Particle Accelerator

John Rennie

04:33

← 92 messages moved from Problem Solving Strategies

Guru Vishnu

09:00

@JohnRennie: Hi sir. Good morning :-)

Thank you for transferring 92 messages sir.

John Rennie

:-)

Guru Vishnu

That must have been a lot of work, I guess. Thank you for doing it despite your busy schedule.

May I ask one last transfer favour @JohnRennie sir? :-)

John Rennie

@GuruVishnu yes ... ?

Guru Vishnu

Can you please move this and the next message? Then the conversation will become complete. Thank you very much sir.

in Problem Solving Strategies, yesterday, by Guru Vishnu

Sorry for this. From next time onwards, I'll post it here to prevent such huge transfers.

John Rennie

← 4 messages moved from Problem Solving Strategies

Done! :-)

Guru Vishnu

09:07

Thank you sir :-) I'm sorry for this. I'll not do this again. Are you free now for a small doubt?

John Rennie

Yes, what's the question?

Guru Vishnu

For the force $F=2x^2-3x-2$, I was asked to choose one of the following options:

(a) x=-1/2 is position of stable equilibrium

(b) x=2 is the position of stable equilibrium

(c) x=-1/2 is position of unstable equilibrium

(d) x=2 is the position of neutral equilibrium

My approach:

In the potential energy versus displacement graph, the points of stable equilibrium are minima and unstable are the maxima. If we apply this here, we get two options as correct ones (b) and (c). But I'm allowed only to choose one.

I first found the critical points of $F$. The roots are $x=2$ and $x=-1/2$

This is the basis of my understanding sir - physics.stackexchange.com/a/547078/238167

John Rennie

If you take the potential $V(x)$ then the equilibria are at $dV/dx = 0$ i.e. at $F = 0$, and you've found those positions.

Guru Vishnu

Yes sir.

John Rennie

Then the stable ones are the minima i.e. $d^2V/dx^2 = dF/dx > 0$ and the unstable ones are when $d^2V/dx^2 = dF/dx < 0$. Yes?

Guru Vishnu

09:14

Yes sir.

John Rennie

$dF/dx = 4x - 3$ so at $x = 2$ this is positive and the equilbrium is stable.

And likewise at $x = -1/2$ the equilibrium is unstable.

Hmm, I agree with you.

I'm just going to draw the graph ...

Guru Vishnu

Ok sir.

John Rennie

You are definitely correct.

Guru Vishnu

Ok sir. Thank you for the clarification. Do you wish to see the author's solution?

John Rennie

There must be a misprint in the question.

Yes

Guru Vishnu

09:25

Ok sir.

$F=0$ at $x=-1/2$ and $x=2$. Therefore these are the positions of equilibrium.

Now, $dF=(4x-3)dx$

At $x=-1/2$, $dF=-5dx$ or $dx\propto-dx$

i.e., $x=-1/2$ is stable equilibrium position.

At $x=2$, $dF=5dx$ or $dF\propto dx$

i.e., $x=2$ is unstable equilibrium position.

And thus he chose (a).

John Rennie

Aargh! $F = -dV/dx$ not $+dV/dx$.

My potential graph is inverted.

Guru Vishnu

Ah. Fine. Now understood the error. I thought it was a misprint. Now realised it's my error.

Thank you sir :-)

(removed)

Guru Vishnu

15:32

@JohnRennie: Hi sir :-)
If you are free after Yuvraj's doubt, can you please ping me?
I hope that you have finished your lunch.

John Rennie

@GuruVishnu I'm free now. What do you want to ask?

Guru Vishnu

Ok sir. Please wait sir. Let me draw the necessary diagrams.

Question:

Three identical, parallel conducting plates A, B and C are placed as shown. Switches $S_1$ and $S_2$ are open, and can connect A and C to earth when closed. +Q charge is given to B, then:

(a) if $S_1$ is closed with $S_2$ open, a charge of amount Q will pass through $S_1$
(b) if $S_2$ is closed with $S_1$ open, a charge of amount Q will pass through $S_2$
(c) if $S_1$ and $S_2$ are closed together, a charge of amount Q/3 will pass through $S_1$ and a charge of amount $2Q/3$ will pass through $S_2$.
(d) All of the above.

My first step:

@JohnRennie: Could you please tell how to proceed from this? I understand earthing a conductor makes its potential equal to that of earth close to 0 volts. But I think we need to determine the potentials of the individual plates for which I don't know any method.

John Rennie

15:49

Your initial diagram is wrong.

Guru Vishnu

Ok sir. I came to such charge distribution by equating the electric field inside the plate B to be zero. May I know why is it incorrect?

John Rennie

The charge on B has to add to $+Q$ and the way you have drawn it the charges add up to zero i.e. you've drawn $+Q/2$ on one face and $-Q/2$ on the other.

Guru Vishnu

Ah! Fine. That was an error while drawing that on my computer. It must be Q/2 on either sides.

John Rennie

Yes.

that obeys the two rules:
- equal and opposite charges on opposing faces
- equal charges on the two outer faces

Guru Vishnu

Yes sir.

John Rennie

15:53

The third rule is that when any plate is earthed the charges on the outer faces are zero.

Guru Vishnu

Ok sir. I remember you mentioned about the third rule yesterday. Is it possible to tell why is that so? Or is that a pure experimental result?

John Rennie

@GuruVishnu if I'm honest I don't know why it is the case. I even asked about it on the main site.

5

Q: Why are the two outer charge densities on a system of parallel charged plates identical?

One of the ways examiners torture students is by asking them to calculate charge distributions and potentials for systems of charged parallel plates like this: the ellipsis is meant to indicate any number of additional plates could be inserted where I've placed the ellipsis. The plates are ass...

So shall we work out the charges when S1 is closed?

Guru Vishnu

Thank you for sharing that question. I think I can read it when you're not available. It looks like a big post.

@JohnRennie Yes sir.

John Rennie

OK if S2 remans open the charge on plate C cannot change i.e. it must be zero.

Guru Vishnu

Yes sir.

John Rennie

15:58

And we know the charge on the outer surface has to be zero because we have one plate (A) earthed. That means the inner face of C must also have zero charge.

And that means the charge on the right face of plate B must also be zero. OK so far?

Guru Vishnu

Ok sir.

John Rennie

Guru Vishnu

Ok sir. So far I understood.

John Rennie

The total charge on B is $+Q$ so the charge on the left face of B must be $+Q$ and therefore the charge on the right face of A must be $-Q$.

Guru Vishnu

Oops. You were faster :-)

John Rennie

16:04

:-) But yes you're correct. So now you know the charge on A and therefore you know how much charge flowed through S1 when it was closed.

Guru Vishnu

Yes sir. It's Q. Or in other words, option (b) is justified.

John Rennie

$-Q$ ...

Guru Vishnu

Ok sir. $-Q$ from the ground to the plate is same as $+Q$ from the plate to the ground. I usually like to use positron picture rather than the electron picture as I like to be positive :-)

John Rennie

Yes, true, the question doesn't specify the direction the charge flows.

The trouble is that if we leave S1 open and close S2 we get exactly the same result so both (a) and (b) will be true.

Guru Vishnu

John Rennie

16:09

Yes :-)

Guru Vishnu

:-)

Ok sir. Now the problem is so easy. It seems the key fact is:

John Rennie

Actually (c) is going to be true as well isn't it.

Guru Vishnu

15 mins ago, by John Rennie

The third rule is that when any plate is earthed the charges on the outer faces are zero.

John Rennie

Yes

Guru Vishnu

John Rennie sir's third law analogous to Newton's third law :-)

John Rennie

16:10

Sadly it isn't my law :-)

I think it was a JEE student taught it to me.

Guru Vishnu

Fine sir :-)

John Rennie

Can you see why (c) is also true?

Guru Vishnu

@JohnRennie I think I need to verify that.

John Rennie

There's an extra factor to consider when both switches are closed. Can you see what it is?

Guru Vishnu

I think as per our previous method, the outer surface of plates A and C must be zero. Does it count as an extra factor sir?

Or do you refer to the differential capacitance?

John Rennie

16:13

Plates A and C must be at the same potential because they are both connected to earth. Yes?

Guru Vishnu

Yes sir.

John Rennie

And the two faces of plate B must be at the same potential because B is a conductor.

Guru Vishnu

@JohnRennie Fine sir. I get it. Are we going to assume the capacitors AB and BC are in parallel?

And we could make use of this $Q/C=V=$constant

John Rennie

The potential change between two surfaces is $V = Ex$ where $x$ is the distance between the surfaces and $E$ is the field between them.

And the potential changes shown by the two red arrows have to be the same.

Guru Vishnu

Yes sir. I understood that after seeing your diagram (it was so clear even without explanation) :-)

John Rennie

16:18

:-)

And now I'm going to call it a day.

I'll be around tomorrow as usual.

Guru Vishnu

Ok sir. Good bye :-)
Even though I completely got the answer, now, I wish to discuss the other method with you tomorrow.

John Rennie

OK, see you tomorrow.

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♦ Particle Accelerator