in the field itself, at density

$$u = \frac{1}{2}\varepsilon_0 E^2,$$

which multiplied by the gap volume $Ad$ reproduces $\tfrac12 CV^2$ exactly.