How does a Leyden Jar work?

studiot

Joined Nov 9, 2007
4,998
The whole point of the Leyden Jar is that there were no batteries or field generators available in those days.
Further, it's mode of operation enabled the early pioneers to correctly figure out the very basics of electricity.

So studying how it works or worked is of worthwhile interest to many.
 

alfacliff

Joined Dec 13, 2013
2,458
even I study the history of electricity and electronics. I have studied how measurements were made in the old days too, it helps gain an understanding of how our field works. but too many attribute magic to things that are easily explained today. the leyden jar was used to study "electrical fluid" and storage of "fluid" electricity. these days who conciders electrical current or charges as fluid?
 

nsaspook

Joined Aug 27, 2009
16,370
these days who conciders electrical current or charges as fluid?
Well, the water analogy is still commonly used today to explain electricity to beginners, very poorly IMHO as most people don't understand hydraulics either at that point in their education and make assumptions that have to be relearned later.
 

alfacliff

Joined Dec 13, 2013
2,458
I even use that water and pipe analogy when I teach ham radio classses to people with no background at all in eectricity. there are even analogies for teaching propagation, antenna theory, and such. sometimes its a little hard to dumb down stuff enough to get people to understand, but its still posable.
 

nsaspook

Joined Aug 27, 2009
16,370
Teaching by analogy is a good thing but and this is a biggie to me the analogous system must be something they already understand at some depth so they can also see the differences between systems clearly. I think it's better just to say that electrical energy is a unique force, start with charge, how it interacts with other charges with fields and build on that if the group will deal with electricity in ways more complex than home wiring where water or marbles in a tube will suffice.
 

studiot

Joined Nov 9, 2007
4,998
Yes I agree one should start with charge as the fundamental.

But, of course, static charge first. so water and other analogies are not needed.

Unfortunately, too many courses these days fail to give a suitable mechanics grounding before that so seriously hampering any half ways decent study program or progression.

I also like to emphasis the idea of a model.

Much of what we do is a model of part of the full reality of what we are modelling.
So long as we acknowledge that our 'truth' is just a model that follows certain laws, rules or equations in its behaviour within the constraints designated we will not need to worry that our model is not compelete.
We can, after all, always seek out a better model if we want or need to go beyond the original constraints.

In the case of the Leyden Jar (or other capacitor) this is evident in the dielectric model of the glass. We normally model the dielectric as if it had a surface charge equal to the polarisation. We know this is not actually the truth and we have more accurate models, but always at the expense of complexity. The polarised fictitious surface charge serves well in many applications.
 

nsaspook

Joined Aug 27, 2009
16,370
Yes I agree one should start with charge as the fundamental.

But, of course, static charge first. so water and other analogies are not needed.

Unfortunately, too many courses these days fail to give a suitable mechanics grounding before that so seriously hampering any half ways decent study program or progression.

I also like to emphasis the idea of a model.
I agree, the problem is I see today is the model of electricity some are being taught in the beginning is actually a Potemkin village that looks solid and polished up front with limited examination but when you need it to hold up under a little strain you find it's really just cardboard with sticks holding it up and the whole framework collapses leaving you with little to build on. Just about everyone gets past this hump later but it seems a large waste of resources.

http://www.chinanews.com/sh/news/2009/07-06/U177P4T8D1763067F107DT20090706164724.jpg :D
 
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alfacliff

Joined Dec 13, 2013
2,458
and then when you are teaching a class of policemen, housewives and such, analogies are pretty much al you have time to do without confusing the %%%% out of them. they arent going to be echnicians or engineers, ZI just hope I can get a little into them so they arent too unsafe, and can operate their radios properly. a couuple of hours a week for six weeks isnt enough time to explain how a mosfet works at the molecular level.or even a plain old bipolar transistor for that matter. some times it is real dificult to dumb it down enough to keep their intrest.
 

studiot

Joined Nov 9, 2007
4,998
and then when you are teaching a class of policemen, housewives and such, analogies are pretty much al you have time to do without confusing the %%%% out of them. they arent going to be echnicians or engineers, ZI just hope I can get a little into them so they arent too unsafe, and can operate their radios properly. a couuple of hours a week for six weeks isnt enough time to explain how a mosfet works at the molecular level.or even a plain old bipolar transistor for that matter. some times it is real dificult to dumb it down enough to keep their intrest.
So yes it's all down to an appropriate level of model.
 

davebee

Joined Oct 22, 2008
540
Steps 1-3 and 5 look fine to me, but not step 4. If the same charged object were brought up to the charged Leyden jar a second time, I don't think any additional charge transfer would take place.

A van de Graaff generator will accumulate charge, but only because charges are physically carried through a repulsive field into the inside of the sphere. The Leyden jar would not work like that.

Aside from that, I think you're on the right track.
 

Thread Starter

superconductor

Joined Jun 27, 2010
11
Steps 1-3 and 5 look fine to me, but not step 4. If the same charged object were brought up to the charged Leyden jar a second time, I don't think any additional charge transfer would take place.
Thank you! Just to clarify, I of course mean that the charged object that you bring in contact with the Leyden jar must be re-charged each time!

Here's another question on the leyden jar:

Some of my readings imply that the amount of charge that can be stored in a Leyden Jar is significantly greater than the amount of charge that could be stored if the Leyden Jar were to consist only of the inner metal.

Intuitively, this seems to make sense. I mean, why bother with the outer metal if you could store just as much charge without it? (Although perhaps the purpose would be to store both charge on the inner metal as well as on the inner metal.)

I don't exactly see why the presence of the outer metal would affect the amount of charge stored in the inner metal. Do you have an idea?
 

davebee

Joined Oct 22, 2008
540
As one of the earlier posters noted, a Leyden jar is a capacitor.

The charged object used to put a charge on the Leyden jar is also a capacitor.

When you connect a charged capacitor to an uncharged capacitor in parallel, charges will flow until there is no voltage difference between the two capacitors.

Voltage on a capacitor is V=Q/C, amount of charge divided by the capacitance.

A larger capacitance will absorb a greater amount of charge in order to rise to the same voltage as a smaller capacity capacitor.

In practical terms, this means if you bring a charged metal object up to a Leyden jar, most of the charge of the object will pass into the jar. For the same amount of charge, the much higher capacity Leyden jar will result in a much lower voltage of combined capacitors.

So you're right in your previous post that if you repeatedly recharge your object and touch it to the Leyden jar, the jar will appear to keep on absorbing the charge. It will absorb more charge until its greater capacity is finally raised to the voltage of the charging object.

Roughly speaking, if the Leyden jar has ten times the capacitance of the charged object, it will take about ten charges to bring the jar up to the original voltage of the external object.

Anything that increases the capacitance of the jar will increase this effect - a larger jar, thinner glass, glass of a greater dielectric constant. This explains why less outer metal will reduce the capacity of the jar.
 
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