3:00 AM
@JohnRennie good morning ;) I have a question on Fluid Mechanics, with which I need help from you!
I'm stuck on (a) (ii), where I have to find the total pressure at the very bottom of the container
The solution took the weight of the upper liquid as $\frac {HAdg}{2}$ and weight of lower liquid as $HAdg$ - however one important thing to notice is that, the cylinder of length $L$ is already present in the container, hence when it was first added to this cylinder, it must have pushed some of the liquids out..? So why do assume the weight of the top liquid is $\frac {HAdg}{2}$.
I feel the volume of the top liquid is no longer $\frac {HAdg}{2}$, instead some of it was thrown out of the container by the added cylinder. Same arguement is applicable for the lower, denser liquid.

4 hours later…
6:38 AM
@JohnRennie Hey, what is the physical reason behind having large amplitude at the resonance frequency?
This question is pretty simple, yet I cannot figure out anyway to substantiate the intuitive feeling.

6:53 AM
@AjayMishra have you searched undamped oscillator
?

7:08 AM
What?

7:21 AM
Don't mind, I got the answer.

7:33 AM
Actually if you search you will get something good, I mean good content or which answer your question.
No it, s OK.

8:03 AM
@JohnRennie

@pi-π Hi. I'm working for about 20 minutes.

@JohnRennie okay

8:26 AM
@pi-π hi, I'm free now.

@JohnRennie: Hi. Good morning sir :-)

@GuruVishnu hi :-)

@JohnRennie: Could you ping me after pi's doubt?

If pi returns I'll answer his question but he doesn't seem to be here at the moment.

@JohnRennie Ok sir.

8:29 AM
Was it the capacitor question from yesterday?

21 hours ago, by Guru Vishnu
@JohnRennie: Today morning, I asked the following question on the main site. I received a good answer from Farcher, but still I don't understand Why is energy absorbed by the battery when the plates of a parallel plate capacitor are pulled apart? If possible could you please clarify this sir?
@JohnRennie Ah! yes sir.
21 hours ago, by Guru Vishnu
Question : https://physics.stackexchange.com/q/526044/238167

Your question compares charging a capacitor with the change when pulling the plates apart, but there is an important difference.

I've reposted it to avoid confusion. Because we've one more stuff on hold regarding capacitors.

When you are pulling the plates apart the voltage is always the same on the capacitor and the battery.

@JohnRennie Yes sir. I understood that part.
@JohnRennie I think here it must be charging a battery instead of a capacitor.

8:32 AM
If the voltage is the same on the capacitor and the battery then the voltage difference between the capacitor and the battery is zero, so when you move a charge $dQ$ between the capacitor and the battery the work done is $dW = dQ \times 0 = 0$.

> voltage difference between the capacitor and the battery is zero

So no energy is lost. The decrease in the energy of the capacitor is equal to the increase in the energy of the battery.
@GuruVishnu yes?

Is it so sir? I think potential difference is same across both battery and capacitor.
And not zero.

Let me draw a diagram

Ok sir.

8:36 AM

@JohnRennie Ok sir.

So suppose I move some charge dQ off the capacitor onto the battery:
The charge gets moved from a point with potential $+V$ to another point with potential $+V$, so the potential difference the charge gets moved through is zero.

@JohnRennie And $dQ$ must also flow from the negative terminal of the battery to the negative plate of the capacitor to make the charges on the two plates of the capacitor same.

Yes, and in that case the charge gets moved from a point with potential 0V to another point with potential 0V. Again the potential difference the charge has moved through is zero.

@JohnRennie But I think you've not considered some charge $dQ$ which moves from +V to 0V inside the battery.

8:41 AM
The charge flowing through the battery is what recharges the battery.
The point Farchar is making is that all the energy lost by the capacitor goes into recharging the battery.
None is lost moving the charge from the capacitor to the battery. This is very different to the case where we are charging a capacitor from a battery.

@JohnRennie And that's where my doubt lies in. Why should the battery be charged? Instead be liberated as heat as usual? We could consider the internal resistance of the battery.

We're assuming a perfect battery i.e. if the battery pushes current out it discharges and if current gets pushed into the battery it charges.
A real battery would have losses due to the internal resistance of the battery, and possibly also depending on the details of the chemical reaction going on inside the battery.

@JohnRennie Ok sir. Since we are pushing positive charges to the positive terminal and negative charges to the negative terminal, we're charging the battery? If yes, now I completely understood why Farcher told this is similar to a rechargeable battery.

Yes. In principle all batteries are rechargable.
All batteries use a redox reaction to develop the potential difference, and all chemical reactions are revesible.

@JohnRennie Great. But non-rechargeable batteries would liberate heat as I doubted earlier?
@JohnRennie Yes sir. I've learnt that in chemistry (Electrochemistry) and it was also interesting like physics.

8:46 AM
The reason some batteries can't be recharged is that you get physical changes as it discharges. If you design a battery carefully enough it can always be recharged.
So non-rechargeable batteries are not fundamentally non-rechargeable.

@JohnRennie: Ok sir. Now got it. In the above case are the wires having negligible resistance, sir?

But yes, if you did this with a standard non-rechargeable battery then you would get energy lost as heat.

@JohnRennie Ok sir. Thank you.

@GuruVishnu yes, if the wires have a resistance $R$ then a potential difference $V=IR$ develops across the wire, then we get an energy lost $E= QV = QIR$
But note that if we make thee process infinitely slow then $I \to 0$ and we can make the losses arbitrarily small.

@JohnRennie So in reality when we pull the plates of a capacitor connected to a battery - battery gets charged in addition to heat loss unlike what I thought before - only heat loss would take place but no charging of battery. Am I right sir?
@JohnRennie Ok sir.

8:51 AM
@GuruVishnu can you clarify that? You seem to be saying two contradictory things in that sentence.
1. battery gets charged in addition to heat loss unlike what I thought before
2. only heat loss would take place but no charging of batter
In a real battery you get some charging and some heat loss.
How much charging and how much loss depends on the design of the battery and how fast you pull the plates apart.

@JohnRennie I think you've understood that statement. But why should battery get charged in the first place? Farcher sir said "There is nowhere else for the energy to go" Are there any other explanation for this sir? I think I'm restarting the loop.

You said yourself that when you pull the plates apart charge has to flow backwards through the battery.

@JohnRennie Yes sir. Is that an appropriate reason? It was totally my theory and I doubted it.

So the charge $Q$ flowing backwards through a potential difference $V$ does work $QV$ on the battery.
@GuruVishnu charge flows off the top plate and onto the bottom plate. Charge can't just appear or disappear, so it has to flow round the circuit i.e. through the battery.

@JohnRennie So charge passing in one direction discharges the battery and if it passes in the other direction it charges the battery. If yes, then I understood this concept sir.

9:00 AM
Correct

@JohnRennie Then... Thank you very much sir :-)

:-)

@JohnRennie Sir, may I know how did you line break this message?

Press shift-return instead of just return.
That's on a PC. I'm not sure how you'd do it ona phone.

Ok
sir :-)
It works really well
Thank you for the tip.
BTW I'm on a PC and don't have a phone. I use only my mom's.

2 hours later…
10:43 AM
@JohnRennie: Hi sir :-)

@GuruVishnu hi :-)

@JohnRennie I've converted our previous conversation and my understanding into an answer. And the following is an image for that:

It seems it'll be easier for me (and also others) to understand an answer better than the chat transcript codes I use to store our conversations.
@JohnRennie: Sir, Are you free now? If yes shall we continue our discussion on diagonal dielectrics?

@GuruVishnu yes I'm free. I confess I have lost track or where we got to with the dielectrics.

10:49 AM
Let's concentrate on a single dielectric piece. There are infinitely many regions of potentials $V_i$. Now if we swap the other dielectric piece will the $V_i$s change or remain the same?
@JohnRennie: Is this clear or may I add more details to my question, sir?

@GuruVishnu You mean if we consider a single dielectric so all yellow or all blue?

@JohnRennie I meant the top half or the bottom half irrespective of the dielectric. Will the potentials at each and every point change if we change the material of the other half, sir?

Yes.
I can't help feeling you're making this a lot more complicated than it really is.

@JohnRennie Sorry sir. This is not homework. I just made it on my own to simplify another question.

No, I mean the analysis of the situation is really simple.
If you have a single dielectric than the potential changes linearly as we go from one plate to the other.

10:56 AM
@JohnRennie But this doesn't seem obvious to me based on our discussion 2 days ago.
@JohnRennie Yes sir. I understood this point.

If you have two dielectrics with a junction parallel to the plates then it behaves as two capacitors in series, one with each dielectric.

Yes.
Then the potential $V' = V\frac{C_2}{C_1+C_2}$

@JohnRennie Fine sir. That day, I also showed you some graphs I constructed based on my understanding:

Just the usual equation for the voltage.

10:59 AM
@JohnRennie Yes sir.

@GuruVishnu yes, within each capacitor the potential changed linearly, and the discontinuity is where they join - where the potential is $V'$.
Is this all OK so far?

@JohnRennie I understood this point. A small correction the curve is continuous but not differentiable at the interface.
@JohnRennie That's fine sir. I got into trouble when I tried to plot the same for different vertical profiles for two diagonal dielectrics. And that's why I asked the question - whether potentials at all points change irrespective of the other piece.

And to handle a slanted boundary between the dielectrics you split it into an infinite number of capacitors in parallel.

@JohnRennie Yes sir. I did that already. I've no doubt in that.

Each infinitesimal capacitor is treated s two capacitors in series.

11:03 AM
@JohnRennie Yes got it. That day the main reason I asked this is for an alternate approach other than that. I think you might remember why my method of swapping dielectrics failed.

The potential at their boundary will be $V' = V\frac{C_2}{C_1+C_2}$ as before, but this varies as we move sideways because $C_1$ and $C_2$ change.

@JohnRennie Ok sir. Will the plot of potential at interface be something like $1/x$ curve? Or a top left to bottom right diagonal?
in this diagram:

Let me draw a diagram:

@JohnRennie Ok sir.

So our task is to calculate $V'(x)$. Yes?
i.e. the potential at the junction as a function of horizontal distance from the left edge.

11:11 AM
@JohnRennie Yes sir. Now it seems quite familiar :-)

@GuruVishnu so you can do the calculation?

@JohnRennie Yes sir. I'll try on my own. May I share the final graph with you once I'm done?
for PEV :-)

Yes. I'll be here for another hour or so.

11:25 AM
I found $V(x)=xV/a$
where $a$ is the length of the square plates of the capacitor.
So, the graph of potential as a function of distance is a diagonal (from bottom left to top right)
But I don't see, still how to draw a curve like this, sir:
The place where the kink occurs is now figured out, but how to trace the rest of the graph for all profiles?

@GuruVishnu that's obviously wrong because it doesn't depend on the two different dielectric constants. That's the result you would get if the two dielectric constants were the same.

@JohnRennie Ah! yes sir. Please wait. I know where to fix it.

@GuruVishnu note that your graph shows $V(y)$ i.e. the potential if we sit at constant $x$ and move vertically between the two plates.
What you are now calculating is different i.e. the potential at the join as you move sideways.
You wouldn't expect the latter curve to have any discontinuity in gradient.

$V(x)=\frac{yk_1}{dk_2+y(k_1-k_2)}$
where $k_1$ and $k_2$ are dielectric constants of blue and orange dielectrics
$d$ is the plate separation and
$y=xd/a$
@JohnRennie Oops. Again. Yes sir. But I think we could find the position of that kink using this. Am I right?

Looks plausible. There is an obvious quick check, if you set $k_1 = k_2$ does it give the same answer as your previous result?

11:35 AM
@JohnRennie Yes sir. I missed $V$ on the numerator on the RHS. Everything else is correct if we substitute the value of $y$.

I wonder if we're still talking at cross purposes. Your equation gives $V'(x)$ i.e. the potential at the join as we move horizontally.

So final correct version is:
$V(x)=\frac{yk_1}{dk_2+y(k_1-k_2)}V$
where $k_1$ and $k_2$ are dielectric constants of blue and orange dielectrics
$d$ is the plate separation and
$y=xd/a$

Yes.
Well, I haven't checked it, but it looks right :-)

@JohnRennie Now your quick check works great!
@JohnRennie Thank you sir. But I've verified it. So we could build more on this platform (it's quite stable)
@JohnRennie Will we not get the position of kinks using this, sir?

@GuruVishnu no.
The kink appears in the graph of $V(y)$.
What you've just calculated is $V'(x)$

11:40 AM
@JohnRennie I understood in the above graph we'll not have any kinks. Will we see the locus of kinks using this expression, sir?

The kink occurs at a distance $y$ above the bottom plate. Yes?

@JohnRennie Yes sir.

So if you're plotting the distance of the kink from the bottom plate as a function of $x$ it's just $y(x)$ i.e. $y(x) = h \frac{x}{a}$
What you've just worked out is the potential at the kink as a function of $x$

Yes sir :1 and it looks something like the following:
@JohnRennie And now I see, we could not use this for different vertical profiles (layers).

Yes, that looks plausible. Obviously you'll get a family of curves for different values of the ratio $k_1/k_2$.

11:46 AM
@JohnRennie Yes sir. It's interesting when I plugged some negative values for the dielectric constants :-)

Can you have a negative dielectric constant?

@JohnRennie As far as I know - no sir?
I haven't even seen when K<1;

3

What is the physical cause behind a material having a negative real part of its dielectric function? Given the complex permittivity, $\epsilon(\omega)=\epsilon(\omega)'+i\epsilon(\omega)''$, the Drude model gives \begin{align} \epsilon'=1-\frac{\omega_{P}^2}{\omega^2+\omega_{\tau}^2} \end{align} ...

@JohnRennie Sir, I think I haven't seen such big formulas for permittivity. I don't know who is Drude? Maybe a scientist?
Neither do I understand the question or answer completely.

Paul Karl Ludwig Drude (German: [ˈdʁuːdə]; 12 July 1863 – 5 July 1906) was a German physicist specializing in optics. He wrote a fundamental textbook integrating optics with Maxwell's theories of electromagnetism. == Education == Born into a Jewish family, the son of a physician in Braunschweig, Drude began his studies in mathematics at the University of Göttingen, but later changed his major to physics. His dissertation covering the reflection and diffraction of light in crystals was completed in 1887, under Woldemar Voigt. == Career == In 1894 Drude became an extraordinarius professor at the...

11:50 AM
Thank you sir. Could you please tell in short whether negative dielectric constant exists or not?

I don't know.
I know that the refractive index does strange things when you get near an absorption line.

@JohnRennie Absorption line - did you mean absorption spectrum of atoms?
Or a different one?

i.e. suppose you have some material with an electronic excitation corresponding to a wavelength $\lambda$, then if you graph the refractive index as a function of $\lambda$ the refractive index goes crazy at $\lambda$.
@GuruVishnu yes

@JohnRennie Wow. I didn't know that sir. Is this Drude's law?

No, the Drude model is a simplified model to describe the interaction of light with solids. If I ever learned it I have long ago forgotten it.
Have a look at this answer:
14

You have in fact put your finger on the reason for the refractive index change. It is related to moving electrons in the direction of the fields. NB dispersion is a complex phenomenon, so this is necessarily going to be an arm-waving explanation - do not take it too literally! There is a discuss...

11:55 AM
@JohnRennie Ok sir. No problem. Shall we drift towards plotting this kind of graph for different profiles? I'd like to see how it varies for different profiles.
@JohnRennie Sure sir. Definitely.

But that only shows the refractive index falling below one, not going negative.
Incidentally, the reason materials like glass show dispersion is they have an absorption in the uv. So the refractive index behaves like the left end of my graph i.e. increases slowly with frequency.

@JohnRennie Cool. That graph is really cool. Even we have resonance in light waves much like sound or mechanical waves?

A resonance in light means the photon energy matches an absorption energy in the material. It's not a mechanical resonance.

@JohnRennie So when the applied frequency matches the natural frequency the peak will not go to infinity? It seems yes. If not we'll have negative values for refractive index.

As we increase frequency the refractive index rises to a mximum (not infinite) just before the energies match, then flips to a trough below one (but not negative).

12:00 PM
Also the graph looks quite different from that of mechanical resonance.
@JohnRennie Yes sir. I realised it. It looks much different from that of mechanical resonance. Although the words match.

If you consider the usual graph for the amplitude at the resonance I think this graph is its derivative.

@JohnRennie If that's the case then at $n=1$ slope of the curve I have in mind is not defined. But here if it's the derivative then why is it 1?

I can't remember the details I'm afraid. I'd have to go off and Google them.

@JohnRennie No problem sir.
@JohnRennie: If you don't mind may I vote your ~question~ answer once I understood it properly? I think I need to allocate time separately. I find it quite difficult to manage chat and learning the answer simultaneously.

@GuruVishnu Which question?

12:06 PM

I think we drifted a lot from capacitors :-)
@JohnRennie :-)

Yes :-)
It's all good fun, but these diversions can use up a lot of your time.
By all means do them occasionally, but don't let it detract from the JEE studies.

Yes sir. I agree.

I need to go now. I'll be around later today, or back tomorrow morning as usual.

12:09 PM
@JohnRennie Ok sir. Good bye :-)

3 hours later…
3:14 PM
@YuvrajSingh... Hey!

@AjayMishra hello!

What's up?
What are the subject you are gonna be learning this semester?
Hey @tatan

hi

@AjayMishra have a look!

55 credits in one semester. Huhuh.

3:22 PM
I changed my branch to electrical engineering!

With that low CG?
Seriously?

Yes.

Yes.

@YuvrajSingh... Which college are u studying in bro?

3:24 PM
@tatan IIT R
@YuvrajSingh... What were the topics covered?

oh gr8

@tatan Are you coming in spring fest?

no man... but yusuf is going (I hope u remember him) xD

Yes. How about you? How it feels to relearn biology? :P

@tatan in which college you are?

3:28 PM
more or less good... actually depends on the teacher@AjayMishra
@YuvrajSingh... IISER-K

What stuffs you learned?

@AjayMishra give me second let me upload full-service schedule.

@tatan Cool.

@tatan how was you first semester
@AjayMishra same to you.

3:32 PM
How in what sense? CGPA ?

@YuvrajSingh... What do you mean by same to you?

@tatan @AjayMishra both cgpa and experiencing @tatan I am assuming you are first year, don, t you?

Yaa man.. first year

@tatan The weird thing I noticed is mechanics is taught by physics professors, was the scientific aspects more emphasized.

@AjayMishra is that a problem? I have no idea
Both were decent @YuvrajSingh...

3:36 PM
No, not a problem though, but I guess the substance is quite less if you go through the route of physics.
@YuvrajSingh... ?
@YuvrajSingh... both were bad, too fast.

Have you seen the syllabus? Is is the same as you are learning?

No.

okay...

@AjayMishra why question mark? In second last comment.

@YuvrajSingh... You didn't told me what you have learned.
I need to ask something, I am stuck.
@tatan Our syllabus goes like that ^

3:42 PM
Oops sorry actually I was doing two things simultaneously. OK waves shm and quantum mechanics!

engineering stuff @AjayMishra

What in SHM?
@tatan Indeed, it's crunchier that way.

haha

How is hostel and stuffs?

@AjayMishra sorry for late reply, I have some really urgent work give last 10minutes.

3:46 PM
Okay, it always happens like this though. :P

What does that mean @AjayMishra

Well, are you now?
free*

Yes I am done.

Have you learned coupled oscillators?

OK we have studied the damped oscillation in detail.

3:50 PM
Both are different.

Now they have coupled shm with wave and periodicity.
@AjayMishra Yes!

Oh, no I mean couple oscillators, that's a topic.

What, s the question?
Yes.
I have online notes if you want I can send.

Have you been taught coupled oscillator? That is the first question.
I have the notes too, I can find your college notes too :P

Yes
It was in list.

3:55 PM
2 days ago, by Ajay Mishra
I've literally no idea why I ain't getting the answer.
Can you help me?

Let me try If I can!
@AjayMishra I can answer it just give me second to write the answer on paper.
Hopefully my battery do not die. Till then

@JohnRennie Hello

@pi-π hi

@JohnRennie What are the point of suspension and point of oscillation in case of bar pendulum?

4:03 PM
A bar pendulum is just a rod isn't it?
You could pivot it at any point along its length.

@JohnRennie Yes

@AjayMishra suppose we have two oscillator like given in the question yes!

@pi-π I'm not sure what you are asking ...

With some angular velocity w1 and w2
Yes?

@JohnRennie The point of oscillation and point of suspension are said to be interchangeable in bar pendulum so I wanted to ask how does that happen?

4:10 PM
@pi-π I don't know to be honest.

@JohnRennie It's okay.

@YuvrajSingh... Okay, then?

Let the displacement of each oscillator from equilibrium
be x1 and x2, respectively. Of course, if the two oscillators are uncoupled, that is, do not
interact in any way, each of the displacements satisfies a harmonic oscillator equation
@AjayMishra
Their equation would be d^2x/dt^2=-kx1^2
d^2x2/dt^2=-kx2^2

Don't teach me that stuff, just tell me are you getting the answer?

4:26 PM
Not exactly my final answer doesn so, t match with answer you have given @AjayMishra

What's the answer are you getting?
What the equation of motion you are getting?
Are the amplitude of the basis same?