I've seen the capacitor questions come up a few times, decided to post some information. The formulas were taken from a textbook, the rest is pretty much my own interpretation of everything. Questions or criticism welcome.
Capacitance
Capacitance is a capacitors ability to store an electric charge. When a capacitor is connected to a DC voltage
source the initial current is very high. Electrons from the negative side of the source pour into the negative plate of the capacitor. As a negative charge accumulates, an equal amount of electrons are repelled from the positive plate through electrostatic induction. The electrons flow out of the positive plate to the voltage source. The electrons flowing in and out of the capacitor give the illusion of current flowing between the plates even though they are not connected. As the charges continue to accumulate a voltage is developed across the plates. The voltage continues to increase until it matches the source voltage, and at this point the capacitor is fully charged and the current stops flowing.
The Farad
The Farad is the standard unit of measure of capacitance. One Farad is the amount of capacitance needed to store one coulomb of electric charge under the influence one volt. For those who may not be familiar with it, the coulomb is a measure of a quantity of electrons and the charge they have. The volt, amp and other units of electrical measure are based on it. One coulomb is 6.25 x 10^18 electrons, so a one Farad capacitor connected to a 12 volt source will store 7.5 x 10^19 electrons, or 12 coulombs of charge(coulomb = voltage x capacitance). This isn't very useful by itself since you can't count electrons, but it is used in the next equation, which solves for energy. Energy is measured in joules or watt-seconds. (one joule = one watt-second). The next equation is Energy = coulomb x voltage/2. Therefore, 12 x 12/2=72 watt-seconds.
72 watt-seconds is all you get out of a one Farad capacitor at 12 volts. The intention of the capacitor on the power wires of an amp is to help maintain the voltage at its highest level. If the voltage begins to drop, the energy stored in the capacitor is used to try and stabilize the voltage. If the drop lasts for one second, only 72 watts of power will come from the capacitor before its completly discharged. The power isn't spread evenly over the second, there's a rush at the beginning and it drops off quickly after that. Now the capacitor needs an additional 72 watt-seconds from the vehicle charging system to recharge, which was already strugling to maintain the voltage. The interesting thing here is that if the time of the voltage drops decreases, the power goes up. If the voltage were to drop for only 0.1 second, then 720 watts would be available over that 0.1 second before it was discharged (72 watts / 0.1 seconds). This is using 12 volts, if using 14.6 volts, the energy goes up to 106 watt-seconds for a one Farad capacitor. Over the same 0.1 second, you now have 1065 watts available, which I believe is enough to have at least some help in maintaining the voltage.
I still don't think a capacitor necessary since a proper charging/battery system will never experience a significant voltage drop.
None of the calculations here take into effect the many variables such as ESR of the cap, resitance if the wire,
charge and dischage curves, etc. This is only an overview of the whole thing, there are many other factors to
consider, and pages of additional calculations as well, but should be enough information to form your own opinion.
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