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03:07
@JohnRennie Sir Could you shed some light on above problem ?
04:00
@JohnRennie Do you know any good chemistry simulator?
@Mayank I'm sorry, I don't.
04:13
@AnnamalaiSriram in a circuit the charges on the plates of a capacitor are always equal and opposite. That means if a charge q flows onto one plate an equal and opposite charge -q has to flow off the other plate. In your circuit that mean the charge flowing round the circuit has to be the same everywhere.
Oh!
@JohnRennie hello!
@Anthony hi :-)
Oh this is problem solving strategies, not the hbar.
Sad.
04:43
@JohnRennie OK . Dont they attain a common potential ?
Last Part of Solution
05:06
@Mayank What do you mean by chemistry simulator?
Add 2 reactant
Like Beaker
@Safdar
@AnnamalaiSriram there is no reason why all the capacitors would have the same potential. The only requirement is that if we go round the loop the potentials have to sum to zero as in Kirchoff's laws.
@Mayank IIRC, try using the play store, there was an app for that..
https://play.google.com/store/apps/details?id=air.thix.sciencesense.beaker&hl=en_IN

@Mayank
3 mins ago, by Mayank
Like Beaker
I didn't find it good.
@Mayank That is the only one I've used.. I don't think there are.. Just my opinion..
05:17
@JohnRennie Sir it could be but what made me link it with common potential is this
@AnnamalaiSriram I don't have time to watch that right now but I'll try to find time for it later.
@JohnRennie Ok Sir When you are free ping me about it
05:34
@JohnRennie Hi
@Mayank hi
Does a test charge have inertia?
@JohnRennie
"Inertia" basically just means "mass" so it depends on whether the object carrying the test charge has a mass. Real charged objects do have a mass as even an electron has a mass (though it's a very small mass).
I thought of an example where 2 forces can stop an object.
Can you verify if I am true?
OK ... ?
05:44
A constant force F1 is exerted on an object and after sometime a force F2 is exerted opposite to previous force.
Then Stop exerting F2 when object stop.
I don't know what I am saying.:-(
What situation are you thinking about? Accelerating and decelerating an object?
Yes
@JohnRennie
You could consider a rocket. The rocket motor accelerates the rocket and starts it moving, then when the rocket reaches Mars (or whatever) a second rocket fires in the opposite direction to slow it down again.
Ok
:-)
The velocity is the integral of the acceleration, v = ∫a dt = ∫F(t)/m dt, where the force can change with time.
05:54
Yes
For the object to start and end at rest we just need this integral to be zero i.e. the force F(t) has to have positive and negative values that cancel each other out.
To be honest I don't know Integration
@JohnRennie Can you give your pdf hosting site link?
As you study for the JEE you'll learn more about integration than you want! :-)
:-)
@Mayank do you mean the address for uploading files to me?
06:01
That upload files to you
?
I thought that was a file hosting site.
No.
I have my own server, and it has a facility that allows people to send me files. But it's not a public site.
If someone upload file here, only you will be able to access them.
Correct
@JohnRennie Hello sir :-)
@Jasmine hi :-)
06:04
22
Q: What are the fields produced around a current carrying conductor?

Immortal PlayerIf you consider a current carrying conductor, every instant an electron enters the conductor, another electron will be leaving the conductor. Thus, the current carrying conductor will not be charged (i.e. it would not have any net positive or negative charge). Remember a dipole has zero net charg...

I was going through this
What should be the correct reasoning for it
@JohnRennie Can you give an file sharing website? I want to take some pdf from @RajdeepSindhu .
Why no electric field around current carrying conductor
@Jasmine Do you know Gauss's law?
@JohnRennie yes
Okay
Sio you only get a field if the net charge inside the Gaussian surface is non-zero. Yes?
06:08
So electric field is 0 as charge enclosed is 0
@Mayank just make a google account which both of you know the password to and use drive.. Or just share the link..
@JohnRennie Okay
@Jasmine yes
@Mayank as Safdar says, create a Google account, which is free, and you get 15GB space free to use to share files.
:55639928 true. Gauss's theorem just gives you the total flux so there can be a field as long as it sums to zero over the surface.
But you only get a dipole when there is a charge separation, and this doesn't happen for a current flowing in a wire.
Sorry to interrupt
@JohnRennie but is it really possible to define a Gaussian surface for dipole
06:13
@Jasmine Any surface is a Gaussian surface, it's just a question if how useful the surface is for doing calculations.
With a point charge we can use a sphere and the symmetry means the field has to be the same everywhere on the surface, which makes it very easy to calculate the field.
With a dipole it only has cylindrical symmetry, and I can't think of a way to use a Gaussian surface to calculate the field in this case.
@JohnRennie is it possible to calculate field of dipole using gauss law
@JohnRennie Okay
@Mayank this is 72° i.e. n = 360/72 - 1 = 4.
It looks to me as if it works ...
What kind of electric fields do moving charges produce
At normal speeds the electric field of a moving charge is the same 1/r² field as a stationary charge - just moving.
This changes as the speed approaches the speed of light, but that get's really complicated! :-)
06:20
24° ?
@JohnRennie Okay
@JohnRennie Yes
No observer mode
@Mayank actually with 72° it is possible to find some positions for the observer where one of the images doubles so we get 5 images. We found this yesterday didn't we?
Yes
That's the expected 4 images, but if I move the observer, without moving the mirrors of the object I can get this:
06:24
What is going on?
It just proves that it's complicated ...
Why everything go messy why I try to understand them?
This is a particularly complicated system, but I'd say that was the exception rather than the rule ...
Can you take a look at these comments
I don't think any simple argument is going to work, because as we've found it depends on the position of the observer.
06:35
So we have to apply formula according to case.
@JohnRennie
I don't know. Wait until you have some sample JEE questions and we'll see what exactly they ask.
@JohnRennie They just ask for the blind application of this formula.. They don't go in corner-cases due to the ambiguity.
@JohnRennie When Ø=40, then neither of the two formula holds true.
Give me a moment and I'll try it.
I get 8 images, which is indeed 360/40 - 1
As before, I can get it to show one more image for some positions of the observer:
-Alan Abraham
Great response and visuals. Some things to watch out for is in the cases when 180/θ are not integers, which the answer OP linked for briefly discusses. I'm not sure if this is true, but I tried with θ=40 and got 18 images by continuously reflecting (geogebra.org/calculator/smy6jdqf). I believe it makes sense that if 360n∈Z and v2(n)>2, then the number of images is 720/θ−1. – Alan Abraham
@Mayank that isn't showing the images. It's showing what happens if you combine all possible reflections in the two lines, but not all these positions will appear as images.
06:57
My little brain is gonna explode.
@JohnRennie
What happens at angle bisector?
In 40 degree case
@JohnRennie Are you here?
Nothing special.
I still get 8 images
Thanks
I think I should take a break
:-}
Bye
@JohnRennie
Bye :-)
 
3 hours later…
09:58
@JohnRennie What does OP mean?
@Mayank in the Stack Exchange? If so OP = Original Poster i.e. the person who originally posted the question or answer referred to.
@JohnRennie Why the observer is not able to see all images from some points?
I think the answer must be obvious.
You need to get over this.
The question of what an observer can see comes down to a technique called ray tracing, and this is a complicated business so it's no surprise that there isn't a simple formula that gives the number of images under all conditions.
The time you are spending on this is basically being wasted.
2
@JohnRennie I am not talking about formula.
I was thinking about overlapping of image.
10:14
@JohnRennie Hi
So, a while ago, I made a formula that gives the equation of the reflected ray from a concave mirror if the radius of curvature of the mirror and the equation of the incident ray are known. Do you think it is possible to use that formula to find the point where almost all rays passing through a point (the source) will meet?
By the way, you explained the conservative nature of electric field to me a few months ago using curl.
Here's the link to a graph I made using the formula I talk about above : https://www.desmos.com/calculator/jmzhwztxne
If there is some way to do it, I would most likely be able to derive the mirror formula and a couple of other interesting results...
10:50
aweomse @r
@RajdeepSindhu
Thanks @Buraian
I have more graphs like this linked on my blog, https://the-epsilon-delta.blogspot.com/
They're in the "About me" section.
https://the-epsilon-delta.blogspot.com/2020/09/mathrmthevarepsilon-deltamathrmblog.html
11:18
I too write

https://medium.com/@brianbabu890/a-brief-introduction-to-taylor-series-47416e7b9123

you may like this xD
@Buraian It looks awesome
Looks interesting @Buraian
 
2 hours later…
12:52
according to the free body diagram of the foot(sending after this), there is no equilibrium in the horizontal direction, so the foot should slide forward no matter what. But that doesn't happen, we can keep our foot on the ground while pushing it(tried it myself) and that it would imply that there is work done by friction or the friction is kinetic which is not true in walking. What is wrong in the free body diagram
I think the problem is more related to rolling friction because I had the same confusion
@Satwik Where is reaction force exerted by ground?
@Mayank, I have broken it into two components( the normal and friction)
 
3 hours later…
16:04
@JohnRennie Sir When I worked with 2 Capacitor, Finally I got that the equipotential method actually comes from Kirchoff's Laws.
@JohnRennie Sir So I think that it clarifies me as you stated. Is there anything wrong?
16:21
@AnnamalaiSriram when you have only two capacitors then the two capacitors are in parallel and therefore the voltage across them has to be the same.
The problem here is we have four capacitors.
Yes Sir But it works
For two capacitors it works.
Sir Actually my main doubt is when two caps are connected they become equipotential only due to the capacitors itself not due to external emf so likewise why can't the 4 capacitors act in the same way?
In your question you have four capacitors arranged like this:
But you could equally well draw then like this:
16:35
And now A is in parallel with B + C + D.
Or you could draw them like this:
So now A + D is in parallel with B + C.
There are lots of ways of drawing different combinations of the four capacitors in parallel so there is no simple way to use the fact that the voltage on things in parallel is the same.
With only two capacitors we don't get this problem because there is only one way they can be in parallel.
Yes Sir I got the fundamental point.
Thank You Sir
I viewed as individual potentials so they need to be equal as of two capacitors case
16:43
@JohnRennie Hi sir!!
@satan29 hi :-)
That made me to puzzle
7 hours ago, by Rajdeep Sindhu
@JohnRennie Hi
So, a while ago, I made a formula that gives the equation of the reflected ray from a concave mirror if the radius of curvature of the mirror and the equation of the incident ray are known. Do you think it is possible to use that formula to find the point where almost all rays passing through a point (the source) will meet?
By the way, you explained the conservative nature of electric field to me a few months ago using curl.
Here's the link to a graph I made using the formula I talk about above : https://www.desmos.com/calculator/jmzhwztxne
@JohnRennie
Sorry for bothering you again, sir, but do you think there's a way to do this?
sir, I have a small doubt regarding siphon
@RajdeepSindhu I guess so if your incoming ray comes from the object i.e. it passes through the point where the object is. The reflected ray must then pass through the point where the image is. So if you calculate two reflected rays they will intersect at the position of the image.
@satan29 yes?
16:48
But how do I take the paraxial approximation into consideration?
As far as I know, if the object is placed at a considerable height from the principal axis, the rays won't meet exactly at a point...
Yes, for a spherical mirror the rays intersect along a surface called the caustic not at a point, but if you stay near the axis you'll get intersection at approximately a single point.
We were told to calculate maximum flow velocity possible in a siphon
OK ... ?
so i had a look on the wiki article on siphons
Use the pressure difference and the Darcy Weisbach equation I guess ...
16:51
darcy...whatsthat?
@JohnRennie Even if we don't stay close to the axis, though, the intersection of any two rays should still give a pretty good approximation of the image's position, right?
In fluid dynamics, the Darcy–Weisbach equation is an empirical equation, which relates the head loss, or pressure loss, due to friction along a given length of pipe to the average velocity of the fluid flow for an incompressible fluid. The equation is named after Henry Darcy and Julius Weisbach. The Darcy–Weisbach equation contains a dimensionless friction factor, known as the Darcy friction factor. This is also variously called the Darcy–Weisbach friction factor, friction factor, resistance coefficient, or flow coefficient. == Pressure-loss form == In a cylindrical pipe of uniform diameter D,...
wikipedia gives a purely mathematical proof
pressure at the top point of the tube >=0
Or for slow flow rates where the flow is laminar there's another equation called ... erm ... I can't remember ...
@RajdeepSindhu yes
and after using Bernoullis theorem, we can get it easily
16:54
Ah, OK, yes that is the maximum rate assuming not losses due to viscosity.
Basically the Torricelli equation.
but i dont understand it physically: why exactly is there an upper limit to the velocity>
@JohnRennie One more question.
I think that trying to make a mirror formula for parabolic mirrors using a formula similar to the one I've talked about above would be better as compared to doing the same for spherical mirrors as the rays actually do meet at a point in parabolic mirrors.
@satan29 Because to get water to flow through the siphon you have to accelerate the stationary water in the vessel to the flow velocity in the siphon. That means you get a rate of change of momentum dp/dt = ρ A v², where A is the area of the siphon and v is the speed of the flow. OK so far?
And rate of change of momentum is force. That means to get the water to flow at a velocity v you have to apply a force F = ρ A v²
And that force has to come from the pressure difference between the water surface and end of the siphon.
So the speed is going to be proportional to the square root of the pressure difference.
16:59
@Binod Here too??
@RajdeepSindhu yes, that's true.
@JohnRennie Thank You :)
flow velocity of siphon is sqrt(2gd)
which doesnt seem to be bounded
@satan29 right, and that's exactly the same as the Torricelli equation:
Torricelli's law, also known as Torricelli's theorem, is a theorem in fluid dynamics relating the speed of fluid flowing from an orifice to the height of fluid above the opening. The law states that the speed v of efflux of a fluid through a sharp-edged hole at the bottom of a tank filled to a depth h is the same as the speed that a body (in this case a drop of water) would acquire in falling freely from a height h, i.e. v = 2 g h {\displaystyle v={\sqrt {2gh}}} , where...
17:02
no but
Yes ... ?
it seems to imply that ican simply increase v to whatever value i want
by increasing d
Yes, you can keep lowering the bottom of the siphon all you want. Well until it hits something.
Or you could keep raising the water vessel.
so then there shouldnt be any limit to the velocity.
I think you're going to struggle to make a siphon that is infinitely long.
Or even a few light years long.
17:04
no no thats not the point
like
say the "max velocity" derived by the wiki article occurs at depth d
what will happen at if i make the depth say.. d+1 meters?
"max velocity"?
There is a single flow velocity in the pipe that depends on d
where d is the vertical distance between the top of the water and the bottom end of the pipe.
A siphon (from Ancient Greek: σίφων, "pipe, tube", also spelled nonetymologically syphon) is any of a wide variety of devices that involve the flow of liquids through tubes. In a narrower sense, the word refers particularly to a tube in an inverted "U" shape, which causes a liquid to flow upward, above the surface of a reservoir, with no pump, but powered by the fall of the liquid as it flows down the tube under the pull of gravity, then discharging at a level lower than the surface of the reservoir from which it came. There are two leading theories about how siphons cause liquid to flow uphill...
Ah, OK.
If you make the siphon longer than 10m the pressure difference exceeds 1 atm and the water will leave a vacuum at the top of the siphon.
So the maximum usable length would be that 10m.
17:10
so but what happens if a vacum is formed?
like whats "bad"?
Don't know. I'd have to think about it and I'm getting tired now.
sure sir. I guess we can talk tommorrow. 'night!
Bye :-)

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