The complex, highly technical formula for capacitors is
Q = C V
Charge = (capacitance) (voltage)
Charge = (3 F) (24 V)
Charge = 72 Coulombs
The positive plate of the capacitor is missing 72 coulombs worth of electrons. They were sucked into positive terminal of the battery stack.
The negative plate of the capacitor has 72 coulombs worth of extra electrons. They came from the negative terminal of the battery stack.
You should be aware that this is a humongous amount of charge ! An average lightning bolt, where electrons flow between a cloud and the ground for a short time, is estimated to transfer around 15 coulombs of charge !
The scenario in the question involves a "supercapacitor". 3 F is is no ordinary component ... One distributor I checked lists one of these that's able to stand 24 volts on it, but that product costs $35 apiece, you have to order at least 100 of them at a time, and they take 2 weeks to get.
Also, IF you can charge this animal to 24 volts, it will hold 864J of energy. You'd probably have a hard time accomplishing this task with a bag of leftover AA batteries.