I have had several students who have used super cap chargers. They provide a lot of nice features like variable current limit on the charging, the particular one that I linked to will also make it so you could use 2 caps with lower voltage on each one and it will make sure you don't over voltage one.
The only down side is the chips for these are VERY small and can be difficult to solder if you aren't very experienced.
You can also solve the capacitance of a capacitor using the capacitor charge equation.
\$V = V_p{e^{-t/RC}}\$
Based on what you can see in the equation, you have to wire the resistor and the capacitor in parallel, and then put a voltmeter in the capacitor. Then, all you have to do is to have a supply voltage (Vp), the resistor value (R), find the voltage of the capacitor (V) at time (t), and then you'll get the Capacitance of the capacitor. Assuming that you have a very accurate resistor reading, an accurate voltage source that does not sag when a load is connected to it, and an accurate voltmeter, given time (t), you will get an accurate enough value of the capacitor.
Here's an example:
![enter image description here](https://i.stack.imgur.com/MjyJU.png)
Let's say we have an unknown capacitor just like in the schematic. I grab my stopwatch, and start it the moment the switch is closed. I will then check the Voltmeter and take note of the time that passed to reach the capacitor voltage to, let's say 5 Volts. The capacitor reached this voltage in roughly 14 minutes and 30 seconds. Converting that to seconds and we will get 870 seconds. So we now have, V = 5V, Vp = 12V, t=870s, and R which is 20 Ohms. Plugging that into the equation will give me 49.69 F, rounding it up will give me 50 F.
Since you're dealing with a super capacitor here, charging times will be large enough to get a good reading as what I have demonstrated in the example above. However, if you need to measure a small capacitor, you can use a larger resistor that will charge the capacitor slower to prevent sub-second stopwatch readings, since we humans are limited to our ~20ms response time :)
Best Answer
Overview
In general capacitor lifetime (including supercapacitors) is dependent on three things:
If you want the capacitor to last a long time, limit the applied voltage, keep it cool, and limit the output current. All this should be in the datasheet of your capacitor.
Electrolyte Life
If the capacitor is, say, ceramic or tantalum, the electrolyte is a solid and the cap will basically never go bad. If it's an electrolytic, then it contains fluid which will evaporate and eventually cause the cap to fail. In an electrolytic double-layer super-capacitor, the electrolyte is a combination of a fluid and of activated carbon, so it is mildly vulnerable to evaporation.
Voltage Derating
More important, though, is the voltage derating of the capacitor. If the voltage burns the cap up on the first use, you won't need to worry about evaporation. A capacitor is a carefully constructed device which separates two conductive films with a thin, thin layer of insulating material over a wide area (folded or rolled into a package). Decrease the separation, and you've got higher capacitance with the same area. This thin separation is vulnerable to high voltages; that's why capacitors have specific voltage maximums, often printed on the case. In a supercap, this barrier is often just nanometers thick, and the dielectric will not insulate high voltages across this short distance.
Using a cap at close to its maximum voltage will cause it to fail more quickly than using it at a lower voltage, this tradeoff is known as a derating curve. It should be available from your capacitor manufacturer.
Temperature / Power Dissipation
A capacitor is negatively affected by heat. It causes the electrolyte to evaporate more quickly, causes the dielectric to be weakened, and it can damage the thin conducting elements in the capacitor. Both environmental heat and self-heating effects should be considered. If the capacitor is discharged very rapidly, the small resistance of the foil and leads will be inconsequential compared to the square of the current.