Problem Solving Strategies

Ashish Ahuja

5:26 AM

@JohnRennie hi

John Rennie

@AshishAhuja hi :-)

Ashish Ahuja

I was studying electric flux, and for a sphere

we can easily show that the flux is $\frac{Q}{\epsilon}$

We can extend this to all closed surfaces

and an intuitive way to think about this would be to think of electric flux sort of like air, and since this is a closed surface, the amount of total air exiting would be constant. However, would it possible for you to prove it a bit more formally, or give me a hint?

John Rennie

@AshishAhuja Formally this is the divergence theorem:

Divergence theorem

In vector calculus, the divergence theorem, also known as Gauss's theorem or Ostrogradsky's theorem, is a theorem which relates the flux of a vector field through a closed surface to the divergence of the field in the volume enclosed. More precisely, the divergence theorem states that the surface integral of a vector field over a closed surface, which is called the flux through the surface, is equal to the volume integral of the divergence over the region inside the surface. Intuitively, it states that the sum of all sources of the field in a region (with sinks regarded as negative sources) gives...

But there is a simpler way of looking at it.

Ashish Ahuja

yeah that theorem looks extremely complex

John Rennie

In electrostatics every field line has to begin and end on a charge

Ashish Ahuja

5:37 AM

yes

John Rennie

So if there are no charges inside a surface every field line that enters the surface has to exit the surface again because it cannot end inside the surface.

Ashish Ahuja

yup

John Rennie

So loosely speaking the amount of field entering the surface has to equal the amount of field exiting the surface and the integral over the whole surface must be zero.

Ashish Ahuja

hmm but the electric flux is not zero for a closed surface is it? I'm talking about the situation where a charge $Q$ is placed inside the surface, sorry if I wasn't clear before

John Rennie

Conversely, if there is a charge inside the surface then field lines begin at the charge, but since they cannot end on the same charge they must exit the surface. Hence there must be a net outwards (or inwards) flow through the surface.

Ashish Ahuja

5:40 AM

Ohhh neat

John Rennie

Field lines are a bit vaguely defined, but you can consider each field line associated with a constant amount of field i.e. we have N field lines per coulomb of charge.

Ashish Ahuja

Aren't there an infinite amount of field lines?

John Rennie

Then the amount of field lines exiting (or entering) the surface will be proportional to te charge inside the surface.

@AshishAhuja A field line isn't actually a physical object. It's actually just a direction. It's the direction of the force the field produces on an infinitesimally small test charge.

Ashish Ahuja

@JohnRennie for a sphere we can prove that flux is $\frac{Q}{\epsilon}$ and here you essentially proved that it has to be constant for all closed surfaces, so doesn't that complete the proof?

John Rennie

But we tend to think of it as representing some small part of the field.

Ashish Ahuja

5:44 AM

@JohnRennie hmm, I know that but considering a finite amount of field lines has got me confused..

John Rennie

@AshishAhuja I'm not sure this counts as a "proof" :-)

It's more a argument about plasusibility.

The proof is the divergence theorem.

Ashish Ahuja

yeah true :D But it seems logically sound, so for general purposes I guess that's enough. The divergence theorem looked scary.

John Rennie

:-)

satan 29

6:03 AM

@AshishAhuja once you understand what divergence means physically, youll realise that the theorem is actually pretty intuitive

Ashish Ahuja

ahh

satan 29

interestingly, a heuristic "intuition" for the divergence theorem is similar to your earlier argument. its tough to prove the theorem rigorously though

Ashish Ahuja

i.e treating electric flux similar to air flowing out?

satan 29

just outflow of any quantity

like water for eg.

Ashish Ahuja

got it.

satan 29

6:08 AM

@AshishAhuja how I understood it was

consider a charge inside any arbitrary surface

draw a small sphere of radius delta with delta--->0

Ashish Ahuja

ok

satan 29

then the flux through the sphere= q/epsilon

now if you interpret the flux as the measure of the no. of field lines, then

Ashish Ahuja

ohh got it

satan 29

the flux through the sphere=flux through the surface obviously, since the all the field lines piercing the sphere pierce the surface too

Ashish Ahuja

yup

satan 29

6:11 AM

and voila

Ashish Ahuja

this is pretty neat too

thanks

satan 29

very similar to what you did

Ashish Ahuja

Yeah but mine lacked any formality at all. It isn't intuitively obvious why air = electric flux, so I like explanations such as yours and JR's.

satan 29

without vector calc, this is as far as you can go I suppose. A "proof" with all the rigour would require the understanding of the divergence , etc

Sarabsri mt

6:43 AM

Hi guys. I am back.

Ashish Ahuja

7:10 AM

If we have two point charges $Q_1$ and $Q_2$ with fields $\vec{E_1}$ and $\vec{E_2}$ inside a closed surface we have $$\phi = \oint{(\vec{E_1} + \vec{E_2}) \cdot \vec{dS}} = \oint{\vec{E_1} \cdot \vec{dS}} + \oint{\vec{E_2} \cdot \vec{dS}} = \frac{Q_1 + Q_2}{\epsilon}$$ Now we can extend this to as many charges as we want. Is this a valid proof of Gauss's Law?

@JohnRennie

John Rennie

I wouldn't say it was a proof of Gauss's law.

Ashish Ahuja

Why not? Assuming that we know that flux for a point charge is $\frac{Q}{\epsilon}$

John Rennie

@AshishAhuja right, but you need Gauss's law to tell you that the flux for a point charge is Q/ε

Ashish Ahuja

wait what

John Rennie

So all you're saying is that if Gauss's law applies for a single charge then it applies for any collection of charges, which is true.

Ashish Ahuja

7:14 AM

Ohh wait does divergence theorem require gauss's law?

John Rennie

The divergence theorem is Gauss's law.

Ashish Ahuja

But wiki has two separate pages for them?

John Rennie

The Wikipedia article for the divergence theorem starts: "In vector calculus, the divergence theorem, also known as Gauss's theorem or Ostrogradsky's theorem ..."

Ashish Ahuja

hmm okay, thanks.

John Rennie

Oh, OK, Gauss's law is a special case of Gauss's theorem/divergence theorem.

Though whether it really deserves its own page is debatable.

Buraian

10:04 AM

Greetings, could I get some help on this problem:

I have seen the solutions of this on the internet but I do not understand them

once we do FBD on the block we get are supposed to get , mg-N=ma, then set N=0 and say a=w^2 A where A is amplitude

I don't get the logic here

Ashish Ahuja

@Buraian Do you agree that the most probable point of losing contact is when the piston is at it's highest position?

Buraian

well I considered that but how would you proof that?

but I think there is an arguement for it

so the least frequency such that it loses contact msut be when it is shot at top

if it is more frequency than that, it loses contact before hitting the top

Ashish Ahuja

yeah that's correct

Buraian

right, now how do you do the force balance

Ashish Ahuja

Ok let's just talk about the highest position now

Buraian

10:11 AM

right

Ashish Ahuja

normal force = zero as you said

so acceleration of the washer = g

Buraian

Hmm right so here is where my doubt begins

what exactly does normal force mean here?

teh spring is supporting it, so it should be spring force, no ?

Ashish Ahuja

The washer rests on the piston, the normal force I'm talking about is between the washer and the piston

Buraian

but the piston is acting like a spring here, no?

John Rennie

@Buraian you've probably done problems with a man on some scales in a lift. Yes?

If the lift accelerates downwards the man's weight decreases

Buraian

10:13 AM

yea

John Rennie

Well the acceleration of the piston in this problem is given by a = -kx or some constant k. Yes?

Buraian

hmmm I'm not sure there, that'd assume you treat the lift as a spring

John Rennie

Sorry I meant piston not lift.

Buraian

yes yea

John Rennie

So the downward acceleration (i.e. most negative acceleration) occurs at the greatest value of x i.e. at the top of the motion.

So that's when the weight of the ~~man~~ washer is least.

i.e. that's when the the normal force between the bottom of the washer and the top of the piston is least.

Buraian

10:18 AM

normal force = spring acceleration?

Ping back when replyin

John Rennie

@Buraian Go back to the lift problem.

Buraian

yes

John Rennie

The man's weight is mg.

Buraian

yyes

John Rennie

If the acceleration of the lift downwards is a, then the weight of the man decreases to m(g - a). Yes?

Buraian

10:22 AM

yes due to the pseudo force

John Rennie

And that "weight" m(g-a) is the normal force between the bottom of the man's shoes and the lift. Yes?

Buraian

hmm yes

John Rennie

If the lift accelerated faster than g the man would actually take off and lose contact with the floor of the lift.

Well in this problem the piston is the lift and the washer is the man.

Buraian

Ohkay I think I get it now, we can think of the man in elevator as the washer and piston

John Rennie

@Buraian Yes. The washer will lose contact with the top of the piston if and only if the piston accelerates downwards faster than g.

Buraian

10:26 AM

so the difference between case is that, here the acceleration of the 'lift' is time varying, correct?

John Rennie

Yes.

Buraian

like a time varying pseudo force

John Rennie

Yes

Buraian

I have one question about pseudo force and normal, for accelerating frames, to find normal is it neccesary to use pseudo force?

John Rennie

If the lift position is x(t) = A sin(ωt) then a(t) = -Aω² sin(ωt)

Buraian

10:28 AM

for example, in this case unelss you put in pseudo force it's impossible to get correct acceleration. So, is it impossible to get the correct normal by some inertial frame outside?

John Rennie

No, that's wrong.

Consider the situation as seen from outside the lift.

Buraian

yes

John Rennie

We have a man with a gravitation force of Fg = -mg (let's take upwards to be the positive direction).

Buraian

yep

John Rennie

We observe the man to have the same acceleration a as the lift, so we know the net force on the man must be given by Fnet = ma. Yes?

Buraian

10:32 AM

If I were not to observe physically would it be possible to conclude that?

John Rennie

I'm not sure what that means. The man's acceleration is a regardless of whether anyone is watching him or not.

That is, if he holds an accelerometer it will have a reading of a.

Buraian

precisely what reason do you give for justifying that both lift and him must have some same acceleration? It is intuitive for me but I'm trying to think of a justification for it

but then he'd have to use some external experiment to know that (if he uses accelerometer)

John Rennie

Because the man and the lift are in contact so their positions are the same i.e. x_lift(t) = x_man(t)

If we differentiate twice we get a_lift(t) = a_man(t)

user586228

0

Q: Please explain me the last term in the given statement

Having doubt in understanding the last term in the expression.Please givea detailed explanation to the calculation methodology.

Buraian

Previously you had described case where the lift moves in such a way that man floats

user586228

10:35 AM

@JohnRennie Sir please help me with this

Buraian

then we no longer have a contact between man and lift

John Rennie

Yes, if the man loses contact with the lift then the only force acing on him is the constant Fg = mg.

Buraian

like if lift was moving downwards with greater acceleration than g

John Rennie

So now he accelerates at a constant g.

Buraian

@JohnRennie so if man loses contact with lift, we can not say that net force sum on man= ma? where a is acceleration of lift itself

@JohnRennie based on this and the current point

John Rennie

10:38 AM

correct. The acceleration of the man is:
1. a if a <= g
2. g if a > g

Buraian

oh damn

that is actually a very interesting point

user586228

@Buraian You may also try the uestion I posted..

question*

John Rennie

@user586228 if you have n electrons then they all repel each other. Yes?

user586228

ok

Yes

Buraian

Thank you sir for the help, it's a bit difficult for me to fully understand the implications but I have understood the idea required for the question @JohnRennie

John Rennie

10:42 AM

So if the distance between electron number i and electron number j is r_ij, then the potential energy due to that repulsion is U_ij = +ke²/r_ij. Yes?

That's just the usual equation for potential energy of two charges.

@Buraian OK :-)

user586228

Correct..

John Rennie

So to get the total potential energy due to the electron repulsion you sum U_ij i.e. you do Σ_i Σ_j ke²/r_ij

Only you need to be careful not to double count i.e. don't sum both U_ij and U_ji

So the sum is only over j from zero to i.

user586228

I couldn't get the last explanation..

Please help me out

John Rennie

Suppose we have two charges, so both i and j can be 1 or 2

user586228

ok

John Rennie

10:47 AM

So Σ_i Σ_j ke²/r_ij would give us:

ke²/r_00 + ke²/r_01 + ke²/r_10 + ke²/r_11

We obviously exclude i = j because a charge doesn't have potential energy due to itself, so we are left with:

ke²/r_01 + ke²/r_10

OK so far?

user586228

ok then..

John Rennie

But the PE between two charges is just ke²/r

And in our sum we double counted it.

Because we counted both i=1 and j=0 as well as i=0 and j=1.

So we eliminate the duplication by making the second sum only over j < i

user586228

Could not get this

now I got it

Buraian

Sir if you have some time, could you tell me what you think the answer for this question I posted is?

user586228

11:03 AM

Ok then how is the second term different from the first term please explain this one

John Rennie

The second term is the same as the first term because r_01 = r_10. Both are just the distance between the two electrons.

user586228

Not that one ..the second term in my expression...That is what I am asking..

John Rennie

There are two sources of potential energy:
1. the electrons are attracted to the nucleus and this gives a negative PE
2. the electrons are repelled by each other and that gives a positive PE

The first term is the electron-nucleus PE and the second term is the electron-electron PE.

user586228

Thanks a lot for helping..

John Rennie

:-)

user586228

11:09 AM

And what about the first term in the expression?

John Rennie

The first term is the kinetic energy of the electrons.

user586228

Ok can you justify slightly..

What the terms signify....???

John Rennie

You mean what $\nabla^2$ means?

user586228

Laplacian operato but nothing more than that

operator*

I really want to know physically what it means..

John Rennie

I don't think there is a simple physical interpretation ...

Sarabsri mt

11:20 AM

Just a Q , is quantum realm or time travel in real life possible ?.

John Rennie

Quantum mechanics is real but time travel is impossible.

Sarabsri mt

Ohk.

user586228

What is the origin of kinetic energy of electrons with reference to the laplacian operator.

Ashish Ahuja

@JohnRennie while deriving the electric field at a point a distance $d$ away from an infinitely large thin uniformly charged electric plate, I used a closed cylindrical surface with one end on the plate and the other at a distance $d$ from the plate. I was able to do it, but first I had to show that the electric flux for the curved surface is zero, and I used an informal (yet logically sound) argument to do that. Could you hint me towards a formal way to show that electric flux for the...

curved surface = zero?

John Rennie

@AshishAhuja use symmetry.

The plate is infinite, so is move parallel to the plate in any direction the field cannot change.

Ashish Ahuja

11:27 AM

Is that really a formal way? I was looking for something rigorous if possible.

user586228

@JohnRennie Sir please justify the kinetic energy for electrons..

John Rennie

@AshishAhuja that is rigorous.

Ashish Ahuja

@JohnRennie hmm okay then, thank you.

Sarabsri mt

Guys . I am going to start making notes for physics and chemistry on IPAD using Apple pen. It saves a lot of time. But it will hurt eyes more. Do you think I should do it ?

My only fear is that people say kids who wrote on notebooks are better in remembrance than kids on digital media.

Sorry to disturb . Please answer whenever anyone’s free and can help me.

I can arrange things better and write things that I may have left on the same using iPad.

Ashish Ahuja

Maybe just try it out and see if it works for you? I personally don't even make extensive notes, maybe I will have to for chem in 12th, and I seem to remember stuff just fine, so I don't think a general one-fits-all approach exists.

Sarabsri mt

11:36 AM

Ohk.

Thanks @AshishAhuja