I'm learning about parallel plate capacitors, in series and in parallel. I just don't understand it conceptually very well. I know what capacitors are, they store energy by separating a charge and keeping them at a distance apart but I really have no comprehension of how/why, when in series or parallel, they would have the same charge and/or different voltage across them and vice versa.

I'm told that when in series, the capacitors will have the same charge but different voltage and when in parallel they will have the same voltage but different charge?

Quite possibly I won't understand anything yet until later topics I need to learn?Havent hit circuits or electric current stuff yet. But, I don't learn well without visualizing and conceptualizing in my brain, and I just can't do that with this stuff.

Capacitors in series or parallel are just the opposite of resistors in series or parallel. Think about in a capacitor the area of the plates and what happens in series or parallel assuming just two identical capacitors.

Start with basic DC theory and simple laws and then move on to more complex things. It's a very big pool and the deep end is not where to learn to swim.

In the real world you almost never see capacitors in series. With DC, the voltage divides unpredictably depending on leakage resistance, and the total capacitance is reduced.

With AC signals the capacitive reactance (impedance), which is inversely proportional to capacitance, sums-up like resistance in series.

When you see capacitors in parallel you'll usually see different types of capacitors. Electrolytic capacitors don't "act like" capacitors at high frequencies so in a power supply you might see a 1000uF electrolytic (the main filter capacitor) with a 0.1uF ceramic capacitor to prevent high frequency oscillation.

You can analyze/calculate the charge on each capacitor separately. In series the voltage gets divided (unpredictably in the real world) so with less voltage across each capacitor the total charge is less (than when in parallel).

Q =C*V => V = Q/C | C = capacitance
| V = voltage
| Q = Charge in Coulombs

Charge: Q = CV where C is the capacitance in Farads, V is the voltage across the capacitor in Volts and Q is the charge measured in coulombs (C). Energy stored: W = ½ QV = ½ CV2 where W is the energy measured in Joules

footnote: When I worked at Compulaser (formerly KORAD Lasers) in the early 80's they had a laser spot welder they sold to Los Alamos Atomic Labs for spot welding fuses on H-bombs. It was the most reliable
spot welder Los Alamos Labs could find. It was old school. Had two High Voltage caps that were size of a
5-gallon gas can (and heavier). I made the company name by spot welding paperclips shaped into letters
using needle nose pliers and a pair of dyke wire cutters. The Marketing guy found it and absconded with it
and had it gold plated to put on his desk. I bet he still has it.

However, the whole topic is a bit abstract as part of learning the fundamentals of electricity.

If you ever saw an old radio, with an aluminum vane tuning capacitor with trimmers, you’d have come quite close to a practical application of this, that is plate capacitors and capacitors in parallel. However, as has been pointed out, capacitors exhibit different properties in DC and AC applications.

One thing that might not be obvious if you're new is that capacitance is more analogous to conductance
than to resistance (conductance is the reciprocal of resistance, ie current divided by voltage).

capacitance is charge divided by voltage in fact. (at any particular AC frequency that implies capacitance
is proportional to current/voltage).

This why the series/parallel laws are opposite to resistors. Its an arbitrary human convention that we
use charge/voltage rather than voltage/charge for the value of these components.

Whenever components are in series the current is the same, but the voltage sums, so the overall
voltage/current increases.

Whenever components are in parallel the voltage is the same, but the current sums, so the overall
voltage/current decreases (since current/voltage increases).