For those who might not be sure how a capacitor flash gun circuit works, I post below the circuit diagram of my Leica CEYOO. When you plug in a flash bulb, it completes the outer circuit and a small current, limited by the 10K resistor to just 0.00225 amp (V=IR and assuming the resistance of the bulb is effectively nil compared to the 10K resistor), so the bulb does not fire. This current charges up the capacitor. So how long do you need to leave the bulb in for the capacitor to charge? If my maths is correct and I have remembered my electricity from physics 50 years ago, at 0.00225 amps, a 100 μf capacitor would require just 0.04 seconds to charge up. When the flash contact switch is closed, the battery is effectively out of circuit and it is the capacitor only which discharges rapidly through the flash bulb. The current rise rate of the capacitor would be much faster than a battery, particularly an old fashioned zinc carbon multi cell small battery, which was all that was available in the 1950's. If you have a leaky capacitor and leave a bulb in the flash, the battery will run down quite quickly, which is why it is always a good idea, if you are leaving a bulb in a flash for display purposes, take out the battery.
I have now found a source for 100 μf axial wire capacitors and should have 2 waiting for me, when I get back to the UK next week.