Lets say I want to know how many mA per hour my supercapacitor can supply and assuming my capacitor is rated 2.7V and has 2F:

```
Farad = (Ampere per second) / Volt
Farad * Volt = Ampere per second
so
2F * 2.7V = 5.4 Ampere per second
5.4 Ampere per second = 0.0015 ampere per hour //dividing it by 3600
0.0015Ah = 1.5mAh
```

edit: this part is wrong

~~but. this assumes constant voltage. Actually there is a linear drop in voltage until the capacitor reaches zero.~~
~~So the actual energy it can provide is 1.5mAh/2~~

~~Answer: ~~
**A capacitor with 2F and 2.7 volt can supply 0.75mAh** (asuming I can somehow use all the energy to the last bit.)~~*~~
~~Is this correct?~~