Why does the voltage goes away in my leyden jar?

Question

This question is similar to this one in the sense that I have my amateurish Leyden jar and I'm supplying ~7.5v from 5 batteries (~1.5v each) in series.

Some details and facts after experimenting:

• The brown wire is touching the water with salt inside the bottle and is connected to the lower voltage side of the battery
• The red wire is touching the aluminium and the higher voltage side of the battery
• If I use a multimeter touching the aluminium and the water, I get my ~7.5v, as expected
• If I remove the wires touching the battery and measure the voltage, there is none
• If I remove the wires touching the battery while the probes of the multimeter are connected I can see the voltage drop very very quickly.
• Making some calculations I estimate that the jar has around $ ~1.1pF $

Then I decided to connect the circuit with the led that you can see in the figure. A similar circuit with a capacitor $ 220 \mu F $ attached makes the led fade-away instead of switching off when I remove the battery. So I've been trying to see the same effect using the Leyden jar to convince me that it actually holds some charge.

However as soon as I separately the brown and red wire connected between the bottle and the batteries the led switch off without fade-away effect.

Does anybody know what could affecting the performance of my Leyden jar?

Answer 1

The problem with your Leyden jar is that its capacitance (like that of all Leyden jars) is very small (probably in the order of 1000 pF). So if its leakage resistance (through moisture films on the glass and so on) is 100 M$\Omega$, the time constant of the circuit will be $$\tau=CR\ \ =\ 10^{-9}\ \text F \times 10^8\ \Omega\ \ =\ 10^{-1}\ \text s.$$ This is the calculated time for the capacitor's charge to fall to 37% of the original!

There is obviously some guesswork involved in the calculation above, and the resistance could be out by one or two orders of magnitude, but the calculation certainly reveals why the Leyden jar capacitor discharges so quickly. With a voltmeter connected across it the discharge will be even quicker, because the voltmeter resistance is likely to be a mere 20 M$\Omega$ or so!

Answer 2

As has been pointed out using a digital voltmeter of input resistance$10\,\rm M\Omega$ will result in a discharge time constant of $200\times 10^{-12} \times 10 \times 10^{6} = 2 \,\rm ms$.

So for a direct measurement of voltage you need a voltmeter/electrometer of much higher input resistance which you may not have access to.

A cheap/easy to build high resistance voltmeter is a (gold) leaf electroscope but such a device will only register hundreds of volts so to see if the Leyden jar is storing charge you need to connect it to a high voltage dc supply which is likely to be very dangerous.

So build yourself an electrophorus (this one has included a leaf electroscope in disguise).

Dry all the bits of apparatus with a fan heater / hair dryer as surface moisture is a relatively good conductor of electicity, and note that the insulation on your connecting lead might not be as good as you think so have as much of the non-earthed lead in the air (ie not touching anything except the the two ends) as you can.

Connect the leaf electroscope to the inside of the jar and the outside of the Leyden jar and the electroscope together (and to earth) and repeated charge the Leyden jar.
When there is a deflection on the electroscope stop charging and observe the deflection over a period of time. The slower the dropping of the leaf the better is the insulation.

If you charge the Leyden jar sufficiently you may be able to generate a spark when the inner and outer metal parts of the Leyden jar are connected but be very cautious when you do this but as you may get a nasty shock. The use of a plastic ruler might be considered?

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I have just found a ridiculously sensitive electric charge detector which may enable you to test your Leyden jar charged from your bank of $1.5\,\rm V$ cells or even just one cell and a milliamp meter might also be included instead of the led. The circuit might well work as the JFET gate resistance is very high.