[Coda]

Short-circuiting batteries ARE dangerous. The smaller the battery, the less of a problem it is, but the current flowing without any sort of load to slow it down generates heat, and the heat causes the battery to degrade. The chemicals inside the battery start to react more vigorously, generating more heat. The heat causes the battery casing to weaken, AND the chemistry starts to produce gas (either as a result of the chemical reaction or as a result of the liquids boiling), making the battery swell. If the casing ruptures at all, then it leaks out caustic chemicals that can burn skin and corrode plastics and metals.

For lithium batteries it's even worse, because an external short-circuit runs the risk of creating a bridge of lithium metal deposits INSIDE the battery, at which point there's no stopping it -- it becomes a runaway cascade, building up more metal that makes the short circuit worse, and the whole time it's making heat and hydrogen gas until the whole thing melts down, catches fire, and destroys everything nearby.

[Potironette]

So batteries do burn--and do worse things.
Thanks for answering! I'll remember not to try short-circuiting batteries if I ever happen to obtain large batteries.

[Potironette]

Is there much of a difference between uniform fields and electric circuits? I keep hearing how there's a nice field and current running opposite the direction of the electrons in a circuit, and it feels like a uniform field to me..

Does E = ΔV/d work for circuits?

[Coda]

Hmm... I can see where you're coming from. At first I was going to say that an electric current isn't a field at all, but then I realized that from the perspective of the circuit itself it IS sort of like the anode and cathode are generating a uniform field between them.

I couldn't find a simple answer so I manipulated it for a while to see if I could come up with something myself, and it turns out that the answer to your question is "yes."

E = ΔV/d

Electric field strength is force divided by charge:

F/q = ΔV/d
Fd/q = ΔV
Fd = ΔVq

Force times distance is energy:

E = ΔVq

This is a well-known formula for finding the energy required to move a charge against a uniform field without acceleration, or the kinetic energy of a charge after being accelerated by such a field.

Divide both sides by time:

E/t = Vq/t

Energy per unit time is power (P), measured in watts. Charge per time is current (I), measured in amperes.

P = I * V

So it turns out that uniform electrical fields are equivalent to one of the most fundamental equations in electrical engineering. This is basically the F=ma of electricity, and it gets used ALL THE TIME. Need to know how big of a fuse your circuit needs? P=IV. Need to know what kind of power adapter will charge your device? P=IV. And so on.

I worked on this for a LONG time. I did a TON of research to see if there was some exciting correlation to see if E = ΔV/d might be a meaningful description of the behavior of a length of superconducting wire connected across a battery. Unfortunately, things just weren't making any sense at all. The units weren't working at all and I was coming up with contradictions and absurdities.

Then I realized my mistake... I was treating E in that formula as energy, but it's electric field strength. That was a HUGE WASTE OF TIME. ^^() Once I fixed that, coming up with the rest of the derivation was easy.

[Potironette]

Thank-you very much! And sorry for the time lost '';;

Umm, so in a uniform field electrons aren't going anywhere, but if an electron were to be placed in the field there would be. In a circuit, electrons are moving, but there does exist some sort of potential pushing(?)/pulling(?)/moving those electrons. And E = ΔV/d works as being equivalent to P = I * V but in the end an electric current isn't an electric field anyway?

Also, why is V = IR not ΔV = IR ?

[Coda]

Technically it IS ΔV = IR, because it's measuring the voltage drop across a resistive load, but when you're working with electric current you never have to deal with absolute voltage, so nobody bothers writing the delta.

The equivalence of the uniform field equations to Ohm's Law implies that having a circuit connected across a potential difference (such as the terminals of a battery) creates an electric field constrained to operate mostly inside the wire. (It's not perfectly contained within the wire, just like a field between two plates still has a field OUTSIDE the plates that isn't uniform; this turns out to be very important because that's how inductance works, which is how we can build transformers that can trade current for voltage while keeping power constant.) But BECAUSE it's constrained like that, it's really not worth TREATING it as a field for the most part. One of the biggest reasons it's not worth it is because distance really doesn't MATTER and resistance is the relevant proxy for that.

And the electrons are moving because they're present in the circuit and there's a potential acting on them, so in that sense it's not different.

Remember, it's both pushing (negative repels electrons) and pulling (positive attracts electrons), and that's why the force is uniform throughout the field / circuit.

[Potironette]

Why is distance important in a uniform electric field?

I'm not actually sure what induction is. In class, there was an example of induction as charging one object negative, then moving it nearby another object, then grounding the not-charged object, then moving the negatively charged object away so that the grounded object ended up with a positive charge. In the end, all I got from that was that induction was some indirect way of making an object charged...what's "induction"?

One of the biggest reasons it's not worth it is because distance really doesn't MATTER and resistance is the relevant proxy for that.

So..E = ΔV/d is like I = V/R ?

[Coda]

Originally Posted by Potironette

Why is distance important in a uniform electric field?

Because the farther apart the plates are, the weaker the field is.

I'm not actually sure what induction is. In class, there was an example of induction as charging one object negative, then moving it nearby another object, then grounding the not-charged object, then moving the negatively charged object away so that the grounded object ended up with a positive charge. In the end, all I got from that was that induction was some indirect way of making an object charged...what's "induction"?

Current traveling through a wire creates an electric field. Another wire inside that electric field will have a current induced in it. It's as simple as that.

So..E = ΔV/d is like I = V/R ?

Yup. The resistance of a wire is directly proportional to its length, even. And there's a similar formula for measuring the energy wasted as heat through a resistor (and wire is a resistor) that's just like E = Vq.

[Potironette]

Ohh, so that's what the distance means!

I see! I'd been confusing "charging by induction" with "induction."

There's a formula for that :o. energy lost = current * something?

[Coda]

Originally Posted by Potironette

Ohh, so that's what the distance means!

I see! I'd been confusing "charging by induction" with "induction."

There's a formula for that :o. energy lost = current * something?

Actually, it's P=IV again! A current of I amperes passing through a resistor with a voltage drop of V volts generates P watts of heat -- that is, P joules of heat per second. When you talk about a 1000 watt heater or a 100 watt incandescent light bulb, this is exactly what it's talking about.

There's another way to express it: P = I²R, where R is the resistance in ohms. These are related by Ohm's Law, I = V/R, which means that resistance reduces the current produced by a given voltage.

[Potironette]

Watts of a incandescent light bulb measure heat..? As in when something gets more heated with energy, it gives off more light energy?

So..power is energy, except by second?

Ohm's Law, I = V/R, which means that resistance reduces the current produced by a given voltage.

So that's what the division means! Come to think of it, does that mean every time I have division of A = B/C, such as m = F/a means that acceleration reduces mass by a given force (actually, I'm not sure that makes sense)?

[Coda]

Yes. Remember our discussion a while back about blackbody radiation? An incandescent light bulb is exactly that. Most of the light it gives off is in the infrared spectrum, which is to say, it's felt as heat instead of seen as light.

Power is energy per unit time, yes.

The division doesn't QUITE mean that. It doesn't strictly imply a cause-and-effect relationship, only an inverse-proportional one. If you have m = F/a, then that means that if you hold the force constant, then increasing the mass decreases the acceleration, or vice versa it means that if you hold the force constant and you observe an increase in acceleration then the mass must have decreased.

[Potironette]

Looking back at the blackbody posts, I remember that black bodies emit photons...are heat and light observations of the same thing? Or just that light can be felt as heat and humans can't see some lights? Actually, is light just "photons" and heat is atoms moving around or something, so whatever the light is from the bulb that people just happen to not be able to see is, will transfer energy to say, a human hand, to be felt as heat? Or is it just that in the case of a lightbulb, energy just gets transferred into making a lightbulb's glass move around and that is felt as heat? Or a combination @_@?

Oh, that makes more sense. Can I generalize all equations as describing an observed effect that may or may not actually happen?

[Coda]

Originally Posted by Potironette

Or just that light can be felt as heat and humans can't see some lights?

This is exactly it.

Actually, is light just "photons" and heat is atoms moving around or something, so whatever the light is from the bulb that people just happen to not be able to see is, will transfer energy to say, a human hand, to be felt as heat?

Another correct deduction!

Or is it just that in the case of a lightbulb, energy just gets transferred into making a lightbulb's glass move around and that is felt as heat? Or a combination @_@?

It's a combination. The glass is still transparent to most of the light passing through it, even the infrared light, but obviously it absorbs SOME of the energy itself because you can't unscrew a hot lightbulb.

Oh, that makes more sense. Can I generalize all equations as describing an observed effect that may or may not actually happen?

Yes, exactly. Physics formulas are models that mathematically describe observed relationships. They don't assert that any particular thing is possible, only that if it were possible you would expect to see that result.

It's particularly important to remember that they're models, too, because they can be wrong -- especially at the level you're working at, the formulas are approximations that work out within an acceptable margin of error within the domains you're likely to be able to observe. They also describe ideal interactions, but no real-world experiment happens in a truly closed system with 100% perfect components and 100% efficiency.

[Potironette]

I had a question on a quiz today asking for how the electric potential (I think) changed in a circuit of battery and a lightbulb with copper wiring that did have resistance. The problem is, I forgot what electric potential is...and so I'm not sure what potential difference is either.
What's the difference between electric potential and potential difference?

[Coda]

Potential difference is also called voltage, because it's measured in volts, and it's almost always what you use.

You can consider absolute electric potential to be the potential difference between the source of the field and a point at infinity, but you never use absolute electric potential when discussing circuits.

And as I described a couple posts ago, all elements in a circuit cause a voltage drop across them. If you have a circuit with a battery and a light bulb, then you'll measure a greater voltage between the positive and negative terminals of the battery than you will between the light bulb and one terminal.