@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.
@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?
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 $dW = dQ \times 0 = 0$.
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 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.
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.
@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 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.
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.
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?
@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?
@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.
@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.
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.
$V(x)=\frac{yk_1}{dk_2+y(k_1-k_2)}$ 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?
$V(x)=\frac{yk_1}{dk_2+y(k_1-k_2)}V$ where $k_1$ and $k_2$ are dielectric constants of blue and orange dielectrics $d$ is the plate separation and $y=xd/a$
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}
...
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...
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 $\lambda$.
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...
@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.
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 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).
@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.
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