A battery isn't a capacitor!

There is a widespread -- and not too unreasonable -- misconception that a battery is basically a charged capacitor that has a reservior of net positive charge on one side and a reservior of net negative charge on the other side.

Somewhat related to this misconception is a difficulty in understanding why you need a complete circuit for current to flow in. Basically, why can't electrons flow down a single wire since this is what happens when electrons leave the negative terminal of a battery? Why do they have to be able to come back to the positive terminal? Why can't they just go wherever and stay there? For instance, why can't they go to the positive terminal of some other battery, particularly if that battery isn't fully charged?

So let's look at this by looking at some of the numbers involved. To give us those numbers, let's consider a 1.5V D cell battery and treat it as a parallel plate capacitor consisting of two circular plates the diameter of the battery separated by the height of the battery. For a standard D cell, these numbers are approximately 30mm and 60mm, respectively.

At a reasonable current draw, an alkaline D cell has an energy capacity of about 10Ah (it ranges from about 2Ah to 15Ah depending on current draw and what cutoff voltage is used, so 10Ah is not unreasonable for our purposes). Notice that amp-hours is directly a unit of charge since one ampere of current for one second means that a total of one coulomb of charge has flowed past. Hence the total charge stored on this capacitor would be

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

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 \( \frac{30mm}{2} \)^2 \, = \, 0.156 pC
\)

Notice that the ratio between the amount of charge that needs to be stored in order to provide the 10Ah capacity of the battery and the amount of charge that is needed to result in the 1.5V terminal voltage of the battery is

\(
\frac{Q_{capacity}}{Q_{voltage}} \, = \, \frac{36kC}{0.156pC} \, = \, 230 \times 10^{15}
\)

Or more than 17 orders of magnitude! Another way to look at it is that if the battery really did consist of reservoirs of charge, the terminal voltage would start out at 345 million billion volts!

Now, you say that you wouldn't make a battery with a vacuum between the plates, you'd use a dielectric with a really big dielectric constant. Fine. Note that this wouldn't change the charge associated with the 10Ah capacity at all. But it would increase the charge associated with obtaining the terminal voltage by a factor equal to the dielectric constant.

But how big can the dielectric constant get? Most materials used for capacitors have dielectric constants in the range of 3 to 5. Many liquids, such as water, are in the 50 to 200 range. Some glasses get into the 1000 range. Some exotic materials, such as calcium-copper-titanate get into the 10,000 range and, under some conditions, have been claimed to get into the 250,000 range. But even if we could get 6cm think chunks of this stuff for our battery, we are still twelve orders of magnitude off and would be looking at a battery with a fully charged terminal voltage of over 1 million million volts.

So I think we can forget about the notion that a battery is some reservior of stored charge.

Before we move on, let's ask another question. Using our vacuum capacitor model again, how many electrons have to be moved from the positive plate to the negative plate in order to achieve the 1.5V between the terminals?

The amount of charge that has to be moved from the negative plate to the positive plate is 0.156 pC, which is

\(
Q_{voltage} \, = \, 0.156 pC \cdot \frac{1e^-}{-1.603 \times 10^{-19} C}\, = \, = -974ke^-
\)

Since the number of electrons that have to be moved from the negative terminal to the positive terminal is negative, that means that 974 thousand electrons -- roughly one million -- actually have to be moved from the positive terminal to the negative terminal.

[RANT] For those that insist that conventional current ignores the fact that its the electrons that actually move and therefore is not physically correct, perhaps now you can start seeing that this isn't the case. It is completely consistent as long as you recognize that there is a difference between talking about charge and talking about charge carriers -- something that proponents of electron current never seem to do and hence end up with all kinds of inconsistencies in their own descriptions. [/RANT]

So we only need a charge separation of about a million electrons to give us the 1.5V. A million sounds like a lot, but not when we are talking about electrons. A D cell battery is designed to operate at currents of about 200mA to 2A. So using 1A, how long does it take for 0.156 pC of charge to flow?

\(
t \, = \, \frac{0.156pC}{1A} \cdot \frac{1A}{1\frac{C}{s}} \, = \, 0.156ps.
\)

For comparison, the typical propagation delay of the fasted 74-series logic families are about 1.5 ns, or 10,000 times this long.

The take-aways from the discussion thus far are the following:

1) It takes very little charge, relatively speaking, to result in the kinds of voltage differences that we see in circuits.

2) It takes very little time for this very little amount of charge to move around at the kinds of currents that we see in circuits.

So with this in mind, let's take a look at how a battery really works. At the risk of oversimplifying things (okay, not a "risk" so much as a "certitude") you have two materials forming your terminals and some chemicals in between. Because of the nature of the materials and chemicals, ions are formed that want to bond to the terminals and, in the process, essentially move a few electrons from the positive terminal to the negative terminal. But in doing so, a charge separation develops and a voltage difference starts to build. At some point, this voltage difference creates a strong enough electric field to prevent ions from continuing to collect on the terminal materials and the chemical reaction stops. In fact, if the electric field were to become any higher, the tendency would be to drive the electrochemistry the other direction (assuming it is possible) and detach ions and effectively more electrons from the negative plate back to the positive plate. This is essentially what happens when a rechargeable battery is recharged.

We can now appreciate that the number of ions that it takes is so small that there is effectively no accumulated charge on the terminals and that the timescale on which it happens is so small that it is effectively instantaneous.

Instead, what happens when we connect the battery into a circuit is that an electron is removed from the negative terminal and, on the other side, added to the positive terminal (not the same electron!) which reduces the electric field within the battery enough for the chemistry to produce one more ion (or pair, in some cases) that accumulates on the terminals and replenishes the charge that flowed out into the circuit.

But if we just remove an electron from the negative terminal and don't add one to the positive terminal, the electric field isn't reduced much at all. To see this look at the voltage at the same distance from a uniformly charged sphere with 0.156 pC of charge on it. And, in removing this electron and not replacing it, we have a net electric charge on the battery as a whole that wants to pull that electron back!

So very few electrons are going to migrate away from that terminal if there is no means for electrons to get into the other side. And since this applies to just about everything that we might use to supply that electron to the positive terminal, we quickly (almost instantaneously) end up with a situation in which the tiny amount of current that does flow when you hook just the positive terminal of one battery to the negativer terminal of another battery stops almost immediately -- in fact, things have actually all-but equilibrated before the wire even makes final contact because whatever redistribution of charge is going to happen takes place continuously as you move the wire toward the terminal in response to the parasitic electric fields and parasitic capacitances that exist between everything.

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