Hi everyone. Im an older student to electronics and have been researching how batteries and capacitors hold charges, but dont disipate them when connected to other oppositley charged elements. Could anyone please verify if my analysis is correct



(the following 2 1.5V batteries are connected in series, but the positive end of left battery and negative end of right battery are an open circuit)

On a battery, there is a surplus of positive ions on the positive terminal and a surplus of negative ions on the negative terminal, however overall the battery as an entirety has a zero net charge. Similar to the way an atom with equal protons and neutrons has a zero net charge. Because there is no net charge (no surplus of charged ions) on the battery as an entirety, they do not leave the battery. They are attracted to each other due to the electromagnetic field, but cannot reach each other cause the battery electrode is insulated (similarly to the way the ions on a capacitor are separated by insulation) Connecting 2 batteries together would be like putting 2 net zero atoms together, they wont do anything.

The only way I could think to make the electrons leave the negative terminal of the battery, would be the use a very high positive potential to pull them away from the battery's electromagnetic force using something similar to the highly positively charged anode on an electron gun) to pull the electrons away, leaving the battery with an overall positive net charge.

If this is theoretically possible the battery would lose its potential energy fast as there is no electrons going back in to combine with the manganese oxide and water molecule in the cathode so no new electrons will be made.

Could someone please tell me which statements you would disagree with and why.

Thank you very much in advance! Tom P

 

Welcome to AAC!

Could someone please tell me which statements you would disagree with and why.

If this is schoolwork, we can't just give an answer. If it isn't we can offer more than guidance. Do you really want this treated as homework?

 

I'm not sure where to begin. A battery contains stored chemical energy, which manifests itself as a voltage difference at the poles. It's like a pond at the top of a hill. Any time that current flows, it "falls down the hill", converting some of the chemical energy into electrical energy.

For your batteries in series, the only path for current is through the surrounding medium, presumably air. A small voltage cannot drive current through air and so there is virtually no current and no energy transformation taking place. Connected in series or not makes virtually no difference.

On a battery, there is a surplus of positive ions on the positive terminal and a surplus of negative ions on the negative terminal".

I don't think that's right but I'll let the experts weigh in.

 

Hi everyone. Im an older student to electronics and have been researching how batteries and capacitors hold charges, but dont disipate them when connected to other oppositley charged elements. Could anyone please verify if my analysis is correct

View attachment 302702

(the following 2 1.5V batteries are connected in series, but the positive end of left battery and negative end of right battery are an open circuit)

On a battery, there is a surplus of positive ions on the positive terminal and a surplus of negative ions on the negative terminal, however overall the battery as an entirety has a zero net charge. Similar to the way an atom with equal protons and neutrons has a zero net charge. Because there is no net charge (no surplus of charged ions) on the battery as an entirety, they do not leave the battery. They are attracted to each other due to the electromagnetic field, but cannot reach each other cause the battery electrode is insulated (similarly to the way the ions on a capacitor are separated by insulation) Connecting 2 batteries together would be like putting 2 net zero atoms together, they wont do anything.

The only way I could think to make the electrons leave the negative terminal of the battery, would be the use a very high positive potential to pull them away from the battery's electromagnetic force using something similar to the highly positively charged anode on an electron gun) to pull the electrons away, leaving the battery with an overall positive net charge.

If this is theoretically possible the battery would lose its potential energy fast as there is no electrons going back in to combine with the manganese oxide and water molecule in the cathode so no new electrons will be made.

Could someone please tell me which statements you would disagree with and why.

Thank you very much in advance! Tom P

Welcome to AAC.

You are neglecting the requirement that there is a circuit. The cells have a low resistance path but it doesn’t return. The only path is through the air. The cell voltage is orders of magnitude to low to use that path, so nothing happens.

The reason the cell produces electric current because the two components inside undergo a chemical reaction and exchange ions through the electrolyte. There must be a path capable of conducting at the potential of the cell, from anode to cathode for this to happen. Without this, a small leakage current will eventually cause self-discharge but there is no way for something not between the two terminals of the battery to exchange electrons with the battery using DC.

If you use something that is very high voltage, static electric effects are possible, but they won’t affect the chemical energy of the battery and don’t rely on it.

 

Could anyone please verify if my analysis is correct

A basic rule of electricity is that there can be no continuous current flow without a complete circuit for the electrons to travel.
If you don't have that ,then your further discussion about where the electrons go is superfluous.

 

I don't like to to think of current flow as electrons moving from a place where there is surplus of electrons to a place where there is deficit of electrons.



For your specific questions, yes, there is an instantaneous flow of charge until the electric field is reduced to zero. The time and amount of charge to do so is so short that you would not be able to observe this charge flow.

 

Let's just say a current does flow, with electrons flowing from the battery on the left into the one on the right. That battery would then have a negative charge and push away any more electrons.

 

There is ALWAYS a circuit. It's just a matter of degree. Agreed, a circuit requiring current to pass through the air is effectively not a circuit, but there is a field and a medium and both can be described quantitatively. I thought this concept might be helpful to the TS but maybe not. Just ignore it if doesn't help the understanding.

 

Hi everyone. Im an older student to electronics and have been researching how batteries and capacitors hold charges, but dont disipate them when connected to other oppositley charged elements. Could anyone please verify if my analysis is correct

View attachment 302702

(the following 2 1.5V batteries are connected in series, but the positive end of left battery and negative end of right battery are an open circuit)

On a battery, there is a surplus of positive ions on the positive terminal and a surplus of negative ions on the negative terminal, however overall the battery as an entirety has a zero net charge. Similar to the way an atom with equal protons and neutrons has a zero net charge. Because there is no net charge (no surplus of charged ions) on the battery as an entirety, they do not leave the battery. They are attracted to each other due to the electromagnetic field, but cannot reach each other cause the battery electrode is insulated (similarly to the way the ions on a capacitor are separated by insulation) Connecting 2 batteries together would be like putting 2 net zero atoms together, they wont do anything.

The only way I could think to make the electrons leave the negative terminal of the battery, would be the use a very high positive potential to pull them away from the battery's electromagnetic force using something similar to the highly positively charged anode on an electron gun) to pull the electrons away, leaving the battery with an overall positive net charge.

If this is theoretically possible the battery would lose its potential energy fast as there is no electrons going back in to combine with the manganese oxide and water molecule in the cathode so no new electrons will be made.

Could someone please tell me which statements you would disagree with and why.

Thank you very much in advance! Tom P

Yes, there is a way called EM radiation. If you use a very fast DPDT switch to a set of wires (dipole antenna) from the battery to flip the polarity on the wires at something near the resonant (for efficiency) frequency of the 'antenna' you have a condition where the circuit is the universe. Electrons will be accelerated (to neutralize the charges across the wires) as the polarities are switched resulting in EM fields containing EM energy that move into space (a EM circuit) being extracted from the chemical energy of the battery .

 

An ordinary small rectangular 9V alkaline battery has six tiny 1.5V battery cells in series.
A Name Brand of Energizer guarantees them to last on a store shelf (no load) for 5 years.

 

You are making the common mistake of assuming that batteries are essentially the same as capacitors. They aren't.

A battery isn't a capacitor! | All About Circuits

I don't know if the equations will render correctly for you -- they don't for me any longer, but I haven't been able to find a way to get them to work in blog entries.

 

I don't know if the equations will render correctly for you -- they don't for me any longer, but I haven't been able to find a way to get them to work in blog entries.

Screenshots will be the easiest.

 

You can take two charged capacitors and connect them in series, end to end. The result is the same.
Don't expect to see current flowing between the two capacitors.

 

Screenshots will be the easiest.

There's something missing in rendering the article, I'm not sure what the rendering language is, screenshot is not a great help:

 

Screenshots will be the easiest.

Screenshots of what? The rendering engine isn't rendering the equations properly anymore (and hasn't for years, but it used to). So the screenshots I would take will be the same non-rendered MimeTex code that anyone else sees when they try to view it.

 

There's something missing in rendering the article, I'm not sure what the rendering language is, screenshot is not a great help:

View attachment 302749

The rendering engine is MimeTex, the same engine that is used rendering tex code in forum posts. But, for some reason, it is not working properly in the blogs any more.

For instance, if I cut and paste the contents of the blog here, it renders fine:

\(
Q_{capacity} \; = \; 10Ah \cdot \frac{3600s}{1h} \; = \; 36kC
\)

Yes, thirty-six thousand coulombs!

Now let's see how much charge we would have to have on our two capacitor plates to produce the 1.5V that the battery has, at least nominally, when new. Ignoring edge effects and fringing fields and treating the battery as an ideal parallel place capacitor, we have

\(
E \, = \, \frac{ V }{ d } = \, \frac{ \sigma }{ \epsilon }
\)

The electric field therefore needs to be

\(
E \, = \, \frac{ V }{ d } \, = \, \frac{ 1.5V }{ 60mm } = 25 \frac{V}{m}
\)

To achieve this field with a vacuum between the plates, we would need a charge density on each plate of

\(
\sigma \, = \, E \epsilon _o \, = \, 25\frac{V}{m} \cdot 8.85 \frac{pF}{m} \, = \, 221 \frac{pC}{m^2}
\)

With a plate that is 30mm in diameter, the total charge on the plate would then be

\(
Q_{voltage} \, = \, \sigma A \, = \, \sigma \pi R^2 \, = \, 221 \frac{pC}{m^2} \cdot 3.14 \cdot \left( \frac{30mm}{2} \right)^2 \, = \, 0.156 pC
\)

But in the blog, it simply surrounds the MimeTex code with \( and )\ delimiters.

 

Screenshots of what? The rendering engine isn't rendering the equations properly anymore (and hasn't for years, but it used to). So the screenshots I would take will be the same non-rendered MimeTex code that anyone else sees when they try to view it.

Screenshots from a post on the forums (or preview of one), inserted in the blog post in place of the \( \LaTeX \) markup...

 

Ok so, I've done some more research from multiple sources and I've prepared a new view of viewing this

- Firstly, both the positive and negative ends of a battery are charged, even without any wires connected at all.

- These charges, particularly on the surfaces will not have alot of charge until the battery circuit is connected (either with 1 battery or multiple batteries in series or parallel)

- If I connect the negative end of battery 1 to the positive end of battery 2, any surface charges will equalize around the wire (creating a small current) until the potential from the battery 1 negative to battery 2 positive is 0. (however the potential from battery 1 positive to battery 2 negative will be doubled assuming the batteries are the same potential) During this process of equalization some chemical equalization and of course electrical equaization will occur.

And I have read, that removing this wire from the 2 batteries will actually loose a small amount of electrical force at the terminals as the charge is chained through the wires from terminal to terminal.

"When you remove the wire, you take away some of those electrical forces that act on the charges near the terminals (there were charges on the wire after all, held there by a chain of charges leading back to the other battery where there are chemical forces)

I'm not sure exactly what he means here regarding the taking away of some electrical forces near the terminals. But I understand because of the fact that the current was moved during the equalization means there was energy used and it came from the charges on the ends of the batteries

- The reason the current stops and causes an equilibrium is because it is restricted to whatever surface charges are in battery 1 from the negative terminal as without a completed circuit, battery A cannot maintain the surface charges with chemical to electrical force to make current flow in the wire.

 

I have to disagree.
When the wire is removed, nothing else happens since all terminals are in equilibrium.

Another point to note, if the connecting wire were connected to earth ground, there would be no excess or deficit of charge at both terminals. When the connection is removed, again nothing changes. There is still no excess or deficit of charge at both terminals.

 

Screenshots from a post on the forums (or preview of one), inserted in the blog post in place of the \( \LaTeX \) markup...

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Ah. I see what you are getting at. That's a possibility. Would be nice if the forum software simply rendered tex code in blogs properly.