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3:29 AM
@ZeroTheHero thanks for sharing.
@RyanUnger yo
 
123
3:42 AM
Yo..
 
\o @BalarkaSen
 
4:02 AM
2
Q: Why is computer science hard?

Daniel R. CollinsData pretty regularly shows that computer science programs have among the highest failure and dropout rates of any college program. A number of sources all echo the finding that roughly one-third of incoming CS majors do not progress to a second year, higher than most other majors. Computer scie...

 
123
4:13 AM
If two capacitors charged separately with batteries then disconnect batteries and attach both capacitors in series combination circuit which is not closed (circuit is open) Also no battery/load is connected with circuit and no effect of environment.
Does this system isolated or not. Charge conserved or not???
Total Charge before series joined $\neq$ Total Charges After Series Joined
$q_1 + q_2 \neq q_1' + q_2'$
$C_1V_1 + C_2V_2 \neq 2\times \frac{C_1\times C_2}{C_1 + C_2}\times (V_1 + V_2)$ {$\therefore q_1' = q_2'$
 
 
1 hour later…
5:37 AM
@user6232128 It's easy! Pretty easy!
 
@ACuriousMind idk something about AdS/CFT
@SirCumference unless you've been learning from Hans, probably
 
@RyanUnger i have eric poisson's gravity as a textbook
it's one of the worst books i've read
 
5:53 AM
yeah its not great
you should look at aretakis' notes on his website
 
6:10 AM
^ I second this, although he uses more mathematically-inclined diff-geom notation
 
6:23 AM
@123 suppose we separately charge the two capacitors, then disconnect them and place them in series like this:
The +q₁ charge on the top plate of C₁ cannot change because the charge has nowhere to go i.e. there is no route for it to flow away. Likewise the -q₂ charge on the bottom of C₂.
So when we connect the capacitors as shown on the right the top charge +q₁ and bottom charge -q₂ have to remain the same. And that means we need equal and opposite balancing charges on the other plates. The end result is that all the charges remain unchanged when we connect the capacitors in series.
 
123
7:21 AM
Thank you so much @JohnRennie sir. But why answer in series gives us different total charge before and after series combination. In my opinion it should be conserved because of isolated system.
 
I would have to see the book to understand exactly what calculation it is doing.
 
123
What I think in my situation if we charge capacitor before put in series and then attach to series. It should follow some different formula or equation. Not the usual one we used.
Sure @JohnRennie you are awesome and this community will always help. Really thank you so much because internet and book not give me answers to critical situations.
You and community always help.
There must be some additional term in equation to compensate remaining/additional charge in series. Which make final equation to show charge conservation. Or Might be entirely new formula.
 
7:38 AM
So I had a doubt. We can also define an "epsilon-delta limit" definition for any metric and not just the cartesian metric, right? And in all those cases, I will obtain the same limit, right?
 
8:00 AM
1) Yes, 2) No
 
8:30 AM
@RyanUnger I actually don't know a lot about that - most CFT I know is 2d (worldsheet theories in string theory) and beyond the buzzwords I don't have a good understanding of what goes on in AdS/CFT specifically
@FakeMod All norms on finite-dimensional vector spaces are equivalent (generate the same topology and hence the same notions of convergence), but not all metrics are.
 
9:07 AM
@ACuriousMind @NiharKarve What if I only consider those metrics which are continuous, i.e. not taxicab metric? Could I then obtain the same limit from all of them?
BTW @NiharKarve, if you don't mind sharing, where do you study?
 
@FakeMod A "non-continuous metric" does not make any sense - in a metric space, you're using the metric to define the topology, and you can show that every metric is continuous in the topology it induces
 
Additionally, all metrics induced by p-norms in finite-dimensional spaces are strongly equivalent
@FakeMod for legal reasons, I'm gonna have to say Eurasia
 
9:34 AM
@FakeMod Hi, do you have some free time right now?
 
@ACuriousMind @NiharKarve Thank you for the responses, but I have not yet studied this stuff, so it is almost incomprehensible to me as of now, but I have bookmarked this convo for future references.
@RajdeepSindhu Yeah, sure.
 
@FakeMod Nice. I have some queries regarding ALLEN and I think it would be best to clarify them in another room. Hence, chat.stackexchange.com/rooms/116293/…
 
@NiharKarve From your name, I figured out that you might be Indian. And I am from India as well, so I thought that you might be studying at a place/college/university that I know of :)
 
123
9:48 AM
@JohnRennie Any progress about my problem of capacitors
I have already shown usual calculation above comments.
 
@123 you need to post a picture of the book so I can see what it is claiming.
 
123
I have numerical/problem of the same situation does not have any text to read the deta about this specific situation.
 
Hello everybody
 
123
Then I calculate the charge on both parallel and series situations which is fine as per book result. But in my mind I decided to check whether charge conserved or not. In parallel it is conserved. But in series it did not.
 
It's asked me to find the voltage between points a and b
 
123
9:54 AM
Then I thought system is isolated why charge is not conserved in series.
 
I found the currents across left mesh as $I_1=3A$ and right mesh as $I_3=2A$. And the current thorught the middle branch is $I_2=I_1+I_3=2+3=5A$
 
@FakeMod Still a few years before I go to university :)
 
So the voltage of $2\Omega$ resistance is $V_{2\Omega}=2I_1=6V$ and voltage of $9\Omega$ resistance is $V_{9\Omega}=9\I_3=18V$
How do I find the voltage between these 2 points?
 
123
Problem : If two capacitors are charged separately with batteries one is C1 = 10microF, V1 = 600v and another C2 = 20microF, V2 = 1000v then batteries removed. Find the charge on each capacitor (a) when it joined in parallel (b) also in series.
 
@ICCQBE if we take the negative terminal of the battery to be our zero then the voltages at a and b are +18V and +6V respectively. Yes?
 
10:01 AM
@JohnRennie thank you for drawing. Yes, i'm listening to you
 
Kirchoff tells us that if we start at b, go down through the 3Ω resistor, up through the 6Ω resistor to a then back to b the total voltage change muct be zero. Yes?
 
Yes, I understand.
 
I've labelled the bottom point as C. From B to C the change is -6V and from C to A the change is +18V so the total change along B-C-A is +12V.
That means the voltage change going from A to B must be -12V. Yes?
 
Hmm, I got it. Could we find the same result if we would make our calculation using each resistor's voltage?
I mean the voltage increases from C to positive terminal of source, +24V
From that middle point to A point, there would be voltage drop which is +24V-6V
 
Yes. You just have to be careful about whether to add or subtract the potentials across the resistors. The nice thing about using Kirchoff is that it always works.
 
10:09 AM
Hmm okay, I understand the idea
Thank you for your help! Have a good day @JohnRennie!
 
:-)
13 mins ago, by 123
Problem : If two capacitors are charged separately with batteries one is C1 = 10microF, V1 = 600v and another C2 = 20microF, V2 = 1000v then batteries removed. Find the charge on each capacitor (a) when it joined in parallel (b) also in series.
@123 I get the initial charges to be 0.006C on the 10μF capacitor and 0.02C on 20μF capacitor. Yes?
 
10:25 AM
Two years! That's an extended tantrum!
 
apparently they cared so much about it that they forgot about it for 2 years, yeah
 
at least in German it's both an adjective and a noun and I suspect in some other languages too, so probably just the native tongue slipping through
 
10:49 AM
Two great picture of bundles
 
@Charlie @bolbteppa Any decent resources on principal bundles?
 
123
@JohnRennie yes you are right. Add q = q1 +q2
 
The notes the last picture came from set up principal bundles pos.sissa.it/323/002
 
@123 in parallel you get a charge 0.026C spread over both capacitors.
 
123
Check it on parallel and series. Then add both charges in parallel and series. q' = q1' + q2' in parallel q = q' but in series not equal
 
11:35 AM
@NiharKarve see even today this s-matrix stuff is still coming out
 
11:47 AM
@ACuriousMind do you happen to know why this profile says last message 914d ago?
 
@user6232128 not sure how that's a reply to the message you replied to, but no, I don't know :P
note that that's not a "real user", but a feed bot
 
true, but the recent messages are much less than 914d ago
 
yeah, no idea what the "last message" for a feed bot refers to
 
ok, thanks
 
Does anyone know in the vanishing gradients problem, why do the earlier layers (in a neural network) get to have less and less impact (and therefore lower and lower gradients) on the cost function as training happens?
nooo lol, it's a gradient descent issue
that has something to do with the activation function
I'm asking this in a physics chatroom, that might be problem
i'll research more and as a last resort, ask in AI SE
 
12:02 PM
@NiharKarve I will take any possible opportunity to plug Frederic Schuller's lectures on the geometric anatomy of theoretical physics, this is where I started, I also think Frankel's "The Geometry of Physics" is a good textbook.
 
nvm I found the vanishing gradients explanation that I needed
 
coolio
 
12:42 PM
I have another question with electric circuits
I found the node voltages using node analysis (node-voltage method)
It's asked me to find the voltage of $15\Omega$ resistor
I don't know how to keep my solution. Could you help me, please?
A2A @JohnRennie
I thought that the voltage of $15\Omega$ resistor is simply $\frac{V_a-V_c}{15}=26,25V$ but the answer page tells that the result is 30V.
 
@bolbteppa @Charlie thanks so much for the links :)
 
1:17 PM
Sometimes Empty Pockets, Hungry Stomach and Broken Heart gives the best lessons of life
2
 
Even simulation gives 26,25V but somehow my book tells that voltage of that $15\Omega$ resistor is 30V...
 
1:38 PM
So an axial vector is invariant under inversion. But how does it transform under a reflection? Does it depend on the direction of the reflection? The axial vector I am considering spin
 
@B.Brekke In odd dimensions, a reflection is just an inversion + a rotation.
so since you're probably looking at 3d, it just ignores the inversion part and transforms under the corresponding rotation
In even dimensions, it's either trivial (d=2), or your "axial vectors" are more complex objects than vectors and transform non-trivially but not in a straightforward way.
 
1:54 PM
@RewCie did you read the one answer?
Talk about being overly pedantic :-/
 
Yes. It's true.
Comtupar Science Engynars are not programmers.
They are kinda, expert systems, network analysts, hackers, testers, they design embedded systems, reverse engineering and all
That's the reason Comtupar Science Engynaring is easy for me.
 
@user6232128 who do you think is being pedantic there?
 
Although most of them can progam too, in multiple languages, use frameworks and with best algorithms, we leave those to programmers.
 
@ACuriousMind Thanks, this is very helpful.
 
@ACuriousMind the person answering
 
2:03 PM
Do you know how does a CSE Course looks like?
No, he isn't being pedantic
 
@B.Brekke I think in effect this will just end up in the "reflected" axial vector being collinear to the reflection it would have if it were a normal vector, but having the opposite direction (since the "ignored" inversion would turn this into the regular image under reflection)
@user6232128 ...and in what way do you think they're being pedantic?
 
@RewCie it should look like applied science
 
@user6232128 I don't know about Applied Science...
No, CS = Pure science (BSc.) I gues
 
@RewCie I'm curious. You always use the term "computer science engineer", but what exactly do you mean by that? Usually I see "Computer Science" and "Computer Engineering" as different topics, but "Computer Science Engineering" sounds like you would engineer courses on computer science...
 
you can do CS as a purely theoretical exercise, like math, producing mostly papers as output, and you can do it as an applied exercise, producing software and papers about that software as output
 
2:07 PM
@JMac Computer Science is only study of systems (how do they work and how they are used), CSE deals with (How to design them)
It's like Cosmology and Astrophysics bit thing....
 
the borders to explicitly "applied" fields like software engineering are fluid, and often software engineering is considered part of the math/CS department even if it is much more concerned with architectural principles or process designs than traditional scientific knowledge
 
yup
 
@RewCie So are you talking about software engineering? Because that's typically the term I hear, while there is also "computer engineering" which involves actually integrating the hardware so you generally have a knowledge of electrical engineering too. Saying "computer science engineer" seems pretty vague in respect to which you mean.
 
@JMac Yes, Software engineering and Computer Science Engineering are same.
 
The question asks ""What makes it hard"
Answer the question.
 
2:10 PM
@JMac yeah, to me a computer engineer would be primarily someone who designs microchips
 
@RewCie I would disagree. "Computer Science Engineering" to me at face value implies that you would engineer computer science courses. "Software Engineer" has a clear meaning. "Computer engineer" has a clear meaning. "Computer science engineer" seems like a confusing mix of the two.
 
In Computer Science they learn How to use processors, assembly language, programming, compiling, linking, loading In Computer Science Engineering it is How to design processors, write in different assembly lanugage, machine storing format, Compiler design, reverse engineering
 
Like I wouldn't call myself a "Mechanical Science Engineer" for example.
 
@JMac Computer Science Engineering is most similar to Computer Science..... But projects are difference...
@JMac Only if you engineer Mech Science
 
@user6232128 Frame-challenge answers are accepted on many sites, and the question explicitly cites drop-out rates as primary evidence for the assertion that CS might be unusually hard. I'm not sure what you have against someone offering alternative explanations for the drop-out rates. If CS isn't unusually "hard", then there is no question.
 
2:12 PM
10 hours ago, by user 6232128
2
Q: Why is computer science hard?

Daniel R. CollinsData pretty regularly shows that computer science programs have among the highest failure and dropout rates of any college program. A number of sources all echo the finding that roughly one-third of incoming CS majors do not progress to a second year, higher than most other majors. Computer scie...

 
@RewCie So are you agreeing that "Computer Science Engineer" is a bad term unless you are actually "engineering" the science? (which seems like a kinda nonsense phrase)
 
@user6232128 The answer is correct. Most of the students do join his course because they do love pogramming. But the focus is different here.
@JMac No, Computer Science Engineers do engineer stuff like processor, assembly, designing compiler, different system, data structures, algorithm, and all. They make them. So, that's why it's engineer
 
Many places seem to call courses that study both CS and computer engineering "CS&E". In Germany we have Informatik (~computer science) which has technische Informatik (~computer engineering) as a subfield/cross-over field to stuff like electrical engineering.
 
Yes, that's what I'm doing, CSE is just short course for CS&E.
 
The principles of compiler design - and the underlying theories of formal language - belong pretty firmly into the realm of traditional computer science, while building a processor is pretty firmly computer engineering. No one uses these terms like you do.
 
2:18 PM
@ACuriousMind Yeah this is where it's getting confusing for me. "Computer Science & Engineering" is a pretty common topic, the term "Computer Science Engineering" just confuses me.
 
The traditional computer science course doesn't has Compiler design they only are taught about principles of compiler design.
 
So basically just software engineering?
 
where is software engineering here?
You've to see and learn about processors to design compilers.
 
yeah, anything that refers to details of the implementation process (such as the use of design patterns, version control, etc.) would be called software engineering by most people
 
@RewCie But you aren't designing processors; which is what you typically call "computer engineering".
 
2:23 PM
See, this data sheet for x8086 16-bit processors they'll give you processor, BIOS Design, Motherboard or Bare machine, and will tell you to design an OS. This is Computer Science Engineering.
@JMac No. designing processors is semiconductor engineering.
 
Just because a particular task may involve all of computer science, computer engineering and software engineering doesn't mean that there exists an overall category where you just mash all these together and call it "computer science engineering"
 
Yes, Computer science engineers can't design a full fledge OS.... but they can do it at masters :P
 
there isn't even a Wiki article for that term, it just redirects to "computer science"
 
which term?
 
@RewCie "computer science engineering"
 
2:26 PM
@RewCie ... Semiconductor engineering is probably best described as a specific aspect of "computer engineering".
Or really just electrical engineering.
 
@ACuriousMind Same as CS&E
@JMac electrical engineering doesn't deals with semiconductors
 
@RewCie yes, because there's an "and" in there - "computer science" and "computer engineering" are both well-established terms on their own
 
yep
CSE = Small version of CS&E
 
but "computer science engineering" is not, and I really don't understand why you seem to keep using it after several people have told you it doesn't really mean anything to them
 
Different countries call it differently
 
2:28 PM
& is removed, most of the time..
I'm a Computer Science Engineer and not computer science & engineer
That's why I omit &
 
@RewCie I'm pretty sure electrical engineers learn about semiconductors... They likely don't learn enough theory to actually make CPU chips, but that's probably because learning something like that is more advanced than just a bachelors degree.
 
In Masters course, things start to merge, computer scientist become computer science engineer and semiconductor engineers become electrical engineers too.
Knowledge wise^
 
@RewCie Where are you seeing bachelors courses in "semiconductor engineering"?
 
@RewCie What about pay wise?
 
@JMac I've seen that term multiple times... No idea where they are taught.
@user6232128 Same. Since they have equivalent knowledge, you can take example of Machine Learning experts and Data Science Experts.... both have almost same pay.
After masters ofc, not bachleors
 
2:36 PM
@RewCie People engineer semiconductors, but you don't just take "semiconductor engineering". You would likely study something like electrical engineering or computer engineering, and then go on to specialize in semiconductors and get into the "semiconductor engineering" field. Saying "electrical engineering doesn't deal with semiconductors" just seems totally off base...
 
@JMac Electrical Engineers might not be totally off from semiconductor engineering
They may know it, but semiconductor engineers must know more.
 
So then why try to tell me "electrical engineering doesn't deals with semiconductors". It's hard to believe anything you say when you're also always saying things like this.
 
Electrical engineering doesn't deals with semiconductor in the sense that they don't design semiconductors although they might have courses to learn about some basic and standard chips and like that.
It's the job of semiconductor engineers to make them, they are paid for it.
 
And so what, a company calls themselves "semiconductor engineering". Doesn't mean it's like a separate thing from electrical engineering. This link suggests Semiconductor engineering itself would be a masters program, with a background in electrical or computer engineering... kinda like I said.
Like I could be a "boiler engineer" if my job had me working on boilers mostly, but my education is still a "mechanical engineer".
 
No idea, if it's a different thing. I've heard semiconductor engineering multiple times. Maybe, in bachleors they aren't taught about it.
 
2:40 PM
So then next time maybe don't try to tell me electrical engineers don't deal with semiconductors?
 
Because a couple of days, I met a Electrical engineer on SE about designing physical processors, he told that they are taught in semiconductor engineering courses.
Most semiconductor engineering courses are offered at master’s and doctorate levels.
they are not
type
*typo
 
@RewCie Processors are a very specific type of semiconductor that require a lot of specific knowledge, so yes, to design processors, it's not surprising that someone gets a masters in "semiconductor engineering". You told me "electrical engineering doesn't deals with semiconductors" in the context of talking about semiconductors, not specifically processors.
 
@JMac Okay, then they might not be taught enough to design processors
at bachleors courses
 
@RewCie Right... and saying that would have made no sense in the context of the situation where I only mentioned semiconductors in general. chat.stackexchange.com/transcript/message/56178806#56178806 I'm just saying man, you need to think about if you actually know what you're talking about when you say things like this. It's hard to believe anything you say when I see you post obviously wrong things so often.
 
No idea about Electrical/Semiconductor engineering.
Modern education is complicated.
 
2:50 PM
It's alright to just not know things and not say anything about it instead of just saying wrong things when you don't have any idea about the topic.
 
Well, those were some most probabilistic assumptions.
 
3:34 PM
What is the value of current pointed with question mark?
It should be $I=\frac{V}{R}=\frac{V_1}{28}=\frac{170,4}{28}=6,09A$
My book shows that $V_1=170,4V$ which is correct in my calculations but it shows that I=2,04A but I find it as 6,09A
Is my book showing incorrect results?
 
Jim
4:19 PM
@JMac I don't have any idea what this is about, but I think JMac is correct
also, it's usually phrased as Better to keep silent and let them think you a fool than to open your mouth and remove all doubt
 
1
Q: Does spin really have no classical analogue?

AkobenIt is often stated that the property of spin is purely quantum mechanical and that there is no classical analog. To my mind, I would assume that this means that the classical $\hbar\rightarrow 0$ limit vanishes for any spin-observable. However, I have been learning about spin coherent states rece...

This was discussed a while ago in here
 
123
Hi @JohnRennie Any progress
 
Jim
4:36 PM
@RewCie This sounds like a department at a university and not a job title. Is the case better described that you are in Computer Science and Engineering? A student in such a program would not be a "computer science engineer", there would be a more specific term to describe any one student. My university had a CSE department too, it's a broad umbrella but there were no "computer science engineers". I can't imagine anyone going by such a job title. It's too vague for an engineer to want to use
 
4:58 PM
@Jim Computer Science Engineering students are not computer Science Engineers?
 
Jim
@RewCie Apologies, but I am thrown by the omission of the "&" in your statement. I worked in a CSE department for many years and never once heard people say "Computer Science Engineering". People always said it with an &... If the title includes "&" in its proper form, then I'd answer with "generally, no". If it does not, then it is something new to me. In that case, I'd say "probably no, but I can't be sure without context"
 
@Jim with & and without & are almost same -> shiksha.com/engineering/computer-science-engineering-chp
 
Jim
@RewCie true, but when you get used to hearing it a certain way, it throws your mental processes off a bit when people use it differently. Also, removing the & does have some meaning. One is saying "here is a collection of sciences and engineering" the other says "here is the engineering of science". However, as a departmental title, there is no significant difference
 
Okay. I'm only Computer Science Engineering student then. Taking a BTech Course. I
 
Jim
having read the link, it's clear that the students are "CSE students", but not computer science engineers. Everywhere they would refer to job titles, they use "computer engineer" or "computer scientist" or some other similar and familiar title. My best guess is that they dropped the & in order to make it sound fancier to prospective students, but that it teaches the same content. I would doubt that the graduates call themselves comp sci engineers
 
5:09 PM
I've heard people using Computer Science Engineers.
 
Jim
@RewCie strange, but not unbelievable
 
Although Computer Science and Computer Science Engineers are different.
3 hours ago, by RewCie
@JMac Computer Science is only study of systems (how do they work and how they are used), CSE deals with (How to design them)
 
Jim
@RewCie no doubt. The engineering side (computer engineering) is different from the science. The link you provided lists the differences and refers to "computer engineers" in the process. The beginning of the page seems to highlight the distinction of CSE, but the latter half seems to focus more on CS or CE, using CSE only as an umbrella term for the two together. This is why I guessed that they are insisting CSE is its own thing mostly to attract prospective students
 
Oh okay... Whatever, then what should I call myself?
 
Jim
what do you specialize in?
 
5:15 PM
BTech in Computer Science Engineering.
 
Jim
nothing more specific than that? Did you tend more toward engineering or science? Any particular kinds?
 
engineering.
 
Jim
software or hardware?
 
both... actually... more towards software.
 
Jim
I suppose, depending on the fanciness and portentousness you want, you could call yourself a computing systems engineer, a CSE major focused on firmware, a core computer systems engineer,.... the list goes on
 
5:19 PM
ok
I'm Comtupar Systems Engynar!
 
Jim
I had a program in my undergrad called "Earth and Space Science and Geodesy". It was a dumb name if you ask me. The students wouldn't dream of calling themselves "Earth & Space Science and Geodesists". They called themselves planetary scientists, even though that wasn't on their degree. It was more helpful for potential employers than the degree name
 
Oh Ok
 
6:05 PM
@RewCie After it goes through the program structure it gives some "job profiles", which seems like what you would call yourself career wise. Like I have a mechanical engineering degree but I don't work as a mechanical engineer (though that is an actual job as well, albeit really vague on it's own).
 
How can I apply mesh-current method to this circuit?
That current controlled current source on the left side mixes my calculations...
 
6:37 PM
@ICCQBE I think you would need to explain more than that. It seems fairly straightforward to me.
 
 
1 hour later…
Jim
7:44 PM
I just watched a TEDx talk where the speaker referred to himself as a computer science engineer.... Coincidences, amirite?
3
 
123
8:01 PM
If an object fall in air it achieve its terminal velocity. What if we apply additional constant force in the direction of gravity does it achieve new terminal velocity??? Also its increases terminal or not?
 
Jim
drag force is proportional to velocity squared. For a constant increase in downward force, there will absolutely be a new, higher terminal velocity
 
123
Thanks @Jim
What if we apply potential difference across conductor. The free electron which is random before attain acceleration or drift velocity???
 
Jim
8:51 PM
yes? If you know the term "drift velocity", then it sounds like you already know the answer to that question
 
123
9:13 PM
But book days after connect battery across conductor random electron accelerate in one direction. Then it says move with drift velocity. Why they used the word electron accelerate??
 
123
9:39 PM
Does electromagnetic field in space can be deflected by electric field and magnetic field separately?
Direction of Force experienced by electric field and magnetic field is same???
 
fields don't "experience force", I don't understand what you're talking about
 
123
I meant to say if placed another charge in path of electromagnetic field. Does it experince force?
 
I don't know what the "path" of the electromagnetic field is, either
 
123
Also the direction of direction of Force of electric and magnetic field is same.
 
It's a field. It has a value at every point in space, it does not have a path.
and since the magnetic force depends on the velocity of the charged particle, I don't know what you mean by the direction of it being the same as the electric force, either
 
123
9:54 PM
@ACuriousMind yes my bad explanation. Same as you said. If a placed electron in way of electromagnetic field. Electron has electrostatic field. Does it deflect by electric field component of electromagnetic field.
Deflect or experince force
 
I have no idea what you're trying to say/ask, sorry
 
123
I want to say in electrostatic electric field of one charge apply force on another charge? Is it right
The same electric field has electromagnetic field. By this analogy electromagnetic field component of electric field should apply force if charge is placed where EM has.
 
An electric field does not "have" an electromagnetic field. The electromagnetic field is just a name for the electric and the magnetic field together. You're not making a lot of sense here.
 
123
@ACuriousMind I think I did not explain my question well. Let me try again. Pls help
Charge has electric field which has capability to apply force another charge.. Is that true. It is due to electric field when charge at rest. It has no magnetic field.
EM has both electric and magnetic fields. The electric field part of EM should have same property as I said in above comment which apply force if I placed charge at rest (which has own electric field). By means the electric field of charge and electric field of EM.
Does it happened? Does charge experince force by EM. Because EM has electric field part.
 
The electromagnetic field is not some entity like a particle that "has" an electric or magnetic field. It's just a name for the tensor $F_{\mu\nu}$ which has the electric and the magnetic fields as some of its components. If there's an electric field then a charged particle will move due to that electric field, it doesn't matter whether you talk about "the electromagnetic field" or not.
 
123
10:12 PM
I agree with you @ACuriousMind pls explain. If electric field component of EM interact with electric field of electron. Does electron experince force or not?
 
it's not about "interacting with electric field of electron" - what does that even mean? If you place a charged particle in an electric field, it will move due to that electric field according to $F = qE$, I don't know how much clearer I can be.
 
123
No matter how much intensity of EM we needed to move a electron. Does it move or not by interaction.
@ACuriousMind do you now understand what I am trying to say? My question is clear to you.
 
no, I honestly have no idea what troubles you here
you're using a lot of strange phrasings but I do not understand what your actual question is
 
123
:-) electric field has capability to move charge. The electric field component EM has. Does EM has capability to move charge same way as in electrostatic?
 
there is no difference between an electric field in an electrostatic context and an electric field in an electrodynamic context with respect to what an electric field does to a charged particle
 
123
10:18 PM
You are right. Can I understand EM can also move charge same as in electrostatic. Is the right?
 
i think so?
(I'm not entirely sure what you're saying still but I think the answer is 'yes' :P)
 
123
What is in my mind to EM. It leaves the source charge let from antenna.
 
oooh, you're talking about an electromagnetic wave
 
123
Hahahaha...
Yes yes EM means electromagnetic waves
 
you always just said "electromagnetic field". Now the whole "path" stuff makes a bit more sense
yes, an electromagnetic wave will move electrons due to its electric field component. That's how the current in receiving antennas is generated
 
123
10:23 PM
Because EM has both components electric and magnetic that's why I said electromagnetic field. Actually I meant to say EM wave
Sorry @ACuriousMind my fault used bad naming.
 
@123 The problem is that "electromagnetic field" is a lot more general than "electromagnetic wave". A wave is a very special configuration of the generic electromagnetic field
 
123
It means moving charge also experince force by EM wave
Due to magnetic component of EM wave
One more question is that EM has sinusoidal wave pattern of electric and magnetic field. Static electron has linear as in electrostatic. How they interact?
Electric field Sinusoidal waveform of EM and static electron Electron field
I have confusion, direction of Force due to electric and magnetic force is same? Because F = q(v x B) it is perpendicular to both v and B due magnetic field. In diagrams of EM, electric field vector is drawn where magnetic field force vector is.
I think I figure answer where I asked about waveform. May be I am wrong. It depend on source charge field. Not depend on the field the charge itself has. It's own field effect another another charge
 
11:05 PM
@ACuriousMind outer worlds?
 
@RyanUnger a resounding "meh" :P it's a competent game, but nothing about its mechanics is new or interesting, the story didn't really excite me and while some of the companions were fun we've done "ragtag group of weirdos on a spaceship" before in KOTOR 1 & 2
 
@ACuriousMind sounds like a "wait for sale"
 
yeah
I feel like Obsidian is just rehashing their greatest hits with their last few games
 
did you like PoE
 
Same feeling, really. PoE was pretty nice because there hadn't been a game like that in a long, long while, but it also suffered from the same "overbalanced" uninteresting mechanics and not really doing anything new that BG2 hadn't done
 
11:17 PM
overbalanced?
 
PoE 2 was...just more of the same with a naval minigame tagged on. Still not a bad game, but...not exciting to me
 
I just started PoE but I haven't had enough time
 
@RyanUnger like, they wanted to make sure that there's no clear "best" build that beats all others and so all the stats and items just give boring tiny percent bonuses to everything
but it's got a nice story, even if it gets a bit too wordy for its own good at times
 
@ACuriousMind I've noticed this already
do I need to read the soul thingies for everyone
 
not really...I think they're mostly in there for flavor and I think a lot of them were written by/for kickstarter backers. Few if any relate to the main story iirc
 
11:23 PM
ah
 
kinda plays against the "named NPCs are important" theme every other RPG trains us to expect :P
I'm hearing good stuff about BG3 but I won't buy it until it leaves early access
and at this point I wonder whether Cyberpunk is ever gonna come put or if they're pulling a Half-Life 3 on us :P
 
@ACuriousMind right
@ACuriousMind it actually never existed
@ACuriousMind I havent played any BG games
 
for shame! (well, the first one isn't actually all that good from a modern perspective, but I think the second still holds up pretty well)
you can find many of the recurring elements (e.g. companions with banter, upgradable player base) of later Bioware games starting there
 

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