If you are new at this, start with a 0.022uF cap or thereabouts, as it is easiest to see the difference between the two orientations. If you cannot see a large enough induced AC signal by holding the capacitor between your fingers, place the capacitor on top of an AC line cord (that is plugged into the mains wall socket, of course!) instead of holding it between your fingers and you will see a larger signal on the scope. Mark it, and connect that side of the cap to the lowest impedance point in the circuit, typically the driving source plate when used as a coupling cap, or the grounded end if used in a shunt position. The orientation with the lowest induced signal is the one you want, and the ground lead of the scope is connected to the outside foil in that position. While still holding the capacitor tightly, reverse the scope leads and you should see a dramatic difference in the amplitude of the induced AC signal. Grab the capacitor tightly with your fingers, and note the amplitude of the induced 60Hz AC signal (or 50Hz if you are on the other side of the pond). Set the scope up to the most sensitive vertical scale (20mV or less, preferably) and connect the scope probe across the capacitor (ground to one side of the cap, probe tip to the other). However, if you have access to an oscilloscope, you can do a simple test to determine which is the outside foil terminal. If the capacitor has no banded end, the outside foil connection could be on either end, so there is no easy visual method to determine the best orientation of the capacitor. This marking of the outside foil was very common in the "good ol' days" of electronics, but, sadly, most capacitor manufacturers nowadays do not bother to mark the outside foil, so we're left to fend for ourselves. What if the capacitor doesn't have a banded end? Tubes with lower internal plate resistances, such as the 12AT7, will have even lower output impedances. If the cathode resistor is unbypassed, the output impedance is a bit higher, around 68K or so, depending on the value of the cathode resistor, but still well below the input impedance of the next stage. The plate, on the other hand, has an impedance equal to the internal plate resistance of the tube in parallel with the plate resistor (assuming the cathode is bypassed), which for a typical 12AX7 is around 38K total. Because of this, it would make a very poor choice for electrostatic shielding. #Bi polar caps in signal path seriesThe grid of the tube itself is very high impedance, and it is usually shunted by a high resistance of 220K to 1Meg, and also usually has a large series resistance as an interstage attenuator as well. This is completely wrong, because the grid circuit is a very high impedance point in the circuit. I have seen where a well-known guitar amplifier "guru" said to connect the banded end to the grid of the next stage because it is at ground potential. For this reason, it is sometimes a good idea to bypass electrolytic capacitors with a smaller value foil or other type capacitor. As a side note, electrolytic capacitors have an internal resistance that tends to rise with frequency, which can make the capacitor less than ideal as a bypass at higher frequencies. For all practical shielding purposes, connecting the outer foil to the power supply rail is just as good as connecting it to ground. This means that the large electrolytic bypass capacitors in the power supply are effectively "short circuits" to AC signals above a certain very low frequency. Since reactance is effectively a measure of the "AC resistance" of the capacitor, the capacitor will exhibit a very low resistance at higher frequencies, while looking like an open circuit for DC and frequencies low enough to make the capacitive reactance significant. As you can see from the above equation, the frequency term is in the denominator, so as the frequency increases, the capacitive reactance decreases.
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