Problem Solving Strategies

John Rennie

4:08 AM

@Abcd Morning :-)

Abcd

@JohnRennie Please see

Anonymous

m.hindustantimes.com/science/…

Anonymous

Coming up: Solar eclipse today, longest lunar eclipse in century and date with Mars later this month

Anonymous

But the solar eclipse will not be visible in India, lunar one will be.

John Rennie

@Abcd that's an Ampere's circuital law problem isn't it?

There's going to be a discontinuity at the cylinder because as you move inwards the current enclosed by your loop falls as you pass the cylinder.

So it's either B or C. I'd have to actually calculate the field to tell which of B or C it is.

Abcd

4:19 AM

@JohnRennie Not getting how to use it

John Rennie

Consider a circle of radius $r$ centred on the axis of the cylinder. The field is tangent to this citcle, so the integral of the field around the cylinder is just $2\pi r B$. OK so far?

Abcd

@JohnRennie How big will that circle be

John Rennie

The circle has a radius $r$, so the distance from the axis of the cylinder to the circle is $r$.

Abcd

@JohnRennie No I mean is it larger than cylindrical circle?

John Rennie

To answer the problem we need to consider $r$ both greater and less than the radius of the cylinder.

Abcd

4:24 AM

@JohnRennie Okay

John Rennie

Call the current in the wire and cylinder $I$ so the total current is $2I$. Outside the cylinder, i.e. $r > R$ we get $2\pi r B = 2\mu_0 I$

So $$B = \frac{\mu_0 I}{\pi r} $$

Abcd

@JohnRennie Yes

@JohnRennie Dont we have to care about the direction and stuff?

like we did in the case of rectangular sheets

direction of line integral

John Rennie

Once we get inside the cylinder, i.e. $r < R$, the current enclosed is just $I$ so $$ B = \frac{\mu_0 I}{2\pi r} $$

The nice thing about this approach is that the field B is always tangent to our circle.

Because the field lines around a straight conductor are circles.

That makes the integral trivially easy.

Abcd

@JohnRennie how?

John Rennie

The field lines around an axially symmetric conductor are circles ...

@Abcd for example see this

Abcd

4:30 AM

@JohnRennie why?

@JohnRennie not sure why it should be a circle

John Rennie

@Abcd the system is axially symmetric, so the field has to be axially symmetric too.

I guess you could use Biot-Savart to show that the field lines are circular, but it should be obvious.

Abcd

@JohnRennie Hmm, but why would they be circular because of the cylinder too?

@JohnRennie $$\int \vec{B}.\vec{dl}$$

Won't B be perpendicular to dl that way?

SO it should be 0?

John Rennie

@Abcd A cylinder is axially symmetric

@Abcd let me draw a diagram

Abcd

@JohnRennie Kay

John Rennie

Abcd

4:45 AM

@JohnRennie Oh Okay

John Rennie

The field lines are circular i.e. the direction of the field on the circle is always tangent to the circle.

Abcd

@JohnRennie Now what about the graph?

John Rennie

Well what is the field for $r$ infinitesimally greater than $R$?

@Abcd infinitesimally greater than $R$ not infinitesimally greater than 0.

21 mins ago, by John Rennie

So $$B = \frac{\mu_0 I}{\pi r} $$

Abcd

For r<R, $B = \dfrac k r$, for r>R, $B = \dfrac k{2r}$

@JohnRennie $\to \dfrac{k}{R}$

$k = \dfrac{\mu_o I}{\pi}$

John Rennie

Yes, and what is the field for $r$ infinitesimally less than $R$?

Abcd

4:50 AM

@JohnRennie $\to \dfrac{k}{2R}$

so its lesser

John Rennie

Right, so as we move from just outside of the cylinder to just inside the cylinder the field halves.

Abcd

@JohnRennie But how to decide between options B and C?

John Rennie

C shows the field falling to about a third of the value outside, so that's wrong.

Actually, wait a moment ...

B looks as if the field is going to infinity i.e. the line goes straight up. I wonder if that's bad drawing.

It's a pretty rubbish figure really, but C is definitely wrong because the field only halves at the cylinder.

Abcd

@JohnRennie won't field be $\to \infty$ at $r= R$

John Rennie

@Abcd No

Abcd

4:54 AM

@JohnRennie so what would it be at $r= R$?

Function is discontinuous at r=R , but it can have some value there

John Rennie

@Abcd yes. Discontinuous but finite.

Abcd

@JohnRennie One more question, do you have time?

John Rennie

The discontinuity is of course only there for the ideal case of an infinitely thin cylinder wall

Yes, I don't have to start work for an hour yet.

I'm just staring out of the window and drinking coffee :-)

Abcd

nice.

I just drink coffee to stay awake at night during exam times.

Because it does help stay awake

John Rennie

I drink it to wake me up in the morning :-)

Abcd

4:57 AM

But I dont like it

I just drink it forcefully

it gives me headache when I wake up

John Rennie

I used not to like coffee when I was younger, but I love it now.

I suspect it's effect is mostly psychological

Abcd

May 16 at 5:35, by Abcd

@JohnRennie And what about this one? A solid conducting sphere of radius 10 cm is enclosed by a thin metallic shell of rafius 20 cm. A charge q= 20 $\mu C$ is given to the inner sphere.*Find the heat generated in the process*. The inner sphere is connected to the outer sphere by a conducting wire

@JohnRennie this one^

@JohnRennie When they are not connected what will be the situation like?

Shouldn't field inside the solid sphere be 0?

John Rennie

It's a spherical capacitor

The field inside the solid sphere is indeed zero. You need only consider the surface of the sphere.

Abcd

@JohnRennie But when they aren't connected how will the solid sphere make field inside itself zero?

John Rennie

The zero field inside the sphere follows from the spherical symmetry and Gauss' law.

Abcd

5:03 AM

@JohnRennie But charge is enclosed.

So how can Field be 0

(In inital situation, when they are not connected by wire)

John Rennie

The field inside a conductor of any shape is always zero.

Abcd

@JohnRennie I know but how

For initial condition

It cant just loose its charge

John Rennie

Initial condition is just a charged conducting sphere. The charge will all be on the surface of the sphere because the electrons repel each other and going to the surface is the way of maximising the distance between the electrons.

So there is no charge inside the sphere. If you draw a Gaussian surface inside the sphere the enclosed charge is always zero.

@Abcd that's what I just said

5 mins ago, by John Rennie

The field inside the solid sphere is indeed zero. You need only consider the surface of the sphere.

Abcd

@JohnRennie But what if we take the Gaussian surface passing through the surface of the sphere then?

John Rennie

Why would you draw a Gaussian surface like that? It's not symmetrical so it's not use for doing calculations? The field outside the sphere is non-zero, because there will be enclosed charge inside your Gaussian surface.

Abcd

5:09 AM

@JohnRennie Okay, what;s the next step?

John Rennie

I wonder if we're talking at cross purposes. Outside the charged sphere the field is just the same as a point charge at the centre of the sphere. That's pretty standard stuff.

Abcd

@JohnRennie I know

1 min ago, by Abcd

@JohnRennie Okay, what;s the next step?

John Rennie

If the sphere and shell aren't connected I don't think the shell has any effect on the field. It will just be the same as if the shell weren't there.

Abcd

I know right

There will be induced charge on inner surface of shell though

John Rennie

No net induced charge

If you draw your Gaussian sphere just inside or just outside the shell then the field has to be just due to the charged sphere because the shell doesn't have any net charge.

Abcd

5:23 AM

@JohnRennie just inside there will be no field though

@JohnRennie Whats the next step in the problem?

John Rennie

@Abcd when you connect the shell to the sphere charge is going to flow from the sphere onto the shell. In fact, all the charge is going to flow off the sphere onto the shell.

Abcd

@JohnRennie Why all :O

John Rennie

Because the shell is a spherical conductor and the field inside a spherical conductor is ... ?

Abcd

@JohnRennie but field could be 0 even if there was charge on it?

Like the case before connecting the two

John Rennie

@Abcd I'm not sure I understand what you're saying there.

Abcd

5:28 AM

@JohnRennie Look when we hadn't connected the two things, still the field inside the shell was zero right?

John Rennie

No

Abcd

:(

John Rennie

Just inside the shell the enclosed charge is non-zero because it's the charge on the sphere

And just outside the shell the enclosed charge is also non-zero because it's the charge on the sphere

In fact the field just inside the shell and just outside the shell is approximately the same because the enclosed charge is the same and the area of our Gaussian surface is approximately the same.

Abcd

@JohnRennie but field should be zero inside the shell because its a conductor

John Rennie

No

Ah, hang on, When I say inside the shell I mean $r < 20$cm

I'm assuming the shell is infinitely thin

I guess when you mean inside the shell you mean in between the inner and outer surfaces of the shell

Abcd

5:31 AM

@JohnRennie ya, so field should be 0 there right otherwise it would violate the principle that field inside a conductor should be 0

@JohnRennie ya

John Rennie

Yes, but the question says it is a thin shell which is exam speak for treat the shell as infinitely thin

i.e. don't worry what happens inside the shell

Abcd

@JohnRennie I am talking about field at P

Shouldn't field there be 0?

Or the "inside" in field "inside" a conductor stands only for the material of the conductor ?

John Rennie

Draw a Gaussian sphere passing through P

Abcd

@JohnRennie Ya, there should be field there

19 secs ago, by Abcd

Or the "inside" in field "inside" a conductor stands only for the material of the conductor ?

Please clear this "doubt"^

I think I have badly misinterpreted "inside"

John Rennie

@Abcd yes, when we sya the field is zero inside a conductor that means actually in the material of the conductor.

Abcd

5:37 AM

@JohnRennie Okay, so if we draw gaussian surface passing just through the material of the outer sphere then?

Sorry for troubling @JohnRennie

John Rennie

@Abcd I'm not sure to be honest. If we consider the shell as infinitely thin then the problem doesn't arise, because an infinitely thin shell has no inside.

I guess with a shell of finite thickness we have in effect two spherical shells, i.e. the inside surface and the outside surface, connected by a wire.

Abcd

14 mins ago, by John Rennie

Because the shell is a spherical conductor and the field inside a spherical conductor is ... ?

@JohnRennie but you used "inside" in above statement^

John Rennie

@Abcd When you connect the sphere and the shell with a wire you have a single connected object. The charge on that object wants to get as far away from all the other charge in that object, so it's all going to flow to the outer surface of the spherical shell.

So all the charge is going to flow off the sphere onto the shell.

Abcd

@JohnRennie I get that. So your 14 mins ago statement was wrong?

John Rennie

@Abcd by inside I meant r < 20cm

My point was that all the charge is now on the shell, so for r < 20cm you are inside a charged shell.

Abcd

5:44 AM

@JohnRennie Oh I get it. So basically the entire thing is a new conductor and field inside the material of the conductor (which includes the vaccuum space between the two spheres) should be 0.

John Rennie

Yes

Abcd

@JohnRennie Nice, so how to find the heat generated?

John Rennie

Just to clarify, is the question asking for the heat generated when we connect the two shells?

Abcd

@JohnRennie yes

John Rennie

So in effect it's the energy change when we let a 10cm radius charged shell expand to be a 20cm charged shell

Abcd

5:49 AM

@JohnRennie yes, how to find that?

John Rennie

Good question ...

I guess you could calculate the self energy of a charged sphere and take the difference between r=10cm and r=20cm.

Or just take the force on the charge on the shell and integrate it from r = 10cm to 20cm to get the work.

I don't think I've ever done that calculation, but it should be straightforward.

Wait a minute. Isn't it just the potential at the surface of the sphere times the charge on the sphere?

Abcd

@JohnRennie why?

John Rennie

If we have a shell of radius $r$ and charge $Q$ then the potential at the sphere is just $-kQ/r$.

And the potential is the work per unit charge needed to move a charge from infinity to the distance $r$.

Abcd

@JohnRennie + not - ?

John Rennie

So multiply by the charge $Q$ and we get the work needed to move the charge $Q$ from infinity to the shell.

Abcd

5:57 AM

@JohnRennie I dont understand why you are brining a charge when the solid sphere is already charged, initially

John Rennie

@Abcd to charge the sphere you had to bring that charge in from infinity to the surface of the sphere, and you had to do work to move the charge.

If the potential at the surface of the sphere is $V$, and the charge of the sphere is $Q$, then the work needed to bring that charge $Q$ in from infinity to the surafce of the sphere is $QV$.

Abcd

Q^2 / (8pi epsilon R)

John Rennie

Isn't the constant $1/4\pi\epsilon_0$ not $1/8\pi\epsilon_0$

Abcd

@JohnRennie we will have to integrate

potential at any instant is q/4piepsilon R

TO bring dq

W = Vdq

John Rennie

So the energy change would be: $$ \Delta E = \frac{kQ^2}{r_1} - \frac{kQ^2}{r_2} $$

Abcd

6:05 AM

$\int_0^Q Vdq $

John Rennie

where $r_1$ = 10cm, $r_2$ = 20cm and $Q$ = 20 $\mu$C

Abcd

I have got the right answer though...

1 min ago, by Abcd

@JohnRennie we will have to integrate

Potential at any instant is $V = \dfrac{kq}{r}$, if charge q is there

$W = \int_0^Q Vdq$

John Rennie

Ah, OK.

Yes, you're right.

Abcd

@JohnRennie I would also like to know why heat was generated

and where was it generated

John Rennie

We know that the energy of the final state is less than the energy of the initial state, so that energy must have turned into heat. Exactly how it turned into heat doesn't matter. Presumably due to the resistance of the wire connecting the two spheres.

Abcd

6:09 AM

@JohnRennie No I really don't think it was due to resistance.

otherwise for different wires we would get different heats

John Rennie

@Abcd no. A higher resistance would give a lower power for a longer time. A lower resistance wire would give a higher power for a shorter time. Both would give the same total energy dissipated.

I suppose you could calculate the current in the wire ...

(if you wanted to :-)

Abcd

@JohnRennie how?

John Rennie

You'd have to do the calculation of the current as a function of time then show the integral of the power dt was independent of resistance. But I have no plans to do that calculation.

Abcd

7:17 AM

@IceInk The plates of a parallel plate capacitor are given charges +4Q and -2Q. The capacitor is then connected across an uncharged capacitor of same capacitance as the first one. (=C). Find the final potential difference between the plates of the first capacitor.

Shouldn't it be 3Q/C ?

Anwer given is 3Q/2C

@JohnRennie If you have any time please see.

John Rennie

@Abcd hmm, my first thought is that it should be 3Q/C

Abcd

@JohnRennie Isn't that violating KVL, the other capacitor would have volatage $Q/C$ but the sum of the two voltages isn't 0.

@JohnRennie Ok, must be an answer key error.

John Rennie

@Abcd the charge would spread over both capacitors, so both capacitors would have charges of +2Q and -Q.

Abcd

@JohnRennie i don't get you

John Rennie

:45627340 you start with +4Q and -2Q on the first capacitor then connect an identical capacitor in parallel.

Abcd

7:28 AM

@IceInk isn't the distribution something like (Q,3Q), (-3Q,Q) , (-Q,Q), (-Q, Q)

John Rennie

I have to work for about 15 minutes now

Anonymous

@JohnRennie Hmm yes, but why would +2 Q charge flow and not just +Q?

Abcd

Okay.

@Ice do you agree or not?

A single bracket denotes two surfaces of a capacitor.

John Rennie

@IceInkberry because that would make the voltages on the two capacitors unequal

Anonymous

@Abcd I didn't get why there are four distributions..

Abcd

7:31 AM

@IceInkberry there are 4 plates

[(Q,3Q), (-3Q,Q)] S [(-Q,Q), (-Q, Q)]

where S denotes series

2 mins ago, by Abcd

@Ice do you agree or not?

Anonymous

Oh, those are plates. Got it

Anonymous

But

Anonymous

Count the charges for yourself.

Abcd

@IceInkberry Net charge is still 2Q

Anonymous

And on the inner one it is becoming zero?

Anonymous

7:34 AM

No

Abcd

@IceInkberry ?

Anonymous

I mean why are you adding 4 + (-2)

Abcd

@IceInkberry ????????

Anonymous

They are on different plates, right?

Abcd

@IceInkberry Why shant I

@IceInkberry Charge flowed

Anonymous

7:35 AM

@Abcd From where?

Anonymous

It is joined to another capacitor

Abcd

@IceInkberry Isn't that the trick, "place half half charge on outermost plates"

@IceInkberry Q charge got induced there

@IceInkberry okay, it didn't flow

@IceInkberry why shouldn't I add them

@IceInkberry Whaz the contradiction

Anonymous

But but, wait. I am posting an image, did you mean this?

Anonymous

Is this the charge distribution according to you?

Abcd

7:39 AM

@IceInkberry No

Anonymous

Got ir

Abcd

My 3Q one is on left side

Anonymous

I didn't understand your brackets first

Anonymous

And got them wrong

Anonymous

Now, I got it

Anonymous

7:40 AM

Yess

Anonymous

Agree

Abcd

@IceInkberry SO my charge distribution is correct na?

@IceInkberry But why is it violating KVL

@Ice Please reply

Anonymous

___Nothing_____

Anonymous

Grrr, I act dumb at times.

Anonymous

@Abcd Yes, I think it is correct by our 'trick'

John Rennie

7:44 AM

But when you connect the second capacitor you have two identical capacitors in parallel. Why would they have different charges on them?

Anonymous

That's what I am wondering

Anonymous

But if we do it by what charge will be induced method

Anonymous

We get that distribution

Abcd

4 mins ago, by Abcd

@IceInkberry But why is it violating KVL

@Ice please reply to this^^^^^^^

I am waiting

Anonymous

@Abcd Please explain me how it is violating KVL. I don't get it.

Abcd

7:47 AM

@IceInkberry two voltages arent equal are they

Anonymous

You might probably be banging your head on your desk as to why you asked me but ¯\_(ツ)_/¯

Anonymous

@Abcd Ah, that's what.

Anonymous

I don't get 'violating law' at times because I just don't know the laws

Abcd

Someone HELPPPP

Anonymous

I think I got it

Anonymous

7:51 AM

Will think for few minutes if it is correct

Abcd

@JohnRennie please tell why this violation is happening?

Anonymous

The charge distribution is incorrect

Anonymous

Your charge distribution is incorrect @Abcd

Abcd

@IceInkberry :O

Anonymous

And JR's reasoning that both the Capacitors will have equal charges is correct but he didn't mention about the distribution

Anonymous

7:54 AM

Posting the distribution

Abcd

@IceInkberry What about our Half-half trick?

Anonymous

@Abcd I think it works when the plates aren't joined together!

Anonymous

What I did is distribute half half charge on both Capacitors and did our trick.

Abcd

@IceInkberry but you just said that our trick isnt valid

Anonymous

7:58 AM

@Abcd But if you consider a single capacitor, it is valid because in individual Capacitors the plates aren't connected in any way. Whereas doing it together to both capacitors, the plates of one capacitor are connected to other

Anonymous

And

Anonymous

And see the distribution

Anonymous

At the ends, it is Q at the ends of your distribution

Anonymous

That means there is 2Q inside the plate. Not possible.

Abcd

Leave all this trick system its confusing

Better to give q1 and q2

then use KVL

and charge conservation

Anonymous

8:00 AM

Arey read my messages again, it is easy.

Abcd

@IceInkberry but they are connected to other plates

Anonymous

I distributed -2Q and Q on each capacitors and separated them and then wrote the charge distribution.

Anonymous

Separating after halving the charges won't make any difference because the capacitance is same and after halving the charge won't flow from one capacitor to another.

Anonymous

@Abcd Connecting means the charge will be distributed, precisely it will be halved. I did that and proceeded.

Anonymous

@Abcd Got it?

Abcd

8:05 AM

not really

Anonymous

Okay, listen.

Anonymous

You have two equal capacitors right?

Anonymous

@Abcd

Anonymous

Quickk, I have to run.

Abcd

@IceInkberry yes

Anonymous

8:08 AM

@Abcd So, if you have 2Q on one capacitor and you connect it to another same capacitor, both will have Q charge, right?

Abcd

@IceInkberry Kk

Anonymous

Similarly, if one capacitor has -4Q and +2Q charge, both the capacitors will have -2Q and +Q charge right?

Anonymous

@Abcd

Abcd

@IceInkberry Huh?

Anonymous

@Abcd The charges will be divided equally

Anonymous

8:14 AM

I have to go

Anonymous

After you divide the charge equally between the capacitors, will there be any flow of charge in the wires?

Anonymous

No, there won't be!

Anonymous

Now, read my messages again.

Anonymous

Bye bye!

John Rennie

8:28 AM

@Abcd I'm still here but I got a bit baffled by the discussion. II seems to have taken half an hour to conclude I was right all along.

Anonymous

@JohnRennie Yes, you were right. My answer is based on that only. But, he isn't understanding how I did the charge distribution after halving the charges. Or maybe he isn't just getting the halving charge thing.

Anonymous

39 mins ago, by Ice Inkberry

Anonymous

38 mins ago, by Ice Inkberry

What I did is distribute half half charge on both Capacitors and did our trick.

Anonymous

And deciding why our previous method was wrong:

Anonymous

39 mins ago, by Ice Inkberry

@Abcd I think it works when the plates aren't joined together!

Anonymous

8:36 AM