Ttyjukl

Is the higher voltage just a safety thing, or can the capacitor actually store more charge?
Also, if I supply it with 24v, will it output 24v or 50v?

I read the wikipedia on capacitors, watched a few youtube videos, and read a few other articles, and they were all very vague about this.

If you exceed the voltage rating of the capacitor, the capacitor may fail.

If you put 24V across a cap, it will charge to 24V.

If it's spec'ed at 22000uF it's 22000uF for ANY voltage.

If I supply 24v to a 50v rated 22000uf electrolytic capacitor, does that mean it will store 44000uf at 24v?

No. The capacitor's capacity is a property of its dimensions.

$$ C propto frac {A}{d} $$ where A is the area of the capacitor and d is the distance between the plates. Note that voltage doesn't appear in the equation so it's still 22,000 μF even at zero volts.

Is the higher voltage just a safety thing, or can the capacitor actually store more charge?

The capacitor has a maximum voltage that the dialectric between its plates can withstand if you exceed this the dielectric will break down, current will flow and the capacitor will overheat and possibly explode.

The amount of charge stored is given by $$ Q = {C}{V} $$ so the higher the voltage on a given capacitor then the more charge is stored. Again, check for nonsense by testing for 0 V: at 0 V the charge, Q, will be zero. This makes sense.

Also, if I supply it with 24v, will it output 24v or 50v?

If you charge it to 24 V, disconnect the supply and measure the voltage you will get a reading of 24 V. Once you start to draw current from it the voltage will decrease.

Figure 1. When discharging through a resistor the time constant, τ can be calculated as RC. After RC seconds the capacitor will have discharged by 63%. After 3RC, 95% and after 5RC by about 99%. Image source: I am Technical.

Finally, a few points: SI symbols for the volt is 'V' and for the farad is 'F'. The symbols are capitalised but units named after a person are lowercase when spelled out.

I'll try to explain with a gradual approach:

A capacitor consists of two plates or conductive surfaces separated by a gap or an insulator (called the dielectric). Voltage is potential difference between any two points. Another name for voltage is electromotive force, which starts to convey the idea that it describes how "motivated" an electric charge is to do work.

I won't try to explain the how, but suffice to say that the positive and negative charges are attracted to each other, and will "stay" on the plates opposite the dielectric, until provided a connecting path (circuit).

A capacitor's voltage rating is based on its ability to function correctly up to the specified voltage. It encompasses things such as properties of the dielectric and physical dimensions of the capacitor. With enough voltage, insulators will eventually conduct (usually temporarily) and break down. Thus, if a capacitor is rated for 50 volts, it means that the dielectric composition, placement, thickness, etc. are only designed to function up to 50V.

The Farad is a unit of capacitance or amount of electric charge. (Technically it's equal to one coulomb (one ampere of current in one second) stored with the potential difference (voltage) of one volt.)

Analogies always fail at some point, but think of a capacitor like a sphere in a water plumbing system which has a flexible diaphragm separating it into halves¹. If you pump water into half of it, you're storing some quantity of water. If the pressure of the water is too high, you'll rupture the diaphragm. You can store some quantity of water up to the maximum capacity of the vessel, and it will be stored at whatever pressure was applied. (Obviously this analogy fails when you think of some small quantity of water not even causing the diaphragm to stretch at all, so you should dispense with hydraulic analogies as soon as possible...)

You can therefore use a 50V 22000µF capacitor to store charge at 1 volt, 10 volts, or 24 volts. In all cases, it will store a maximum of 22000µF, at the same voltage it was applied. Exceeding 50V will cause it to fail. (Applying 51V will probably result in a slow failure over time whereas 100V will cause a much more abrupt and potentially dangerous failure.)

You might be thinking "Well, if it can store 22000µF at 50V, why can't it store twice as much at half the voltage?"

It's tempting to think this if, for example, you're familiar with pneumatic systems: You have a two liter tank and store compressed gas in it. You can store a given quantity of gas at some pressure, and a larger quantity of gas at a higher pressure. (Within the tolerances and specifications of the gas and vessel, of course.) But this is not an analogy to electric charge. In order to understand a capacitor, you have to avoid thinking about it as either a pneumatic or hydraulic system — things which many people have everyday experience with prior to learning about electronics.

A capacitor won't store more than its rated capacitance at any voltage, and won't function properly (read: survive) exceeding its voltage rating. Also be careful to observe polarity. Generally capacitors with such capacity are polarized. Connecting positive and negative incorrectly will also result in failure, often dramatically.

¹Credit to Warren J. Beaty for this particular analogy.

'the voltage rating is just a limit. "halfway full" means it can hold more charge, but only with higher voltages.'

Thank you dandavis, and the rest of you for answering it broadly