Problem Solving Strategies

pi-π

05:47

@JohnRennie Hello

John Rennie

@pi-π hi :-)

pi-π

@JohnRennie 6c?

In C language

John Rennie

That looks straightforward. Where are you running into problems with it?

pi-π

@JohnRennie Adjusting the signs before each term

John Rennie

Can you post the code you've tried so far?

Guru Vishnu

05:53

@JohnRennie: Hi sir. Good morning :-)

John Rennie

@GuruVishnu hi :-)

Guru Vishnu

Are you replying to pi-π @JohnRennie sir?

John Rennie

@GuruVishnu yes.

@pi-π did you try writing some code to do this?

Guru Vishnu

@JohnRennie Ok sir. I'll start after that.

pi-π

@JohnRennie Yes... I could adjust the sign by assigning sign=-1 and sign=sign*(-1).

i was missing that *(-1) part

John Rennie

05:57

@pi-π that's the way I would do it.

pi-π

@JohnRennie okay

John Rennie

Would you like to have a go, and in the mean time I'll answer @GuruVishnu's question. I'll be here for several hours if you run into any problems.

@GuruVishnu what did you want to ask?

pi-π

@JohnRennie Okay. You may continue with @GuruVishnu

user8718165

@JohnRennie Good Morning sir 😊

John Rennie

@user8718165 hi :-)

Guru Vishnu

06:04

@pi-π, @JohnRennie: Thank you :-)

@JohnRennie: I hope the following is a good candidate:

For a real erect image, but it follows after an inversion.

John Rennie

That's two lenses with an intermediate image

Guru Vishnu

@user8718165 Hi. Where did you get that "smiley" from? Did you use any HTML code?

John Rennie

It's two planoconvex lenses with a block of glass between them.

@GuruVishnu unicode

Guru Vishnu

@JohnRennie Yes sir. And I've requested for a direct real and erect image.

John Rennie

Yes, but you specified only one lens

Guru Vishnu

06:08

@JohnRennie But that's still a single device. That doesn't matter. But I'm looking for a direct process.

John Rennie

It's not really a single device.

user8718165

@GuruVishnu I did it with my phone.

Guru Vishnu

@JohnRennie Could you please explain that sir? I interpret it as two refractions = a single lens like something and so a single device.

John Rennie

Let me draw a diagram ...

Guru Vishnu

@JohnRennie Thank you sir.

@user8718165 Ok! Now understood.

John Rennie

06:14

@GuruVishnu If I draw this then it's two lenses producing the erect image via an intermediate inverted image. Yes?

Guru Vishnu

@JohnRennie Yes sir. Even if we have glass between the lenses, we could clearly interpret the inversion as I explained in the reply to the answerer:

Thank you for your answer. I'm glad you understood my definition of a "single optical device". I think the ray diagram in your answer produces real and erect image through an intermediate inversion process (at the middle of the device where blue is below green which is below red). Even though the diagram clearly answers the question, I'm curious to know is it possible to produce a real and erect image as a direct process without any intermediate inversions? — Guru Vishnu 45 mins ago

John Rennie

Ah, OK, then we're saying the same thing.

The point is that it works vi an intermediate image not by directly creating an erect image.

Guru Vishnu

@JohnRennie Yes sir. And the other answer uses something little bit interesting. A material of non uniform index of refraction also using an intermediate inversion:

John Rennie

OK, though neither answer addresses what you are really asking.

Guru Vishnu

@JohnRennie Yes sir :-|

But I'm learning something interesting and new from each answer.

John Rennie

06:24

@GuruVishnu which is good :-)

Guru Vishnu

@JohnRennie: :-)

@JohnRennie: Are you free now sir?

John Rennie

Yes.

I'm doing bits and pieces of work so I may occasionally be a bit slow answering

Guru Vishnu

@JohnRennie Ok sir. No problem :-)

@JohnRennie: I'm asked to find the effective capacitance of the following capacitor, and I found it using regular integration. I'm looking for an alternate solution. Is there any such method sir?

I thought of combining another such capacitor but in an inverted position which will be in series with this one. But I got an incorrect answer.

John Rennie

No, integration is the correct method. It can be subdivided into an infinite number of infinitesimal capacitors in parallel. You integrate to sum up all those infinitesimal capacitors.

Guru Vishnu

@JohnRennie: I did something like the following where I swapped the dielectrics in two different capacitors to give an uniform dielectric. Could you please tell why this method fails?

I just swapped two wedges so that capacitors become uniform.

John Rennie

06:43

Suppose K1 = K2. Your method then gives an answer a factor of two too small.

Guru Vishnu

@JohnRennie Sure. I'll account for it later. The capacitance of the second setup is C/2 means capacitance of the original capacitor is C.

@JohnRennie Fun note: If K1=K2 then I wouldn't have asked this question to you ;-)

John Rennie

@GuruVishnu yes, but when you're attempting to understand a problem always take the simplest options to see what happens. So in this case taking K1=K2 immediately shows your method gives the wrong answer.

This is a very useful general approach to simplifying a complicated problem.

Guru Vishnu

@JohnRennie No sir. I'll just multiply the final result with two after all processes:

2 mins ago, by Guru Vishnu

@JohnRennie Sure. I'll account for it later. The capacitance of the second setup is C/2 means capacitance of the original capacitor is C.

John Rennie

It's a bold assumption that you can just multiply by two even when K1 ne K2 ...

Guru Vishnu

Fine sir. I just wanted to know whether swapping of diagonal dielectrics is allowed? I see the answer is "no" as the final result obtained through this is incorrect. I do wish to know "why"?

Did I explain clearly what I meant by swapping of dielectrics, or may I draw another diagram?

John Rennie

06:56

No, your approach is clear. But the reason your approach doesn't work is because it's nohing like the original circuit. You replace the original circuit with one that's completely differet and wonder why they don't give the same answers.

Guru Vishnu

@JohnRennie And they give drastically different answers. The correct one has logarithm term whereas the incorrect one doesn't.

@JohnRennie Further. I don't think it should make a difference if we change the order of the dielectrics. Probably the diagonal shape is causing the trouble. Again I'm looking for PEV :-)

@JohnRennie: Shall we continue once you're completely free sir?

John Rennie

I'm working now for a while ...

Guru Vishnu

@JohnRennie Can you ping me when you're free sir?

I'll probably be in the main site.

John Rennie

@GuruVishnu OK, though it will be a while ...

John Rennie

09:24

@GuruVishnu hi

Johan Liebert

09:34

4

Q: How does a car gain kinetic energy?

I understand that the engine delivers power to the wheels and friction from the ground causes the wheels to roll. However, given the power (work per time) at the wheels, how does that energy become the kinetic energy of the car, since friction force from road doesn't do any work? Is it simply be...

@JohnRennie Good morning sir! The OP of the given question says that work done is evaluated by the motion of point of contact. Since point of contact is at rest therefore no work is done. Is that reasoning true?

John Rennie

Caculating work done depends on what frame you use. For example in the rest frame of the car the road/tyre contact point is moving.

Johan Liebert

@JohnRennie if the frame of reference is ground then?

John Rennie

In the case of the car the obvious place to calculate the work is at the engine. The rotational equivalent of the work equation is work = torque times angle. You should be able to convince yourself that this is the same as the usual force times distance.

If you're working in the reference frame of the ground then calculate torques about the contact point and you'll find they are non-zero. Then the power is the torque times the angular velocity.

Johan Liebert

11

A: How does a car gain kinetic energy?

There have been several answers given that address the main point that friction serves to convert the energy provided by the engine into kinetic energy of the car, but none seem to address the mechanism behind this transfer of energy. The only force accelerating the car along the road is static f...

But the op is saying that such methods are wrong ( at least that's what he says in comment).

Here

@JohnRennie sir are you there?

John Rennie

09:56

I'm afraid my interest in the question is limited.

Johan Liebert

@JohnRennie ok sir!

1

Q: How can energy have inertia?

How can energy have inertia? To my intuition, inertia is so closely associated with mass that my intuition says "Huh?" Indirectly by mass energy equivalence it works fine, for example: I have a closed system, and add energy. Now it has more mass according to $E=mc^2$, and the inertia associa...

special-relativity energy momentum mass-energy inertia

Guru Vishnu

@JohnRennie Hi sir :-)

John Rennie

@GuruVishnu hi

Johan Liebert

@JohnRennie sir have you seen this question? The answer there seems to use $E = mc^2$ for light but I have made this comment is it right?

John Rennie

@JohanLiebert you are correct

Guru Vishnu

09:59

@JohnRennie Shall we continue after Johan's doubt, sir?

Johan Liebert

@GuruVishnu you can continue now!

Guru Vishnu

@JohanLiebert Thank you :-) I hope I didn't cause a conversation fission.

John Rennie

@GuruVishnu let me upload a diagram I drew:

The actual capacitor is shown at the top left.

Guru Vishnu

@JohnRennie Yes sir. (BTW the diagram is so excellent compared to what I did in paint!)

John Rennie

We do the calculation by considering an array of capacitiv elements in parallel and integrate them. So what I'm showing on the right is what if we approximate the real capacitor by two capacitors in parallel. Then we get the circuit on the right.

@GuruVishnu I use Google Draw. It's really good for this sort of diagram.

Anyhow we get the approximate circuit at the top right.

Guru Vishnu

10:05

@JohnRennie Yes sir. You said this to me previously. I too tried to use that. But it asked to draw a ton of useless stuff and I got bored. So I'm just using paint for faster chat and power point for slower questions.

@JohnRennie Yes sir. Understood.

John Rennie

Then in the bottom half of the diagram I'm showing your suggested approach and the equivalent circuit.

My point is that the two circuits are completely different. There's no reason they should give related answers.

Guru Vishnu

@JohnRennie Fine sir. So we aren't allowed to swap the dielectric pieces in that way? I'm 100% sure the bottom method is incorrect but could you tell the reason behind that?

Sometime ago I tried to calculate the capacitance of the dielectric piece but then realised it's capacitance is undefined. (May be this could help.)

My claim is: We're just swapping the dielectric pieces in the two setup, then why is there a difference (a very huge difference) in the final result?

Did I explain it properly sir?

Do you wish to have a look at the final results, sir?

John Rennie

Guru Vishnu

@JohnRennie Exactly! :-)

John Rennie

OK. Where it's going wrong is that in the right two diagrams the potential on the slanted plate is everywhere the same because it's a conducting plate.

On the left diagram the potential at the junction between the two dielectrics is not the same everywhere along the slanted line joining them.

Guru Vishnu

10:17

@JohnRennie Fine sir. Where did that conducting plate come from on the slanted plates?

John Rennie

By definition a capacitor is two conductng plates with some dielectric (or vacuum) between them.

Guru Vishnu

@JohnRennie Yes sir. But what if we consider the rightmost system as two separate capacitors of uniform dielectric medium?

I think in this case we will not be required to have the conducting plates.

John Rennie

If you do that then the potential everywhere on the slanting plate will be the same because that plate is a conductor.

But this isn't the case for the actual capacitor that is shown on the left. So the two cases are physically different.

Guru Vishnu

@JohnRennie If we're neglecting the conducting plates (in the second and last case), will the potential be different much like the first case sir?

John Rennie

@GuruVishnu take the top left pair, and assume there are no plates on the slanted surface.

Guru Vishnu

10:21

Ok sir.

John Rennie

If you do that you are joining surfaces that have different potentials e.g. the potential at the left side of the bottom bit is fifferent to the potential at the left side of the top bit.

Guru Vishnu

@JohnRennie Yes sir.

John Rennie

I can show what I mean in a diagram if you want.

Guru Vishnu

@JohnRennie Before that shall we consider this fact:

The capacitance of the individual wedges (if they are covered by conducting plates) is not defined. In the lower limit, I'll have to solve $\ln 0$.

$C=a^2\epsilon_0/d \ \ (\ln a-\ln 0)$ is the final expression.

Which is in fact not defined sir.

This is because of the fact we're considering two equipotential conducting surfaces to intersect at the edge tip.

And this explains why we must not be having conducting plates on the slanted surfaces.

@JohnRennie: Ok sir?

John Rennie

Guru Vishnu

10:28

@JohnRennie Ok sir. But the presence of conducting plate (as depicted by a black line) in the above diagram is slightly giving me some trouble as described in my previous messages above this image.

John Rennie

When you rearrange and join as you described you're joining up points that have different potentials.

Guru Vishnu

@JohnRennie Ok sir. So if we aren't having the conducting plates and swapping the dielectrics we're joining non equipotential surfaces and this is what giving a wrong result. Am I right sir?

John Rennie

@GuruVishnu Yes

Guru Vishnu

@JohnRennie Do you agree capacitance of individual wedges is not defined?

5 mins ago, by Guru Vishnu

$C=a^2\epsilon_0/d \ \ (\ln a-\ln 0)$ is the final expression.

John Rennie

I'd have to do that calculation. I can't do it now because I'm answering a question in another room.

In any case it is unrelated to the original question.

Guru Vishnu

10:32

@JohnRennie No problem. Believe me sir. That expression is true to the best of my knowledge.

@JohnRennie I think it's related just because we introduced a conducting plate on the slanting surface.

Else as you said it's unrelated.

John Rennie

The conducting plate is a bit of a sidetrack. I introduced it because it's necessary if you're going to model the system as four capacitors in series, but you weren't doing that anyway.

Guru Vishnu

@JohnRennie Ah! Ok sir. Now got it. However, after another doubt could you ping me again. I've some more to discuss regarding this cap.?

John Rennie

Post the question now and I'll look at it as soon as I have time.

Guru Vishnu

Ok sir.

@JohnRennie: My answer is "no"

John Rennie

In the right two diagrams the potentials will be the same. The potential changes linearly from the top to bottom plate so the potential a distance $d$ above the bottom plate is just $V(d) = V d/D$, where $V$ is the potential of the top plate and $D$ is the separation. Yes?

Guru Vishnu

10:48

@JohnRennie Yes sir.

Is this what meant by "linear" dielectrics sir?

I think I saw this word some time ago in one of the answers on SE.

John Rennie

@GuruVishnu no. A linear dielectric is one in which the polarisation is proportional to the field strength. This proportionality means the dielectric constant is constant. For a non-linear dielectic the dielectric constant change with the applied field string i.e. the dielectric constant isn't actually a constant.

Guru Vishnu

@JohnRennie Thank you sir. I thought the words linear and non linear represented the variation of potential with distance.

John Rennie

Anyhow, on the left diagram the potential does not change linearly with distance because the dielectric constants are different. Within each dielectric the potential changes linearly with distance, but the rate of change with distance is different for the two dielectrics.

Guru Vishnu

@JohnRennie Ok sir. If the slope of variation of potential with distance is different for different dielectric media, why shouldn't the potentials at the second and third diagrams be different?

Contrary to the following message:

8 mins ago, by John Rennie

In the right two diagrams the potentials will be the same. The potential changes linearly from the top to bottom plate so the potential a distance $d$ above the bottom plate is just $V(d) = V d/D$, where $V$ is the potential of the top plate and $D$ is the separation. Yes?

John Rennie

We have the constraint that the potential difference between top and bottom plates is a constant $V$. So as we move upwards from the bottom plate to the top plate the potential must increase from zero to $V$. OK so far?

Guru Vishnu

10:57

@JohnRennie Ok sir.

John Rennie

If the dielectric constant is the same everywhere between the plates then the potential has to change at a uniform rate, so we get the equation $V(d) = V d/D$.

Guru Vishnu

@JohnRennie Fine sir. Got this point. But why is the same not applicable in the first case?

John Rennie

If we have two different dielectrics as in the left diagram then the potential still has to start at zero and end at $V$, but it can change at a different rate in the two dielectrics.

I can show this on a diagram if you want.

Guru Vishnu

@JohnRennie Why that's so? What is preventing the linear variation?

@JohnRennie If it's not a problem for you, then I'd wish to understand this concept :-)

John Rennie

@GuruVishnu let me draw a diagram

Guru Vishnu

11:01

@JohnRennie Thank you sir :-)

John Rennie

@GuruVishnu Suppose we have a capacitor with a spacing $D$ and area $A$ with a dielectric constant $k_a$, and we want to know the potential a distance $d$ in the dielectric above the bottom plate. OK so far?

Guru Vishnu

@JohnRennie Ok. And we do it by using $V'=Q/C_1$?

where $Q$ is the charge on the bottom capacitor.

John Rennie

Yes. With parallel plates the potential along a horizontal line is constant, so we can divide our capacitor into two capacitors along the horizontal line and consider applying the voltage $V$ across the two capacitors in series.

Guru Vishnu

Fine sir. I think I'm comfortable with the parallel plate case, shall we move to the slanted dielectric case?

John Rennie

No, wait. I'm not done.

Guru Vishnu

11:14

@JohnRennie Ok sir. Sorry for the interruption.

John Rennie

Are we agreed we are going to get $V' = Vd/D$ in this situation?

Guru Vishnu

@JohnRennie Yes sir. Agreed.

John Rennie

Now let me update my diagram:

Guru Vishnu

@JohnRennie I'd like to say, the potential within the two individual capacitors vary linearly but as you said previously the slope of variation is not same. This is because we've changed the capacitance of the individual capacitors and thus $V'$ in the second updated diagram is much different than that in the first original case of yours.

John Rennie

Correct!

Guru Vishnu

11:19

Is this the reason why the first case potential is different from the two others in the following sir:?

John Rennie

Yes.

Guru Vishnu

@JohnRennie Ok sir. Let's concentrate on a single dielectric piece. There are infinitely many regions of potentials $V_i$. Now if we swap the other dielectric piece will the $V_i$s change or remain the same?

John Rennie

We can imagine rejoining our two capacitors to get the above.

And now $V' \ne V d/D$

Guru Vishnu

@JohnRennie Ok sir.

So the potential varies something like as follows:

Am I right sir?

John Rennie

Yes, exactly! :-)

Guru Vishnu

11:26

@JohnRennie: Thank you for the compliment sir :-)

8 mins ago, by Guru Vishnu

@JohnRennie Ok sir. Let's concentrate on a single dielectric piece. There are infinitely many regions of potentials $V_i$. Now if we swap the other dielectric piece will the $V_i$s change or remain the same?

@JohnRennie: If possible, could you reply to the above message?

John Rennie

Busy for a moment ...

Guru Vishnu

Oops. Sorry sir.

John Rennie

11:43

I'm running out of time. We'll have to puck this up tomorrow.

Guru Vishnu

@JohnRennie Sure sir. No problem :-) Good bye sir.

skullpatrol

17:13

@JohanLiebert chat.stackexchange.com/transcript/71?m=53250215#53250215

Johan Liebert

@skullpatrol so you know him in real life?

skullpatrol

no, but he's been helping others a long time

all for free on this site

Johan Liebert

@JohnRennie sir I have posted this answer on that question so that our discussion doesn't get lost in this chat and become helpful to future visitors. Even though it's not the best then also I think it might suffice to divert the future visitors over here.

Transcript for

Problem Solving Strategies