2
$\begingroup$

I would like to experiment with leyden jars, but do not want to mess with anything big enough to kill me should I get zapped.

What are some good guidelines as far as size when making homemade Leyden Jars?

Here is a picture of one I created, could it carry a lethal charge? Leyden Jar

Thanks! Ben

$\endgroup$
6
  • $\begingroup$ They have detailed instructions on sciencebuddies, with a warning that "while the amount of current that can flow from this device is low, the amount of charge held in the device can cause a mild to moderate shock". $\endgroup$
    – Conifold
    Mar 22 '17 at 1:54
  • $\begingroup$ Ben: I will actually recommend your question be closed. Not that I don't like it - I think there a lot of physics in there - but your question prompted me to read about what can happen and I just don't think you should get your safety advice from this site. $\endgroup$ Mar 22 '17 at 2:08
  • 4
    $\begingroup$ I'm voting to close this question as off-topic because I do not think Physics SE should be a place to get safety advice. $\endgroup$ Mar 22 '17 at 2:09
  • $\begingroup$ What kind of experiments do you want to do with the device? $\endgroup$ Mar 22 '17 at 3:37
  • $\begingroup$ Safety advice should not be given on this site other than "Do not do it without some professional advice" Generalisations like "It is unlikely to be fatal" should not be made. In this example you do not know about Ben's health or indeed he may not know if he is particularly susceptible to electric shocks. This question and all similar questions should be closed. $\endgroup$
    – Farcher
    Mar 22 '17 at 7:01
1
$\begingroup$

The safety considerations for Leyden jars are the same as for any capacitor; these depend on how you intend to use the device and how much energy you put into it. Leyden jars have small capacitances, but they can withstand very high voltages, which means that they can store a great deal of energy, e.g. if you charge them with a Van de Graaff generator, for example.

Estimate, or better, measure the device's capacitance with an LCR meter or impedance meter.

You should measure the capacitance and check that it agrees with the estimate. A good estimate will be gotten from the formula (which you can derive from the integral form of Gauss's law):

$$C = \frac{\epsilon\,A}{d}$$

where $A$ is the surface area covered by the foil on one side of the jar, $\epsilon$ the low frequency dielectric constant of the glass (can be up to 10, considerably more than the square of the optical frequency dielectric constant) and $d$ the thickness of the glass (distance between the two foils). For 1mm thick glass, you'll get a figure of the order of $10^{-8}{\rm F}$ - about ten nanofarad.

The glass dielectric has a very high breakdown or dielectric strength; typically this is ten to thirty MV per meter, meaning that the device can withstand ten to twenty thousand volts across its electrodes. At that degree of charging, you have stored $\frac{1}{2}\,C\,V^2$ or of the order of one to ten joules. You're unlikely to get to this, but you could approach it if you charged the device with a Van de Graaff generator and this is approaching the realm of defibrillator level energies.

In short it is unlikely to be lethal, but it could be if misused. You need to plan what you want to do with it. You could try measuring the voltage across it in its charged state in a typical experiment with a high impedance voltmeter, and work out the energy stored in it with the formula $\frac{1}{2}\,C\,V^2$ with $C$ set to ten nanofarads. If you come up with a figure of more than a few millijoules, then you're in danger territory.

For example, with $A=0.05{\rm m^2}$ (wine or 1 liter softdrink bottle size), $d=1{\rm mm}$ and a relative permeability of glass, I get a capacitance of $4.4{\rm nF}$. To limit this device's energy to $0.001{\rm J}$, therefore, you need to limit the charging voltage to about $660{\rm V}$. In this state it would still deliver quite a whack to your hand. I would check these figures with a capacitance measurement for your actual device; some plastics can have rather large permeabilities.

$\endgroup$
1
  • $\begingroup$ Thank you for a very helpful and thorough response. This gives me a much better idea on how to gauge the capacitance and safety of these devices. $\endgroup$
    – Ben
    Mar 22 '17 at 11:29

Not the answer you're looking for? Browse other questions tagged or ask your own question.