2
$\begingroup$

Batteries generate fields in wires that essentially cause the movement of electrons. I think of batteries as two charged plates that essentially contain a mechanism between them to move electrons from the positive to negative plate.

If you have two charged plates they would essentially create a field everywhere.

Now consider a circuit that consists of two batteries and a resistor. The first battery would generate a field everywhere. The second battery would also generate a field everywhere.

Now say the first battery is 6V and the second battery is 3V. That means inside the first battery they're is a field that would result in 1C of charge gaining 6J of potential energy if moved from the negative to positive terminal.

But in a circuit, shouldn't the field from the second battery affect the field inside the first battery? In other words, shouldn't the field from the 3V battery affect the field inside the 6V battery? Since fields can be added, shouldn't the field inside the 6V battery change if a 3V battery is brought near it?

So why do we assume the a n-V battery remains constant in voltage regardless of what other batteries are near it?

$\endgroup$
  • $\begingroup$ Usually batteries have metallic covers and electric fields do not pass through metal. The field coming out of the two poles comes from spatially separated charge distributions within the battery, the geometry is not uniform. Even if the covers are not metallic the 1/r^2 drop of the field from the source giving it off will ensure that the interference from close by batteries will be small. $\endgroup$ – anna v Apr 10 '14 at 3:24
  • $\begingroup$ @annav 1)But in circuits batteries are typically close together so, the 1/r^2 drop shouldn't make a huge difference right? 2) Why don't fields pass through metal? I though the electric field is independent of the material its in. Is metal a dielectric? $\endgroup$ – dfg Apr 10 '14 at 3:45
  • $\begingroup$ close is a relative notion. Their centers are at least a few centimeters apart. 1/25 is already a small number . Metals are conductors. Electric field lines do not cross conductors en.wikipedia.org/wiki/Electric_fields $\endgroup$ – anna v Apr 10 '14 at 4:10
  • $\begingroup$ @annav But the field inside a conductor is zero. So the field inside the plates of the battery is zero but that doesn't imply the field between the plates is zero. The field between two conductors can be non-zero can't it? And also, are they're circumstances where the field of other batteries must be taken into account? And if so, could you please direct to the field that entails the study of this phenomena? $\endgroup$ – dfg Apr 10 '14 at 4:43
  • $\begingroup$ The field inside the conducting container of the batteries is non zero because there exist charges inside, from the charge separation. It is only an outside shell that is metal. doitpoms.ac.uk/tlplib/batteries/printall.php , look at the zink acid battery drawing. The two poles that come out of the metal shell create a field around them because of the charges inside the battery. Batteries are usually stacked in parallel and any stray fields will be parallel. $\endgroup$ – anna v Apr 10 '14 at 5:20
2
+50
$\begingroup$

Electrochemical cells (batteries) are not passive components, instead they're active charge-pumps having internal feedback effects which produces a relatively constant voltage at the output terminals.

If an external field impinges on a battery's terminals, this will produce a temporary small change in potential on the terminals. But the battery then actively responds with a very small, brief current which only persists long enough to restore the potential across the terminals to the same value it had before the external field was applied.

So, if you wave an electrically-charged balloon back and forth around one terminal of a D-cell, a very small AC current will appear in the battery terminals and within the electrolyte: exactly the current needed to maintain the ~1.5 volts of potential difference across the battery terminals.

To understand the details of battery operation, look into the physics of "half-cell" electrochemistry. Below is a very brief ELI5 version.

A metal plate dunked into fluid electrolyte will rapidly dissolve: rapidly like sugar or salt in water. But as metal atoms corrode away, they each leave behind one or more outer electrons in the metal's "electron sea." Quickly the metal becomes charged net-negative, and the electrolyte becomes charged net-positive by the population of metal +ions in solution. Electrostatic attraction tends to pull the metal ions back towards the opposite-charged metal surface, while thermal vibrations tend to make the atoms diffuse away into the electrolyte. When the "force of dissolving" equals the electrostatic attraction, the corrosion process slows to a halt. There will be a constant voltage left between the electrolyte and the metal. If an external circuit is used to increase this voltage, then the metal ions are forced back to the metal surface and so we have "charging a battery" and also electroplating. If an external circuit should reduce this voltage, then more atoms can dissolve from the metal surface and so we have "discharging a battery" and also electrolytic corrosion. Note that a single "battery" cell consists of two different metal plates connected by a common electrolyte, and the voltage on the terminals is simply the difference between the half-cell voltages which appear in the thin region between the electrolyte and each metal plate. Note that all components are conductors, and the entire "charge pump" is located in a molecules-thin layer adjacent to each metal surface.

A battery is a chemically-fueled, constant-voltage charge pump.

$\endgroup$
  • $\begingroup$ Thank you. From what I understand, your essentially saying that the battery will always maintain a constant voltage across its terminals - any external fields will be canceled out through an internal current. Is this correct? Also, is the battery only able to counteract changes that decrease the voltage in the battery or is it also able to counteract changes that result in the voltage between the terminals increasing? $\endgroup$ – dfg Apr 13 '14 at 1:37
  • $\begingroup$ Yes, exactly. And in both directions: to maintain the ideally constant voltage, the battery can run forwards or reverse, both discharging as well as charging its stored energy. But note that all of the above this would be a very, very, VERY tiny effect. Crude calculations: empty space is like a pF capacitor, in series with a battery which behaves like a short circuit: like a tenth-ohm resistor. $\endgroup$ – wbeaty Apr 13 '14 at 19:24
  • $\begingroup$ Thanks a lot buddy. I've lost track of how many times you've taken a tough concept and explained in terms that make sense to me... :) $\endgroup$ – dfg Apr 13 '14 at 20:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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