Problem Solving Strategies - 2021-04-13

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RishiNandha Vanchi

2:08 AM

> At jee classes, they taught a result that when a charge is placed inside a conductor, sigma developed outside at a point is inversely proportional to radius of curvature there

is this result true sir? @JohnRennie

I found this: physics.princeton.edu/~mcdonald/examples/ellipsoid.pdf I couldnt find a step that explicitly mentions radius of curvature, but there is a step with divergence or curl which i couldnt understand and mightve been the radius-step

3 hours later…

4:47 AM

@RishiNandhaVanchi Yes it's true. There is a simple proof here.

RishiNandha Vanchi

but the proof only holds when both the original spheres are infinitely far away right?

if not then charge distribution of one affects the potential at other's surface?

Yes, it's not really a proof - more a plausibility argument.

There is a detailed treatment here but it's hard going.

I think the relationship is only approximately true, but it's a good approximation in most cases.

@JohnRennie Hi! Good Morning

@Wolgwang hi :-)

> While developing the Periodic Table, there were a few instances where Mendeléev had to place an element with a slightly greater atomic mass before an element with a slightly lower atomic mass. The sequence was inverted so that elements with similar properties could be grouped together.

@JohnRennie Is this a merit or a demerit of Mendeléev periodic table?

5:02 AM

I wouldn't say it was either a merit or a demerit.

Achievement?

The problem is that in Mendeleev's time it was only possible to measure the atomic mass, i.e. the molar mass, but the chemical properties depend on the atomic number not the atomic mass.

And if you arrange by atomic number then the order is slightly different from arranging by atomic mass.

Mendeleev realised that arranging by mass gave the wrong result in a couple of cases and he changed the order to fit the chemical properties.

We now know he was right to do so :-)

Ohk
Thanks :-)

RishiNandha Vanchi

@JohnRennie thank you sir

@JohnRennie yes sir, thats what I thought it might be, because I couldnt find better proofs than the plausibility argument when I searched for it

@RishiNandhaVanchi the trouble with electrostatics is that it can get very complicated very quickly. Often you have a simple approximation that works pretty well, but when you try to do better the equations quickly get horrible :-)

RishiNandha Vanchi

5:16 AM

XD

3 hours later…

7:55 AM

Hello @JohnRennie sir

@PrateekMourya hi :-)

Sir are you free?

Yes

user image

This question

@JohnRennie sir

Although it isn't obvious from the diagram, the geometry here is like a pair of discs with holes in the centre. So fluid is pumped in at the centre and it flows radially outwards towards the edges of the disks.

I can try and draw a diagram if it will help.

OK so far?

8:03 AM

No i understand

The mechanism

But don't know how to start

You need to find the flow velocity as a function of r because then you can use Bernoulli's equation to calculate the pressure. Yes?

Yes

But pressure is not given at any other point

I think you'll have to assume a pressure P₀ at the centre then give the pressure as a function of r and P₀.

Answer is Q/2πrl

That's the flow velocity ...

Consider the surface a distance r from the centre. This surface is a cylinder of radius r and height h, where h is the spacing between the plates.

8:08 AM

Ok

So the surface area of this surface is A = 2πr h. Yes?

Yes

Then if the flow velocity at the surface is v the volume flow rate is Q = Av i.e. Q = 2πr h v

Then rearranging gives the flow velocity as v = Q/2πrh

But what about pressure

Question is incomplete

Alot

I'm wondering if it's a misprint. The diagram is labelled v(r) not P(r) so maybe it actually means the velocity variation as a function of distance.

8:12 AM

Ok

I can't see how you calculate the pressure without more information.

Ok

user image

1.424

The flow rate through the holes is given by Torricelli's law. Yes?

Yes

Sir i think i should leave this question

Its more of maths

But are these relevant to jee?

That one about the tank might appear in the advanced, though it's probably too lengthy even for the advanced exam.

It certainly wouldn't appear in the mains exam.

8:25 AM

Yes

Approach might be

To first use net force from

Both the holes

Till the level of water becomes low so that water stops from upper hole

And then force from the bottom one

Or is there any other short?

You need to do the calculation in two stages:
1. down to the first hole, then after this there is no flow through the first hole
2. down to the bottom.

And for each stage you need to calculate the total acceleration as a function of time and then integrate it to get the final velocity.

In principle this is straightforward, but it's a messy calculation. I can't think of an easy way to do it.

Ok

Thanks sir

@JohnRennie Hii sir. Free

@SrijanM.T hi, yes I'm free :-)

Great

[![enter image description here][1]][1]

[1]: https://i.stack.imgur.com/aBOhP.jpg

This is the image my textbook uses for finding whether the ratios are +ve or -ve.

Here , i feel they mean to say that tan (+x) = tan x since b and a are +ve . We can tell it by looking at the x and y axis But I want to know if they talk about sin(+x) , how do we know if OP is +ve or not.

Can we also say that x = 1 since it is a radian measure of 1 unit.

This is Q

8:35 AM

I'm not sure what you are asking ...

Ok. We say that OM AND OP are +ve?

Since +ve y and + x axis

Ok ?

That’s why P is a,b

Where is the point M?

user image

I'd guess OP is just treated as the magnitude of the distance from the centre, so it's a positive number.

Ok. But then why is OQ -ve

8:38 AM

OP = √ ( a² + b² )

Hmm k

What about in case of sin(-x)

Is OQ negative? I would also treat it as a positive number.

@JohnRennie Ohk

I got it.

:-)

Thanks

@JohnRennie can we say x = 1 unit

Since it looks like a radian measure

8:43 AM

It's a circle of radius 1 so -1 >= x <= +1

Why not -1=x=1 since whatever is x here , that is radius of circle in terms of radian.

x=-1 only in case of anti-clockwise rotation of x right

Wait, I assumed x meant the point where the vertical line intersects the x axis.

i.e. the equation of the circle is x² + y² = 1

No, the x measure I am talking about

Then isn't that just the radius if the circle i.e. +1 ?

user image

@JohnRennie yes

The x I marked here in circle

user image

Sth like this

8:50 AM

Ah, OK, sorry, I misread the diagram.

OK, so x is the length of the subtended arc.

Yes sir

Again I'd say it doesn't have a sign because it's just a length.

Yes.but it’s mag = 1 right ?

In my book , it’s only x everywhere

The value of x is given by x = rθ where θ is the angle subtended by the arc.

In this case the radius is r = 1, so we get just x = θ.

Ohk.Yes

Thanks sir

8:53 AM

And in the second diagram it looks as if θ = 1 radian, so in that case x = 1.

Yes

Thanks a lot sir

1 hour later…

10:13 AM

@JohnRennie hi

@AshishAhuja hi :-)

let's say we have a non-conducting sphere

and a charge is uniformly distributed across the sphere, and there is a cavity in it not containing the center of the sphere

from the superposition principle, we can find the electric field at any point in the cavity, and this comes out to be a constant

OK, yes I think I remember doing this problem and the field turned out to be constant inside the cavity.

but how can we actually prove that the electric field inside is a constant? I can calculate it at any point and find it to be the same but that's hardly a proof.

@JohnRennie yes, it comes out to be $$\frac{\rho}{3 \epsilon}$$

When I did it I put the origin at the centre of the charged sphere, then the position anywhere inside the sphere can be specified by the coordinates (x, y, z).

10:20 AM

yes that's what I did too

As I recall the problem used a cross section through the void, so in fact there were only two coordinates x and y.

So all you have to do is calculate the field as a function of the position (x,y) inside the hole.

And it comes out to be independent of and y.

And since it's independent of position (x,y) that means it is constant everywhere inside the void.

yeah got it. I did it in a different way which did not lead to the result that it would be same everywhere, it just gave me the answer at a point. I'll try out what you said

if it doesn't work I'll let you know, thanks.

hi

theres a neat method for this

consider any point (x,y) inside the cavity.

next , consider three vectors:

1)joining the cavity centre to the point

2)joining the sphere center to the point

3)joing the sphere centre to the cavity centre

(a,b,c) respectively.

ok

the field at the point is $$\rho (\vec{a})/3 \epsilon + (-\rho)(\vec{b})/3 \epsilon$$

=$$ \rho / 3\epsilon (\vec{c})$$

which is constant.

10:29 AM

ahh nice, I got it.

That is more or less what I did, but I wasn't using such clear notation and I was deriving the rho/3e result as well so I guess I messed it up.

thanks.

np

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