If you know the mass and acceleration of an object, then it doesn't matter what kind of force it is, you can find the magnitude of that force.

As for this particular question, though, you DON'T have the acceleration, and you don't have enough information to find the acceleration without solving the rest of the problem anyway.

But then if I know the mass and the force, I can find the acceleration...right?

------------If I can't then you can ignore this bottom part because then that's probably what I did wrong------------
The original problem I had was: "An electron is released from rest in a uniform electric field of strength 400. N/C that points east. Neglecting friction and gravity, calculate the speed of the electron at a point exactly 1.00 mm due west of ts starting point, given that the mass of an electron is 9.11 x 10^-31 kg."

These were the two solutions I had. The first one I did was the top one, then found it was incorrect by checking it with the solution's sheet. The second one is more or less how the teacher showed on the solution sheet to do it.

Err, these are sort of blurry so if it's illegible I can type it out. It's question #2 (imgur wasn't letting me crop the first picture properly for some reason)

You made a mistake on your division. 6.4x10^-17 / 9.11x10^-31 = 7.03x10¹³ -- you lost the order of magnitude.

Ahh so that was what my mistake was ><'. The answer looks the same for both now.
Thank-you! Now I know that both methods do in fact work.

EDIT: Err, do you have tips for how to check work? I check my work my going through the whole thing, or by being haphazard about where I'm checking things...which means I make the same mistakes a lot.

What is "Wall Street"? I'm researching something for a school paper and came across this phrase:

"The industry’s dilemma is that the average American can only eat so much of that food – about 1500 pounds a year – and the total number of eaters in this country is growing by only one or two percentage points a year. Yet Wall Street demands that food corporations grow at a considerably faster rate."

Does that mean Wall Street literally or Wall Street as a metaphor for some sort of market system thing? Or maybe Wall Street is actually the system. I thought Wall Street was a place?

me·ton·y·my /məˈtänəmē/
noun
the substitution of the name of an attribute or adjunct for that of the thing meant, for example suit for business executive, or the track for horse racing.

Oh, so I guess "Wall Street" means a big, fuzzy blanket term for "the direction these US marketing related things are going."

Metonymies are annoying when I don't see the particular one much ><

Thank-you!

As always, you're welcome! And now you have a word for a thing you already knew about but couldn't label.

I should be doing homework but was trying to do that instead x'D. I didn't figure it out though. ...That question's going to haunt me for a while.

(not that I have a lot of homework, at least)

Don't feel bad. I didn't get it either. It's a troll. 95% isn't an accurate statement. It's more like 99.9999%. The smallest known solution has numbers that are ~70 digits long, and it requires a very advanced mathematical technique to solve it.

70 digits long? A very advanced technique..
..Well, at least in the few hours I was messing around, I practiced multiplication x'D

At one point when I was working on it, I was staring down the cube root of 2/3...

I typed "solve((a/(b+c) + b/(a+c) + c/(a+b), a)" on my school-issued TINspire. The calculator took a much longer time to load than I'd ever seen it load, then gave me a huge line that I decided not to look at all of o_o
----

Why is "electromagnetic" energy called "electromagnetic"?
I've just started learning about currents, and that means lightbulbs! So does electromagnetic energy have a lot to do with electrons?

I remember when I first learned about energy, the teacher was grouping energies with each other, but I forgot those groups :/

Moving electricity creates magnetism. Moving magnetism creates electricity. These two phenomena are so intimately related that they're really just ONE phenomenon, which we call electromagnetism.

Yeah, I was able to get it solved for one variable in Wolfram Alpha, but trying to back-substitute that result into the equation ended up just making a horrible, horrible mess with an x⁶ in the denominator and some stuff with nested radicals. Turns out it's not even theoretically possible to solve it that way, and you have to use some properties of Galois field theory (don't ask, I don't know) to do anything with it.

How dangerous are short-circuiting batteries? Would it be safe to tape a wire from one end of a battery to its other end and leave it there?
(I have this science homework where we have a simulation that shows batteries burning when they connect with themselves..but I've never seen a battery burn in real life)

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.

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.

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?

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.

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 ?

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.

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"?

So..E = ΔV/d is like I = V/R ?

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

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.

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.

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.

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?

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)?

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.

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?

This is exactly it.

Another correct deduction!

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.

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.

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?

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.

I think I was having trouble coming to terms with what voltage dropping means...does "all elements" include the battery?

So far, what I think I understand is that in an imaginary circuit with one resistor, say a lightbulb, and where the wires are perfect conductors, the change in voltage, or the potential difference, across the lightbulb will be exactly that of the battery, because in V = IR, the current doesn't have to be split up among more than one resistor's resistance. It's like the wires didn't exist...so:


On the other hand, when the wires do contribute to the total resistance of that circuit, then the current is affected by all "resistors," which includes the wires. This means voltage keeps dropping steadily all the way to the end.

I'm still confused about the battery though. Setting the negative terminal to zero, and moving across to the positive terminal, the voltage increases.

Although, it has to increase in order for the circuit to move electrons around. I guess chemical stuff inside the battery creates a potential difference from the negative end to the positive end mostly through the wires and mostly not within the battery, and that provides the potential difference required for resistors to turn potential energy of electrons into heat/light/etc. Except isn't it more like potential difference creates some sort of potential energy for electrons --> electrons move and so that's kinetic energy --> electron kinetic energy turns into heat/etc. in resistors?

And I guess that whole thing is related to series circuits (my class started on that immediately after the quiz) which says that the current across the entire circuit is the same, and the circuit's resistance (equivalent resistance) is made up of all the resistors' resistances, and the battery voltage is split up among all the resistors.

Err, so from what I remember when voltage drops it just means electrons came up to the resistor, and slowed down. Now there's one side of the resistor that's more negative, and one side that's more positive, and so now there's a difference..? But how does the battery and overall current fit into this? The difference in voltage across the entire circuit just happens to be that of the battery because that's what's connecting the battery terminals? With the voltage drop being greater on resistors having more resistance because they just take more potential energy to power and slow electrons down more like that so the difference created across those resistors is greater? And all the while the current just accommodates for the resistors?

Also, for parallel circuits, whatever the "equivalent resistance," and the current and voltage for the circuit as a whole is, am I basically treating all the resistors in series as one resistor for the circuit? And it just so happens that because I_b = I_1 + I_2 +..., and I = V/R, V_b = V_1 + V_2+..., it works out that 1/R_eq = 1/R_1 + 1/R_2 +...?

And, do parallel circuits affect everything just because if resistors aren't very resistant (whatever that mean) or incredibly not-resistant (whatever that means), and metals have free moving electrons, then electrons in all wires will be affected?

No. If you INCLUDE the battery, then the net voltage drop across the circuit must be zero (because the battery has a negative voltage drop because it's a source of potential).

This looks correct.

The thing that might be throwing your intuition is the idea that the same resistor is causing a different voltage drop depending on the voltage of the battery. The resolution to that is to realize that it means that the current (that is, I) is affected -- less electricity is flowing through the circuit altogether because of the resistance. What the resistor does is it inhibits the ability for that voltage to be translated into work elsewhere in the circuit. (This is a very desirable trait, because otherwise you're shoving excess power through things that AREN'T designed to deal with it -- connecting an LED directly to 5V is a good way to blow it out, but if you put a load resistor ahead of it in the circuit then the voltage feeding into it will be lower.)

Yep.

This goes back to correlations not being a cause-and-effect relationship, which we were discussing earlier. Yes, you observe that the voltage is higher, but that doesn't explain anything about WHY that's true.

Ideally, NONE of it within the battery -- that would cause batteries to discharge themselves when not connected to anything, which is definitely not what you want to have happen. :P

And yes, you are completely correct about the energy conversions.

Yup. Goes along with what I said above.

Mm... Vaguely, but mostly no.

The entire circuit is already populated with electrons before you ever connect a power source. They're just not really moving. So when you add a voltage, that's applying a force to one end, but you can't really say that the electrons have to slow down when they get to the resistor, because they can't pile up on each other -- there's only so much room for them. This is why the current through the entire circuit is constant: the electrons can only move through the circuit as fast as they can move through the slowest part, because they're always being pushed into the flow.

Think about a circular tube filled completely up with marbles. If you push on one of the marbles, that's applying a voltage, and you can feel the resistance of friction, but all of the marbles move at the same speed. If you were to put a narrower bit of tubing in there somewhere, you'd have to apply extra force to shove a marble through it, but that wouldn't change the fact that they all have to stay at the same speed -- it just slows the whole thing down.

The friction in that narrower tubing has a normal force that acts against your attempts to move the marbles. The net force available to move marbles through the rest of the tube is, correspondingly, exactly that much less. That's the voltage drop of our marble-circuit resistor. And just like the normal force of friction is proportional to the force being applied, the voltage drop across the resistor is proportional to the voltage being applied to the whole circuit!

The battery is the reason that one side is more negative and one side is more positive. The difference in voltage across the entire circuit is equal to the voltage of the battery BECAUSE the voltage is defined by the work necessary to push a charge across the circuit! It's not a coincidence; it's a definition.

But yes, the resistors use up energy, leaving less energy available to do other things.

I'm not quite sure what the question is. Can you try rephrasing it?

When you're mixing parallel and serial circuits, you can treat multiple elements on a branch of a parallel circuit as a single resistor, and you can treat the whole of a parallel group of elements as a single resistor in a bigger serial circuit.

So yes, that means that the effective resistance across the parallel part of the circuit DOES affect the current of the whole thing.

For parallel circuits, can I also pretend that all the resistors in parallel merge together into one resistor? And it just so happens from how the math works out, the resistance for the parallel circuit can be calculated using: 1/R_eq = 1/R_1 + 1/R_2 +...
(I think based on what I learned in class today, I am treating parallel circuits that way. As for the equation, I do remember how to find it, though I wonder why that's the equation on my reference sheet and not (R_1 * R_2)/(R_1 + R_2) = R_eq )

Err..so voltage measures change in potential energy based on current and time, and when voltage drops it means some potential energy got converted into some other energy across a resistor?
And all the while the battery has its own voltage, potential isn't the only thing that matters...so resulting from the potential is electrons moving--which is kinetic energy. But, just like how if I slide a book across the table, the kinetic energy of the book is lost to friction/heat/etc. and it stops, the electrons are sort of "stopping"--except there is a constant "force." Therefore, just as pushing a book across a table vs across a rug, more KE is lost to heat across the rug and so it is slower, so too does more resistance mean the current is slower across a circuit with more resistance?

And is that why Power = Work/time = ΔPE/time? Because somehow, change in PE is sort of Force (and distance)...P = W/t --> W = Fd = ΔPE for electric fields = qEd = qΔV --> P = qΔV/t --> P = VI?
Why doesn't distance matter in a circuit? Does it really just not exist as a value? But at the same time, I can't very well say P = F/t, because then P = IE.


EDIT: Random question-how does a voltmeter measure voltage? I assumed that the greater the current passing through the voltmeter, the greater the voltage measured. But apparently the ideal voltmeter has infinite resistance? Doesn't that mean no current passes through the voltmeter? And if there is no current, how does it measure anything?

In terms of how the rest of the circuit sees it? Yes.

Because they're equivalent (it's pretty simple manipulation to prove it), and 1/R_eq = 1/R₁ + 1/R₂ is less work to evaluate. :P

That's one way to look at it, yes.

Precisely.

Looks like you've basically got it down.

Distance doesn't matter in a circuit for the same reason that current is constant across a circuit: it doesn't matter how far the other end of the wire is, when you push on the electrons on the near end, it forces electrons out on the other end. While the individual electrons don't move very fast, the energy propagates through the wire at an appreciable fraction of the speed of light, so you can more or less disregard it at human scales.

The ideal voltmeter has infinite resistance so that it doesn't interfere with the circuit being measured by using up some of the current. (After all, by applying a voltmeter to a circuit, you're creating a parallel circuit!) The conflict you're trying to resolve HAS no resolution because an ideal voltmeter is impossible. :P You're completely correct: it HAS to have SOME current flowing through it in order to measure anything. From that point, it's just V=IR, and you know the resistance of the voltmeter and you measure the current flow so you can determine the voltage.

Ahh! Mathy stuff! -scurries away frantically-

Is another way to look at it that voltage measures the difference in the joules (of PE?) used to move charge at two different points?

At human scales?

--
Lol, technically it's science?
--
But I do have an English question from writing my history paper:

Is it more proper to write "white middle-class Americans" or "white, middle-class Americans" / "dark, greasy film" or "dark greasy film" or "dark and greasy film" ?

Hmm. I suppose that's not bad as a rough visualization.

At any scale that humans can meaningfully interact with without technological aid, the speed of light might as well be infinite, because any time elapsed for light to get from point A to point B is so small that it's not going to influence the significant figures of your measurements. The speed of electrical propagation is likewise beyond meaningful consideration. The speed of sound is even fast enough to be irrelevant in most cases. (And the speed of sound is a particularly relevant metaphor, because it determines how long it takes for the far end of an object to respond to a force applied to the near end.)

As for the English question: Both are grammatically correct, but they have different meanings. If you can swap around the adjectives without influencing the meaning, include the comma. If swapping them makes the phrase mean something different, then leave the comma out.

So for example, if you have multiple greasy films and you want to refer to the dark one, "dark greasy film" is correct. However, if you only have one film, and it happens to be both dark and greasy, then "dark, greasy film" and "greasy, dark film" are equivalent.

In the case of "Americans" the distinction is particularly significant. Are you referring to the set of middle-class Americans who are white, or are you describing a group of Americans as being both white and middle-class?

Is voltage the joules of PE changed to...some other energy...as charge moves from one point to another?

Is sound how humans perceive kinetic energy?

My exact sentences were:
"After the war, a booming economy that made televisions a household staple made white middle-class Americans more vulnerable to food advertisements."
I guess I was referring to the set of middle-class Americans who were white.
I was told that it was without the comma, but I don't know what the difference is.

"A dark and greasy plastic film called “Saran” was developed by the Dow Chemical Company during the war for soldiers to spray on fighter planes to protect against sea spray."
I was describing one type of film. So, I suppose it's with a comma?
But is there a connection between this and "white, middle-class Americans"?