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4:00 AM
@JohnRennie What are the best universities for studying physics?
 
4:43 AM
@JohnRennie How an atomic cloak works?
 
What is an atomic cloak?
 
Me neither, which means I'm going to struggle to explain how it works :-)
Ah, atomic clock!
 
its okay sir, I asked it out of sheer curiosity
sir, any possibilities of time travel?
@JohnRennie sir?
 
I'll be a moment or two. I'm just fixing a server ...
@Akash.B that article looks pretty good. What parts of it are giving you problems?
 
5:21 AM
2 messages moved to trash
 
5:35 AM
@KingTut any charge distribution can in principle be made up from a sum of many point charges, and for any single point charge Gauss' law obviously works. But the total flux is just the sum of all the fluxes from the individual charges.
So if we consider a surface enclosing some random distribution of charge the total flux through that surface is just the sum of all the fluxes from the point charges within it, and Gauss' law applies to the charge distribution as a whole.
 
6:19 AM
@JohnRennie .. you there...?
 
Morning :-)
 
Good morning ...
I had a query...
 
Yes?
 
@JohnRennie...If 8 water drops falling with terminal velocity coalesce and the large water drop formed fall with terminal velocity , then other than frictional force and weight , should we take buoyancy into account ?
 
You could easily calculate the buoyant force, but I suspect it would be negligible.
 
6:23 AM
@JohnRennie...Yes , my book didn't take it into account , but why ?
 
The density of air is only 1.2kg/m^3 so it's about a thousand times smaller than the density of water.
 
@JohnRennie... Thanks ...
 
6:45 AM
By the way , @JohnRennie ... If I keep the small drops in a freely falling lift , then will the consequence be same , like will they coalesce and fall with same terminal velocity ...?
 
@NehalSamee what matters is the speed of the raindrop(s) relative to the air around them.
Because aerodynamic drag depends on the speed relative to the air.
 
7:19 AM
@john exactly my doubt, all charges obey gauss law is it some postulate
oh i understand its a law, similar to coloumbs law
So it is granted that all charge do obey gauss law, from observing the surrounding, is it true
 
@KingTut Yes. Actually I can't remember how Gauss' law is derived. I think it follows naturally from Maxwell's equatons, but I'd have to Google it.
 
Thanks alot! @John i have another question that can any electric field be created by varying the charge distribution? Conservative or not?
 
If the charge distribution is static then the field is always conservative
If you you have moving charges, i.e. currents, then I think the field is non-conservative though I'd have to check. There is a question and answer on exactly this topic somewhere on the main site.
The issue is whether the field lines can form loops. With static charges every electric field line has to begin on a charge and end on a charge (or at infinity).
But with electric currents you can form field lines that are loops, and when this happens the field is non-conservative.
Aha, found the question:
1
A: Electric field lines forming closed loops

John RennieIf a vector field $\mathbf E$ is the gradient of a conservative potential $\phi$ then its curl must be zero i.e. $$ \nabla \times \mathbf E = 0 $$ because the curl of the gradient of any continuously twice-differentiable scalar field is always the zero vector. However Maxwell's equations tell u...

 
7:34 AM
Yes like in faraday's electric fields?
@John looks like its straight from calc 2
or calc 3
 
Yes, though bear in mind that a constant current produces a static magnetic field and for a non-conservative field we need a time varying magnetic field. I guess that means we need a time dependent current.
 
 
2 hours later…
10:04 AM
@JohnRennie Or Time dependent flux ,if current is independent of time
 
10:50 AM
can anyone please help me understand why (2) is a correct statement? I know a razor blade has small volume, but I also know that whenever any object sinks into water, the height of a water level increases, or remains constant if the object has negligible volume. But how can it decrease?
 
When the blade is floating the water has to provide an upthrust equal to the weight of the blade.
However when the blade has sunk to the bottom of the glass it's the bottom of the glass not the water that supplies (part of) the upthrust.
So the volume of water displaced when the blade is floating is greater than the volume displaced when it has sunk to the bottom of the glass.
 
am trying to digest
but won't the height of water rise due to an additional solid body inside it?
like, if I drop an iron nail in a glass of water, the level rises
i expected in the razor's case for it to at least remain constant if not increase
 
When the blade has sunk the level of the water rises by an amount equal to the volume of the blade. Yes?
 
But when the blade is floating it has to displace a volume of water equal to the weight of the blade. Since water is less dense than metal the volume displaced is greater than the volume of the blade
 
11:00 AM
oh, understood!
thanks!
btw, does surface tension play any role here?
 
No. Surface tension determines the shape of the surface near the blade, but not the total volume displaced.
 
oh, alright!
 
@GaurangTandon The water height indeed increases
I see no reason for decreasing
 
actually the answer is that water height decreases
read JR's responses above @MadhuchhandaMandal
14 mins ago, by John Rennie
But when the blade is floating it has to displace a volume of water equal to the weight of the blade. Since water is less dense than metal the volume displaced is greater than the volume of the blade
 
@JohnRennie How can a body displace Water more than its volume without even fully submerging?
I think it's surface tensions here
Which balances the additional weight
 
11:19 AM
@MadhuchhandaMandal yah good point that
 
@MadhuchhandaMandal Suppose you have a cubic metre of lead. Put it in water and it sinks and dispaces a cubic metre of water.
Now put that lead ina boat floating on the water (assume the boat weight is negligible)
 
@GaurangTandon Hey did you read the question carefully? Reread it !!!
@JohnRennie yeah in that case a support is needed
 
The lead+boat displace 11.35 cubic metres of water
 
But in this case no such things is mentioned
 
@MadhuchhandaMandal the surface tension of the water is in effect acting as the boat
i.e. it's making the surface of the water act like a rubber sheet or a trampoline
 
11:23 AM
@MadhuchhandaMandal the answer is NOT (2); so (2) is a correct statement...
 
@JohnRennie Okay.. say lets remove the boat and keep the Lead block only. Can it displace Water greater than a cubic meter?
 
No, because it will sink
 
@GaurangTandon okay I got it!!
@JohnRennie can you please explain a bit more ?
Okay I got that !!!!!
 
@MadhuchhandaMandal I'm not sure what you're unclear about ...
 
@JohnRennie Do you mean something like that?
 
11:30 AM
Well there won't be air under the blade
 
Yeah ofcourse.. I tried to portray thickness of the blade :-)
The rest thing?
(The dark shade being the metal)
And light shade being the water
 
More like that
 
So the extra volume is the depressed area above the blade.
 
perfect! :D
productive discussion :)
 
11:37 AM
But wait !!! Is the lift on the Blade anyway proportional to the volume of the depressed area above the blade?
 
Yes
 
The surface tension is acting like a rubber sheet, so it is pulling upwards on the edges of the blade.
 
I mean is Flift proportional to Varea?
 
The volume of the depressed area + the volume of the blade = the total volume of water displaced
And the total volume of water displaced times the density of the water is equal to the upthrust on the blade
 
11:42 AM
@JohnRennie how do you make these neat diagrams?
which software?
 
@JohnRennie Yeah I was asking that !!! Is that true?
How can we prove that?
 
@GaurangTandon Google Draw.
 
@JohnRennie oh nice
 
You can export the drawings and download them, but I just copy and paste into Paint then trim them to size
Google Draw takes a bit of getting used to, but once you get the hang of it you can do diagrams really quickly and easily.
@MadhuchhandaMandal that's Archimedes' principle!
Or is it? Hmm ...
 
I think not
 
11:51 AM
I'd have to think about it, but I've been working since 5 a.m. (8 hours ago!) so I'm not going to think about it now :-)
 
I think you'd have to include the forces where the water meets the glass. Ultimately any upwards force must come from the inside walls of the glass.
 
I think the height of the water does not change at all if assumed angle of contact is 90 deg
@GaurangTandon
😬
 
12:29 PM
Okay I worked out the problem rigorously
This is what I obtained
Which is a universal condition as long as the Blade is floating initially
[Note : When the Blade is on the verge of floating and sinking, the change in water Height is 0]
@GaurangTandon @JohnRennie
 
12:46 PM
Or to be more specific " It all depends on the Water and the Blade :-)"
 
12:59 PM
@GaurangTandon Final Conclusion : Thickness of Razor blade is 0
In that case water Height will always decrease
Otherwise it depends on surface tension, it's Density , It's thickness and degree of maximum deformation of the water surface
 
1:21 PM
thanks! @MadhuchhandaMandal
sorry I was away just came back
let me read properly
@MadhuchhandaMandal how did you derive this?
 
Well I will send snaps because it's hard to type here
 
2:14 PM
@GaurangTandon
 
2:46 PM
that is amazing
thanks for sharing @MadhuchhandaMandal !
I'm sure @JohnRennie will be delighted to read it all
 
@GaurangTandon Welcome !!! 😊
 
i need help in this one ^^ I know focal length for both lenses is same, and it is equal to 60cm for each lens (equivalent 30cm). But I don't know how to handle the other part
i totally don't know the concept
so if anyone can just link to an article also, for me to read it, that'll also help!
 
3:08 PM
@GaurangTandon Hope that helps
@GaurangTandon The concept is that : If you cut any lens , only the intensity of the point mage changes not it's position or number
So both the lens can be treated as complete convex lenses placed side by side , just displaced vertically
 
3:41 PM
oh, nice! but what do I do next? :/
 
Okay lemme see
 
3:57 PM
@GaurangTandon
 
got it!
thanks @MadhuchhandaMandal
ok, so the process is (1) find x coordinate of image from first lens (2) figure out its magnification (3) use that to write relation between initial height and final height
and then use data given in question
sweet
:D
 
@GaurangTandon Welcome!!
@GaurangTandon Or maybe some direct formula is available... Which I hate to remember!!!
 
could be, but this approach is easier to do quickly
 
yeah; i think the bottom arrow is supposed to be the second dipole $P_2$
what do you think?
 
4:15 PM
Isn't the Diagram apparently leading to some thing like the above ?
Okay yeah it's not..
 
yeah, the diagram leads to exactly what you drew; but we both know it isn't supposed to be so ;)
 
;-) yeah we know that !!
 
the dipoles are short
and so r>>dipole length
 
Something like that
@GaurangTandon
 
yep, exactly like that
 
4:19 PM
But I found it misleading and didn't attempt :/
 
oh
well
now you can do it, can you?
find electric field, then integrate wrt r
 
Yeah I can now
 
ok good
 
Do we need to find Electric field 🤔
We can do q(-P(r) + P(r+dr) )
 
@MadhuchhandaMandal what does this mean?
 
4:42 PM
ok time for me to go, cya tomo!
 
P is the potential function due to Dipole P2 at the point O1 (at a distance r from O2)
And dr is the length of the dipole P1
So Interaction Energy = q dr ((P(r+dr)- P(r))/dr
@GaurangTandon
Or , P1 * (dP/dr)
 

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