@ACuriousMind I'm still unsure what is the purpose of the power supply. en.wikipedia.org/wiki/Photoelectric_effect#Stopping_potential I'm trying to find out why varying the potential between the cathode/anode is necessary.

Are you allowed to ask questions in this chat. Sorry, I don't know.

@BiggySmallRyeda yes

@Obliv You want to find out how much kinetic energy the electrons have. So you increase the voltage until no more electrons reach the anode, then you compute how much energy an electron would need to reach the anode against that electric field, and that's then (a tiny bit more than) the energy of the electrons emitted by the cathode.

@BiggySmallRyeda but you might get confusing answers from me if I'm having one of those days:P

@BiggySmallRyeda Yes, just ask

@acuriousmind How do you know when there are no more electrons reaching the cathode (I think you meant cathode? unless I'm mistaken)?

I posted it on the questions page can I link it here.

Its an awfully easy question for someone of your caliber

argh

@acuriousmind It's okay ;P

@BiggySmallRyeda pro-tip: don't rush answerers because you have a deadline

I'm back

That was some good tea

Sorry sorry, i'm just so frustrated since I've spent hours without sleep trying to solve it.

@biggysmallryeda also homework-like questions are usually closed if it doesn't pertain to a specific conceptual issue.

My teachers a nuclear scientist and hasn't been able to solve it yet.

your teacher hasn't been able to solve that problem?

@BiggySmallRyeda What exactly is a "nuclear scientist"?

I know it sounds cool, but does that even exist?

@Obliv Wait: I was right. The cathode is what emits the electrons, the anode is what receives them in this case. (Why? Because that means that, in circuit terms, a positive current flows out of the cathode in the other direction, and that other direction is "outside". Yes, it's stupid, let's not talk about this :P)

@bernard @0celo7 is a nuclear scientist engineer of mathematics afaik

@Obliv haha:D

@Obliv No. He is a nuclear engineer, and a mathematician

@BernardMeurer Aha.

She has a bunch of masters related to nuclear physics and all.

@BernardMeurer didn't get my sarcasm even with the edit ;(

@BiggySmallRyeda your teacher can't solve that problem?

Yes, she's having a hard time getting the 0.75 value

@ACuriousMind What?

@BiggySmallRyeda I'm afraid your question is off-topic as homework-like for us - we require that questions posted on the site show some effort and ask a more conceptual question than "How do I solve this?"

@BernardMeurer What kind of tea was it?

Sorry can you tell me how can I do that

@BiggySmallRyeda Are you being recursive right now?

Okay so from what I thought was the electric fields musy be added

@acuriousmind I think our problem is the convention of current. It seems like the cathode is emitting electrons because of the convention (flow of positive charges) when it's really the one receiving them? I have some fundamental issues with circuits as you can probably tell.

The plate with the greater voltage must also exert a force on a particle in position Y. Since Y maybe affected by both the electric fields.

web.mit.edu/preeya/Public/PhadnisPreeya_Photoelectric.pdf @acuriousmind page 2 i'm using that diagram as a reference

How ever I do not know if you can add electric field strength

is it an option to change the teacher?

Nope, final part of my As level exam is tomorrow. No choice.

I just want to get closure from veterans of physics that the value stated is not possible to find.

I'm sorry to hear that:(

@BiggySmallRyeda I'm not a veteran, but I got 0.75

Oh wow. How?!

by solving it

Thank you. Did you add the field strengths.

nope

of course, you could add "the field strengths", but for an appropriate notion of "the field strengths"

@BiggySmallRyeda The value should be completely trivial to find for someone remotely qualified to be a physics teacher. You just take each pair of plates as a parallel plate capacitor and use that the electric field inside such a capacitor is V/d, where d is the separation of the plates and V is the voltage difference.

Yes that is what I would do. However the problem arises with the second plate which is within the electric field of the previous plate.

I refuse to believe that anyone with a masters in physics could be unable to solve that problem.

@AndrasDeak, so I've been thinking about the quantum computer simulator

and I don't think your solution will work.

what's my solution?

the take the user input for both states and then solve solution

oooh, that one

yeah

okay:)

@Obliv Yes, the cathode is emitting the electrons there

so if i take two lists with the words representing the transformations

that's perfectly possible as I don't really see what you want/need to do

and then I go through and run each gate

@BiggySmallRyeda Where's the problem?

I apologize I am just a student who was attempting a question to my maximum effort, also all my given information. We haven't delt with capacitors yet at my level.

I'd still need to go in parallel because at some point I'm going to have to do one gate or the other

@ACuriousMind I think they're trying to superimpose two fields for the plates, rather than looking at the change in voltage vs electric field (cf. what if the charge densities were given)

and so when it comes to the target/control thing, I have a problem.

hmm.

@heather OK, that looks a lot like square 1 to me:D

@AndrasDeak, yeah, good thing I didn't write it out first, right? =D

yup:P

@AndrasDeak if you don't mind can you please explain your solution to get 0.75.

but I'm wondering if I could use some type of if statement, like "if list contains 'control':" type thing

@heather that's already implementation

what do you mean?

I can't shake the feeling that you haven't thought out the abstract logistics yet

but I don't really comprehend your problem, so my impression can easily be wrong:)

@BiggySmallRyeda I'm afraid I don't understand how one is supposed to solve that problem without knowing about the field in a parallel plate capacitor (maybe you didn't call it capacitor, but do you know what the electric field between two (not three) plates with a given voltage difference is?

actually, no, that won't work either, darn

I can't seem to figure this problem out.

=(

@ACuriousMind Unfortunately no, all we were taught so far is the electric field strength = voltage across positive plate / distance between the two plates

What does "voltage across positive plate" mean?

@BiggySmallRyeda was voltage or potential defined as an integral?

Our questions usually have a positive plate with a p.d. Of X volts across it with a positive plate on the other side.

I sense massive disinformation

@BiggySmallRyeda What does p.d. mean?

(I'm being somewhat Socratic here)

Energy lost per unit charge.

No, I mean what does the abbreviation mean?

good luck ACM, and good night:)

What two words does p.d. actually stand for?

Potential difference.

Aha! Difference between which two things?

The positive and negative terminals.

@AndrasDeak night

@AndrasDeak, good night

Point of greater potential and a point of lower potential

@BiggySmallRyeda Correct! (the difference between the potential/voltage of those terminals, to be precise)

Now, if you know that the electric field between two plates is the potential difference between them divided by their distance, what is the problem in solving that exercise?

You compute the field between the first and the second plate and then the field between the second and the third plate. Then you take their ratio, and there's your answer.

@ACuriousMind Ehm

@ACuriousMind Red tea

@ACuriousMind I love you

I am so grateful

I cannot tell explain to you my gratitude at the moment.

@acuriousmind oh my god i'm dumb. I didn't realize the cathode/anode section was within the metal plate. I thought the metal plate was the anode and the cathode was separate from it

@Obliv What?

The metal plate that we shine the light on is the cathode.

I was previously told electric fields are simply between positive points of negative and positive plates

Then we have, somewhere in front of it, the anode, which is another metal ring

in the diagram i linked you the photocell contains both the cathode and anode doesn't it?

@ACuriousMind can I provide with any sort of internet points for your assistance

@Obliv Eh, depends on the specific construction (yes, a full photocell will contain both, but you can easily separate the two parts)

so in all cases, though, the photocell will contain the cathode.

@BiggySmallRyeda That's not necessary, chat has nothing to do with internet point

@Obliv I think you may be confusing yourself with the word "photocell"

@BiggySmallRyeda, may I also recommend, if you have more homework-like questions, to try this new site (still in beta) for physics problem solving. It is not very busy yet, it is a very new site, but it is suited to more homework type questions (though we do expect that you show effort). Glad you figured out the problem!

The basic setup of the photoelectric effect is this: You have a light source, a metal ring, a metal plate, voltage supply, some wires, an ammeter.

You connect the metal ring and the metal plate with the voltage supply and the ammeter in series.

I know the setup (more or less) but I'd like to know the specific interactions between the waves of light and the electrons, the motion of the electrons, what we are detecting, physically, with the ammeter/voltmeter.

So that I can correctly describe why the wave theory of light fails to explain this interaction

Then you put the lamp such that it shines onto the metal plate (the cathode) and put the metal ring (the anode) in front of the cathode. Above a certain threshhold frequency you'll detect a current flowing between anode and cathode.

and so I can describe the whole phenomenon in my report

@Obliv You don't need to know any specifics to understand why the wave theory of light fails.

Thank you all. I try to contribute to as much as I can to thsi website in places I have knowledge over. I will attempt to help those on the new site when I can to lessen the load on all of you.( Though some might not want me to)

You're all great teachers and even greater people.

@Obliv You just need to observe that below a certain frequency, the electrons stop being emitted.

@acuriousmind I have to go for now. Thanks for the help as usual.

Anyone willing to answer some questions about fuzzy logic and complexity theory?

well, really more just about complexity theory, actually

Willing and able are two different things ;)

You know there's Computer Science, right?

yeah, but the chat there is dead right now and it's kind of a more general question that I doubt would be well accepted

I'll look around to see if there's anything helpful there, though, thank you!

some stuff that was kind of but not really helpful came up.

because, the problem is, I don't understand how you can prove classes equal or not equal. I mean, with a mathematical result, you manipulate equations, use inductive or deductive reasoning, or show that there is a contradiction, but how do you prove P=NP or P $\neq$ NP?

Or not necessarily just those two, what about proving stuff about BQP.

A basic way is that you give an algorithm which converts any member of one class into a member of the other class

without changing the computational complexity

I'm sorry, but I'm a bit confused by what you mean. Could you give an example?

Consider the traveling salesman problem, for example

right

There are algorithms to convert any NP problem into the traveling salesman problem

Or at least it's been proven that such algorithms exist

oh, I thought it was just that all NP-complete problems were theoretically equal, all NP problems are theoretically equal?

which means that if you found a polynomial-time method of solving that one problem, you could convert any other NP problem into TSP and thereby have a polynomial-time method of solving any other NP problem

so therefore P=NP?

In that hypothetical scenario, yes

i see...huh. So then, how do you prove something like "all NP problems are fundamentally equal" like you said earlier?

A lot of math, I guess ;-)

I mean, of course you'd have to be precise about what it means to be fundamentally equal