3
$\begingroup$

I just wanted to know if voltage increases or decrease as u move charges away from each other. My understanding of voltage is very basic. From what i know, I see voltage/electric potential as something that increases as pull two unlike charges apart or as you squish two like charges next to each other ( so distance is decreased). So with this being said, does the length of wires in a circuit mainly determine what voltage a battery had? Please help!! I feel like i am confusing so many different concepts.

$\endgroup$

1 Answer 1

2
$\begingroup$

Voltage can be a bit tricky. Sometimes it's easier to split it up into two concepts, one for electrostatics and one for the dynamics we see when circuits are involved. In reality, it's just one thing, but it can be easy to understand how it behaves in the two extreme cases first and then begin the process of tying those explanations together.

The first half of your question deals with electrostatics, because you're talking about charges that aren't moving much on their own. Take, for example, the famous case of rubbing a balloon with wool, which transfers electrons from the wool to the balloon. After the balloon is charged, you can move it closer or further from the wool manually, but because the balloon and the air are both good insulators, the electrons really can't move much on their own. In electrostatics, we are interested in these cases, while the dynamics of electricity is interested in the cases with wires where the electrons can move at a good fraction of the speed of light!

The electric field is the field you are describing when you talk about moving charges towards each-other or moving them away. Electric fields are measured in Volts per meter. Moving a charged particle through an electric field requires work, or does work, depending on the direction of the movement and the charge of the particles. This is why, if you bring the balloon close to the wool, the wool rises up towards the balloon. The negatively charged particles of the balloon attract the positively charged particles in the wool, so the particles in the wool can do work moving towards the balloon. In this case, they do work by lifting the fibers of the wool up against the forces of gravity.

Now when we get to the dynamic situation we see with electricity, voltage behaves a bit different. You still technically have the exact same electric fields you did in electrostatics. All the laws of electromagnetism still apply. However, when it comes to conductors like wires, the shape of the electric fields are different. While it was very hard for a charge to move around on the balloon, it is very easy for a charge to move along a conductor like a copper wire. While the charges had little opportunity to jump off the balloon through the air to the wool, the charges in a wire are rather free to move from one place to another to find the lowest energy state possible.

The electric fields around a 1 foot length of copper wire conducting current from a battery look remarkably similar to those around a 10 foot length of copper wire. Accordingly, while we could still think in terms of electric fields, it's more convenient to think in terms of voltage directly (rather than the electric field and its volts/meter). When we think about the circuit in this way, we tend to say that a wire is at a voltage ("this wire is at 9V with respect to ground"), because many equations to predict how the circuit will work are built this way. Voltage causes current to travel through the circuit.

A very common analogy for this uses a garden hose. The water flowing through the hose is similar to current. The more water flowing, the higher the current. The voltage is similar to the water pressure. The water pressure causes the water to flow from high pressure to low pressure (from the pipes into the open air). If you have an impingement in the hose, like a kink, it acts like a resistor. It slows the flow of water. Raise the water pressure, and you can make more water flow past the kink.

Batteries are where we can start tying the two concepts of voltage and electric fields together. There are many chemical reactions which are capable of pushing an electron towards one side of a battery against a magnetic field. This starts moving electrons towards the negative side of the battery, and positive cations move to the positive side (slowly). Doing this causes an electromagnetic field to form as the electrons get pushed together on the negative side of the battery. As it turns out, this process reaches equilibrium when there is a certain voltage between the terminals (1.5V for alkaline batteries).

At this point, you have a system that can be completely understood by electrostatics. If a battery is 1.5V and 50mm long, there is an electric field generated between the two sides of the battery of a strength 0.03V/mm. We can attach wires to the terminals, which will reshape the magnetic field. If you do this and then bring the free ends of the wires to 5mm away from each other, the electric field generated will be 0.3V/mm, ten times higher than before.

If you keep bringing the ends of the wires closer and closer together, eventually the electric field rises to around 3000V/mm. This is the breakdown voltage of air, and its the point where air stops acting like an insulator and starts acting like a conductor (just like the wire). This is the most complicated part of any electrical system, where you must consider all of the effects of electrostatics and all of the effects of electrodynamics. This can show up as sparks or many other effects. You probably don't need to worry about exactly what happens here, but I wanted to point it out as the transition point where we tend to stop using electrostatics to describe things and start using electrodynamics.

When there's a path for electrons to travel through conductors and resistors from one side of the battery to the other, we have a circuit. When this happens, the electric fields in the wires encourage the electrons to race from the negative side of the battery to the positive side of the battery. The higher the voltage of the battery, the more "force" is put behind these electrons. When they reach the positive side of the battery, they recombine with the cations they left behind at first. Then, the chemical reaction moves them back to the negative side, and they race around the circuit again.

In this situation, we typically ignore the electrostatics because their effects are dwarfed by the massive number of electrons cycling through the circuit. We can ignore the number of charges in a wire because those effects are far smaller than the effects of the charges racing through the wire. At the extreme end of the spectrum, we have the abstract concept of an "ideal wire" which carries current instantaneously, the "ideal voltage source" which generates a voltage on those wires, and other concepts.

$\endgroup$
6
  • $\begingroup$ Thanks for the great answer. although a bit late, what would happen if we had a wire with the length of 1 kilometer that connected the two terminals of the battery? do electrons care if the other end of the wire is the positive terminal OR do they just flow inside the wire no matter what's at the other end of it until they are exhausted? Thanks. $\endgroup$
    – Cypher
    Commented Jun 15, 2019 at 7:00
  • 1
    $\begingroup$ @Cypher The electrons don't care what's connected at the other end in the sense that electrons operate "locally." They act based on the fields at that point. When you connect the wire to the battery, it is locally advantageous to spread out from the negative terminal throughout the wire. Spreading out the charges decreases the potential, so the electrons will always do it. This process will initially go until the charge density on the wire is basically that of the charge density of the negative terminal. At that point, it could become a simple electrostatics problem. $\endgroup$
    – Cort Ammon
    Commented Jun 15, 2019 at 14:19
  • 1
    $\begingroup$ However, if the wire also connects to the positive side, a second set of effects happens. It happens a bit slower. When the electrons in the wire reach the positive terminal, they find a lot of positively charged molecules in the battery which can be made neutral by combining with the electrons coming from the wire. This takes those free electrons on the wire and binds them up into molecules. This changes the delicate balance of a chemical equation inside the battery, which causes the chemical reactions to produce more electrons on the negative side and more positive molecules on the positive $\endgroup$
    – Cort Ammon
    Commented Jun 15, 2019 at 14:22
  • 1
    $\begingroup$ Since the positive side has soaked up some of the electrons, it has a lower electron density than the wire near it, so electrons flow from the wire into the positive terminal, and electrons flow from the negative terminal into the wire. The whole process takes a faction of a second. It's governed by the capacitance and inductance of the wire and the rates of the chemical reaction of the battery and the speed of light. It takes roughly 5us for the electrons in the system to equalize the charge over 1km of wire $\endgroup$
    – Cort Ammon
    Commented Jun 15, 2019 at 14:26
  • 1
    $\begingroup$ (Though you will later learn about drift velocity. While the cumulative movement of all of the electrons is extremely fast, 50% of the speed of light or faster, the movement of any individual electron is actually astonishingly slow. This slow movement is called the "drift velocity." It's not an important detail for the problem you're asking about, but its good to know that each electron moves rather slowly, so you aren't surprised when they tell you about it later) $\endgroup$
    – Cort Ammon
    Commented Jun 15, 2019 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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