Fixed by the hardware, not by the charge

The defining relation is $C = Q/V$, and it invites a causal misreading in which more charge produces more capacitance. It does not. Connect a bigger battery and $Q$ and $V$ rise together, in step, so their ratio holds still.

What actually sets $C$ is construction:

$$C = \frac{\kappa \varepsilon_0 A}{d}, \qquad \varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2),$$

with $A$ the plate area, $d$ the separation, and $\kappa \ge 1$ the dielectric constant of whatever fills the gap. All three are decided by the manufacturer.

Two consequences worth stating: an uncharged capacitor has exactly the same capacitance as a charged one, and doubling the battery voltage doubles $Q$ and leaves $C$ untouched. Capacitance moves only when the hardware moves.

Every problem about pulling plates apart or sliding in a dielectric turns on one question, and it should be the first line of your solution.

1. Battery still connected. $V$ is clamped at the battery's value. Then $Q = CV$ moves with $C$: reduce $C$ and charge flows back out through the wires.
2. Battery disconnected. $Q$ is stuck on the plates, with no path anywhere. Then $V = Q/C$ moves inversely with $C$: reduce $C$ and the voltage rises.

Carrying both $Q$ and $V$ forward from the before-state is the error, and it produces contradictions immediately. So does freezing the wrong one: "the plates are pulled apart after the battery is disconnected, so $V$ stays at $9$ V" pins the quantity the wiring released and releases the one it pinned.

Once you know which is pinned, the field and the stored energy follow without further decisions.

Between two large, closely spaced, oppositely charged plates the field has the same magnitude and the same direction at every interior point:

$$E = \frac{V}{d}, \qquad \text{everywhere in the gap except near the edges}.$$

Both plates contribute at every point, and their contributions add to the same total whether you stand next to the positive plate, next to the negative one, or halfway between. Importing the $1/r^2$ habit from point charges and reporting a field that fades across the gap is the error, and so is picking one plate as "the source" and making the field strongest there.

The payoff is that a charge released in the gap feels a constant force, so it accelerates uniformly. Launch it sideways and you get a parabola. These problems are projectile motion with $qE/m$ in place of $g$.

The energy stored in a charged capacitor is

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

All three are the same quantity; use whichever is built from the pinned variable. With a battery attached, $V$ is fixed, so $\tfrac12 CV^2$ shows what happens when $C$ changes. With the battery removed, $Q$ is fixed, so $Q^2/2C$ does.

The dependence is quadratic in both $Q$ and $V$. Raising a battery from $6$ V to $12$ V therefore stores four times the energy, not twice. That surprises people, so it is worth knowing where it comes from: the charge and the voltage build up together as the capacitor fills, so the average voltage during charging is half the final value, which is also the origin of the factor of one half.