# How does a Leyden Jar work?

Discussion in 'Physics' started by superconductor, May 13, 2014.

1. ### superconductor Thread Starter New Member

Jun 27, 2010
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(If this is not the correct place to post this, can someone please let me know what is.)

In a Leyden jar, I have read that a charged object is brought in contact with the conductor in contact with the metal inside the jar, thus giving the inner metal a similar charge. And the metal outside the jar then gets an opposite charge.

For purposes of discussion, let's assume that the charged object is positive, so the metal inside the jar becomes positive and the metal outside the jar becomes negative.

My question is simply how does the metal outside the jar get the opposite charge, given that there is an insulator in-between?

I'm wondering if what is meant is not that the outside metal gets charge, but rather that polarization occurs within the outside metal, so that the inner surface of the outside metal becomes negative while the outer surface of the outside metal becomes positive, but the outside metal has no net charge?

At any rate, I'd be very grateful for an explanation of what is going on.

Apr 16, 2011
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3. ### crutschow Expert

Mar 14, 2008
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A Leydon jar is just a simple two plate capacitor with glass as the dielectric.

4. ### alfacliff Well-Known Member

Dec 13, 2013
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they work pretty good.

5. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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Howdy:

The charge on the outside of the Leyden jar is RELATIVE to the charge on the inside. They can BOTH have a charge relative to some outside reference....in addition to the differential charge.

6. ### superconductor Thread Starter New Member

Jun 27, 2010
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1. hexreader: thank you but the video doesn't address my question at all.

2. crutschow: Thank you for letting me know that a Leyden Jar is a capacitor. I'm not sure if I made it clear, but I don't know what a capacitor is. If I did and understood how it worked, then I would not have had to post this question. I hope you can see that.

Perhaps you meant for me to read up on capacitors and find the answer there? I read several sources, such as wikipedia and howstuffworks, and a few videos, on capacitors. Most of these sources, which I found difficult to follow, did not address my question. They assume that the plates take on opposite charges, without explaining how that happens, which is my question. I spent about 30 minutes reading on capacitors, and even then, I'm not sure what the answer is.

3. alfacliff: I don't understand how your post helps. Can you please clarify?

4. KL7AJ KL7AJ: You write that the outside plate becomes negative relative to the inside plate. Every single source I've seen states that the outside plate takes on a charge opposite to the inner plate. Not one source spoke of a relative charge.

For example, here is what howstuffworks writes: "The inner tinfoil receives an electric charge which induces an opposite charge on the outer tinfoil." This sounds like there is an actual absolute opposite charge on the outer plate, not a relative charge.

Here's another source: http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Leyden_jar.html. It states: "The inner and outer surfaces of the jar store equal but opposite charges." Opposite charges means that if one is negative, then the other is positive, so that the outer plate is actually becoming positive, not that it is simply positive relative to the inner plate.

So I still do not know what is going on with the Leyden jar. It seems almost certain that no electrons are passing through the insulator. That's the whole point of an insulator! I saw some articles speak of leakage in the insulator, but that seems to be a problematic situation, not the normal mechanism that the Leyden Jar is based on.

So it would seem to me that one of two things is taking place.

1. This video: https://www.youtube.com/watch?v=t9Qwx75eg8w seems to be saying that the inner negative foil repels the electrons of the outer foil, and causes those electrons to actually leave the outer foil into the surrounding air, thus the outer foil is left with an absolute (not relative) outer charge. However, this seems to be discussing a capacitor which is hooked up to a circuit, and thus would not be relevant to the Leyden jar.

On the other hand, this video https://www.youtube.com/watch?v=4k-z5O3Xk4A states explicitly that you give the outer plate the opposite charge to the outer plate by holding it. So this seems to be saying that that's how a Leyden Jar works: you ground the outer conductor, and that's the mechanism that allows a charge opposite to the inner conductor to build up in the outer conductor.

This seems surprising, since this seems to be the only source on Leyden Jar suggesting this mechanism. Most sources on Leyden Jars don't seem to indicate that the outer surface must be grounded.

What does seem certain is that the outside foil is not acquiring charge by being in contact with the surrounding air, since air is an insulator, I believe.

2. However, some sources seem to indicate that what I suggested in my original post is the correct answer, that when we speak of the outer foil taking on an opposite charge, what we mean is, as I stated:

"that polarization occurs within the outside metal, so that the inner surface of the outside metal becomes negative while the outer surface of the outside metal becomes positive, but the outside metal has no net charge."

Here is what wikipedia states on capacitors: "The conductors thus hold equal and opposite charges on their facing surfaces." And here is what wikipedia states on the Leyden Jar: "In capacitors generally, the charge is not stored in the dielectric, but on the inside surfaces of the plates." This sounds like what I'm saying, since it is stressing the inside of the surfaces, implying that on the outer surface not so. But I may be drawing an incorrect conclusion.

So I'm truly confused now, since I now have two different offered explanations, which seem to contradict each other.

I hope that someone respond very specifically and in some detail to the question I'm raising: what exactly is meant and what exactly takes place on the outer metal of a Leyden Jar to cause it to take on a charge opposite to the inner metal, given that there is an insulator between the two.

7. ### alfacliff Well-Known Member

Dec 13, 2013
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the voltage difference between the inner and outter plates is not induced, induction is for inductors, coils, not capacitors. the outside voltage is relative to the inside voltage, and the outside is usually grounded. the reason the outside voltage is relative to the inside is that without a reference, voltage makes no differnce. what would the voltage be compared to? the inside and outside.

8. ### MrChips Moderator

Oct 2, 2009
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1) The Leyden jar is the original capacitor or "condenser".

2) The inner foil is given a charge of either polarity depending on the source of potential. An equal and opposite charge is induced in the outer foil. The experimenter holding the jar in his hand creates a path to ground. If the jar is resting on a table, the table and surroundings act as ground.

Energy is required to charge the inner foil and create the opposite charge on the outside. This energy is stored in the capacitor in the form of positive and negative charges on opposite plates.

3) A capacitor can be created without the glass jar. The plates can be separated by air or even in a vacuum. The material between the plates is called the dielectric and is characterized by its dielectric constant.

The dielectric plays an important role in the capacitor and the resulting value of the capacitance of the capacitor. Even though the dielectric is an insulator, charge separation is induced in the dielectric. This increases the capacitance value.

You can demonstrate the memory effect of the dielectric.
Charge a capacitor, 1μF or greater. Discharge the capacitor by shorting the leads of the capacitor. Measure the voltage across the capacitor with a DMM and you will find the voltage reading gradually increasing.

9. ### alfacliff Well-Known Member

Dec 13, 2013
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and an odd effect is that materials that increase the dielectric constant are the same as the ones that increase difraction in lenses.

10. ### davebee Well-Known Member

Oct 22, 2008
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I'd say that in regard to your original question, both polarization and charge transfer would generally take place.

If the outer conductor is insulated then when charge flows onto the inner conductor, the outer conductor will become polarized. Charges opposite to those on the inner conductor will migrate closest to the inner conductor, leaving the outer part of the outer conductor charged with the same sign as the inner conductor.

If the outer conductor is connected to ground or another conductor, those outer charges will flow to the other conductor, leaving the outer conductor both polarized and with a net charge equal and opposite to the charge on the inner conductor.

Read about the electrophorus; its basic operation depends on these same principles, charge induction and charge flow from one side of a polarized conductor.

https://en.wikipedia.org/wiki/Electrophorus

11. ### Alec_t AAC Fanatic!

Sep 17, 2013
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Think of the dielectric as being a slab full of atoms. Think of each atom as electrons (negative charges) circling around a positive core. Just as magnet poles of the same polarity repel each other, electric charges do likewise. If you place a charged object against one face of the dielectric slab, the electrons and positive cores of all the dielectric atoms will be repelled from or attracted towards the object depending on the polarity of the charged object. The result is a 'strained' dielectric where negative charge accumulates on one face and positive charge accumulates on the opposite face.

12. ### crutschow Expert

Mar 14, 2008
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If it's not clear yet, the charge difference that occurs on a capacitor's plates require a current path between the plates, otherwise the displaced charge has no place to go. Thus when you apply a voltage across the two plates of a capacitor or Leyden jar electrons will be removed from the positive plate and an equal amount of electrons transferred to the negative plate. The amount of electrons moved is proportional to the capacitance of the capacitor and the voltage applied (Q = CV where C is the capacitance if Farads, Q is the charge in coulombs and 1 coulomb equals approximately 6.241×e18 electrons).

13. ### studiot AAC Fanatic!

Nov 9, 2007
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The easiest alternative to the word dielectric is insulator.
Essentially a true dielectric is an insulator.
So a perfect vacuum is a dielectric.

The opposite of an insulator is a conductor.

All matter has some characteristics of a conductor and some of an insulator.
With metals the conductor part dominates.
With ceramics and glasses the insulator part dominates.
But the is no known matter that is a perfect insulator or a perfect conductor under normal conditions.

Aug 27, 2009
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15. ### studiot AAC Fanatic!

Nov 9, 2007
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NSAspooky polarisation is not the same thing as the polarisation of light.

16. ### superconductor Thread Starter New Member

Jun 27, 2010
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alfacliff: Your posts are far too advanced for me to understand them at all, so unfortunately they were of no help to me.

Alec t: Thank you for your explanation of what happens within the dielectric (insulator).

crutschow: I largely find your post very hard to understand, due to my very weak background in electricity.

I want to make sure that I understand your first sentence: "If it's not clear yet, the charge difference that occurs on a capacitor's plates require a current path between the plates, otherwise the displaced charge has no place to go."

Translating that into my layman's terms, are you saying that in order for the outside metal to acquire a charge opposite to the inside metal, there must be a path between the two? That certainly seems obvious: if the two plates were completely disconnected, then obviously neither would have any impact on the other. Are you making some point beyond this that I'm missing?

Also, what exactly do you mean by the term "current path"?

Finally, what do you mean by the "displaced charge"?

Please recall that I'm trying to understand a Leyden Jar, so if your sentence regarding a capacitor is not relevant to the leyden jar, then for now I don't need to understand it.

The rest of your post is way above my head, as my background in electricity is clearly far less advanced than yours. Perhaps at a later date I'll try to understand it, but for now, it would take me far too long.

17. ### crutschow Expert

Mar 14, 2008
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Sorry about that. Sometimes I find it hard to judge the degree of knowledge of the poster.

No. My (somewhat obvious) point is that both plates must have a path for the electrons that go from one plate to the other. Some newbies have a problem understanding that, thinking that you can have a current without a complete path.
That's simply the conductive path for the current (charge) that moves from one plate to the other due to the applied voltage.
That's the charge that's moved from one plate to the other by the voltage difference applied to the plates.
A Leyden Jar is a capacitor so anything I said about a capacitor applies to a Leyden Jar.
The rest of the discussion pertained to the amount of charge that is moved, which depends upon the capacitive value of the capacitor given in Farads and the applied voltage. (A Farad is practically a very large capacitor so typical capacitors have values in the microfarad or picofarad region. For example a typical small Leyden Jar would likely have a capacitance in the 1000-2000pF region.) A capacitor with a larger capacitance (as from a larger plate area [larger Leyden Jar], less distance between the plates [thinner glass], etc.) will move more electrons ("store more charge" as it is commonly called) for a given applied voltage.

Hope that helps some.

18. ### studiot AAC Fanatic!

Nov 9, 2007
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I tried to simplify some of the discussion for you in post#13.

One thing to note about the Leyden Jar is that is was charged in a different way from which we normally charge capacitors these days.

These days we connect to a source of electricity such as a battery or mains derived power supply.
These have two connection terminals.
And all these modern sources derive their electricity in an entirely different way from the original method.

So all the comments relevent external circuitry applies to these, not to the original charging method.

The original charging method was purely mechanical. Electric charges (they didn't then know what it was) are mechanically 'rubbed off' from one body and added to another.
A metal connection leads from the collecting body to the brass knob on the top of the leyden jar. This metal connection provides a path (we say conducts) for the mechanically collected charge to enter the jar and pass down the metal chain inside.

In order to understand what then happens and also why the collected charge would do this it is necessary to understand two things.

1) There are two types or polarities of charge. We call them positive (+) and negative (-).

2) Like charges repel each other and unlike charge attract each other.

All the charges rubbed off onto the collecting body will be rubbed off in the same manner so will be of the same sign.
So they will repel ie they will want to get as far away from each other as possible.
They can't easily move in an insulator but they can in a conductor so they take advantage of any available conductor to spread out.
So they move down the metal connection and into the jar.

Note I haven't said which polarity and it is possible to charge the jar either way.

However according to rule (2) newly arriving charges in the jar are repelled by those already there.
So it becomes successively more difficult to add charges into the jar.

The jar also has a metal coating in the outside. If this has even a weak connection to some source of the opposite charge to what has been placed inside, then these charges will pass along this connection to the outside metal coating as they are attracted to those inside.
The same effect will apply to these ie those already there will make each new one harder to attract, until the point where the outside repulsion balances the inside attraction and no further opposite charges are drawn to the outer coating.

Since the inner and outer coatings are separated by an insulator that does not allow passage of charge, the charges cannot meet and will remain in place for quite some time.

I also recommend you look again at the video suggested in post#2, with the above as a guide.

A bit further on at 27 minutes there is an animation of the Leyden Jar that you may have missed or the above may help make sense of.

Last edited: May 16, 2014
19. ### superconductor Thread Starter New Member

Jun 27, 2010
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The Leyden Jar consists of (a) the inner metal, (b) the insulating glass, and (c) the outer metal. And with those 3 items alone, somehow the outer metal becomes charged as a result of the inner metal being charged. (It becomes oppositely charged.)

So can you please explain how this happens? This is the only question I have asked all along, and as far as I can see, you have not yet answered it.

To everyone else, please note that this is my only question: how does one conductor cause another conductor to become oppositely charged when they are separated by an insulator?

There have been a lot of posts on this thread discussing many issues, but very little of the posts have been addressing this specific question, the only one I am seeking the answer to.

My apologies if I'm coming across as ungrateful or argumentative.

20. ### davebee Well-Known Member

Oct 22, 2008
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You're right, a charged inner conductor by itself can't charge the outer conductor, because as you correctly note, they are separated by an insulator.

But the fields from the charged inner conductor do reach across the insulator, and those fields can polarize the outer conductor; causing charges to move within the outer conductor.

Overall, initially the outer conductor will still be electrically neutral, but once it has been polarized, it will have charge on the side closest to the inner conductor and it will have an equal but opposite charge on the side farthest from the inner conductor, its outer side.

Once the outer conductor has been polarized, it will show a charge on its outer surface. That charge can travel to ground, if a conducting path is made available.

If that is done, then the outer conductor will have an overall charge - the inner half of the polarized set of charges, because the outer half has been bled away to ground.

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