Problem Solving Strategies

John Rennie

06:00

@PinkAura Yes, as long as you stay below the breakdown voltage a zener diode just behaves like a regular diode.

@mo-_- It's a bit odd the question didn't tell us the volume since that is needed for the calculation.

Without that you answer will have to be an equation containing τ₁ i.e. nit just a number.

@Shashaank Hi :-) That isn't really a question for this room. We concentrate on JEE questions.

Pizza

06:14

@JohnRennie Hi :-)

John Rennie

06:39

@Pizza Hi :-)

NOTE Book

@JohnRennie Hi... can we discuss now or after Pizza?

Pizza

@NOTEBook You can discuss now

John Rennie

OK :-)

NOTE Book

Thanks

The infinite sheet formula σ/2ε is only for thin sheet right? Like not for slab with charge on 2 surfaces

John Rennie

It works for a slab as well. It describes the field outside any infinite sheet of any thickness.

NOTE Book

06:46

John Rennie

What is the question?

→ 1 message moved to Trash

NOTE Book

My bad

How is it same won't we get twice the field?

John Rennie

Do you know how we derive that equation for the field?

NOTE Book

Like if I have to use σ/2ε+ σ/2ε

yes but for thin surface like sheet

By taking a cylindrical or cuboidal gaussian surface

John Rennie

NOTE Book

06:53

yes

John Rennie

Yes, we draw a cylinder like this so the total flux is φ = E.A + E.A = 2EA

NOTE Book

yes

John Rennie

And the charge inside the cylinder is σA, so we get σA/ε₀ = 2EA

NOTE Book

Yes

John Rennie

Hence E = σ/2ε₀

So let's do the same for the slab on the right ...

We are going to get exactly the same result, Aren't we?

The charge inside is σ/2 A + σ/2 A = σA just as for the thin plate.

Yes?

NOTE Book

06:57

yes

John Rennie

So we will also get E = σ/2ε₀

There's also another way to understand it. Give me a moment and I'll update my diagram ...

NOTE Book

ok

John Rennie

Assuming the slab is a conductor all the charge is on the two faces, so it is like two thin sheets each with a charge of Q/2 i.e. a charge density of σ/2.

NOTE Book

yes

John Rennie

So E₁ = (σ/2)/2ε₀

And E₂ = (σ/2)/2ε₀

Yes?

NOTE Book

07:01

yes

John Rennie

And electric fields add, so the total field is:
E = E₁ + E₂= (σ/2)/2ε₀ + (σ/2)/2ε₀ = σ/2ε₀

And once again we get the same result as a single thin sheet.

Does this make sense so far?

NOTE Book

yes

oh

John Rennie

The key point here is that for an infinite sheet the field is constant and does not change with distance from the sheet.

NOTE Book

I was taking charge density σ on both sides :(

John Rennie

So for the thick slab we are a different distance from the two faces (where the charge is) but it doesn't matter because the field does not change with distance.

@NOTEBook Aha!

NOTE Book

07:06

Yes

John Rennie

@NOTEBook But the good thing about having made that mistake is now you'll remember it!

NOTE Book

Yes

Thank You ❤️

John Rennie

Is there anything else or shall I answer Pizza now?

NOTE Book

I am done

John Rennie

@NOTEBook You're welcome :-)

@Pizza Are you there?

NOTE Book

07:35

@JohnRennie Can you also explain why is the field due to left side of slab on right side. I can understand for 2 sheets of charges but how does it work when there is a conductor in between?

I just assumed it so if it is not necessary for JEE you can tell me it isn't

Pizza

@JohnRennie Sorry, I'm busy :(, if you have some free time can you let me know more or less what should be done? I'll try to answer as soon as possible. Sorry again :(

John Rennie

@Pizza OK :-)

Just ping me when you're free.

@NOTEBook Hi

NOTE Book

@JohnRennie Hi

John Rennie

The field inside the conductor is zero. Yes?

NOTE Book

yes

John Rennie

07:46

But we have charges around and those charges generate fields in all directions.

So the only way the field can be zero inside the conductor is if the fields from the left and right faces are equal and opposite so the cancel to zero.

OK so far?

NOTE Book

yes

John Rennie

What happens is:
- on the left of the slab the fields from the two faces point point left, i.e. in the same direction, so they add to give a total field σ/2ε₀ pointing left.

- on the right of the slab the fields from the two faces point point right, i.e. in the same direction, so they add to give a total field σ/2ε₀ pointing right.

NOTE Book

yes

John Rennie

- but inside the slab, in between the two faces, the field from the left face points right and the field from the right side points left, so they cancel to give zero.

The point is that a charge generates a field everywhere.

Where we get zero fields it has to be due to a cancellation of the field from two or more charges.

Does this make sense so far?

NOTE Book

Um

I mean since there is conductor in between left slab and right side so field lines of it is zero inside conductor. But its field lines are there on right side hence its field. Since field can't be broken is it like left side causes some induction on inner surfaces?

entropy

07:52

Hi sir.. @JohnRennie

NOTE Book

Um Like this

John Rennie

We tend to think that conductors somehow block or screen charges, but they do not. You cannot block or screen an electric field - the field is everywhere. What actually happens is the conductor has charge induced on it and the field from the induced charge adds to the external field to cancel it.

NOTE Book

Ok

John Rennie

The sheet of charge on the left face produces a symmetrical field on both sides of the sheet, and the same is true for the right face.

NOTE Book

yes

John Rennie

07:55

It's just that in between the fields sum to zero.

NOTE Book

Yes

John Rennie

Incidentally this is a proof that the charges on the two sheets must be equal.

If they were not equal the field inside the conductor would be zero.

NOTE Book

yes

Thank you

John Rennie

OK :-)

@entropy Hi :-)

entropy

NOTE Book

07:57

Thanks for waiting @entropy

entropy

No problem @NOTEBook

In the q, ig we can use derivatives to find the velocity. I wanted to know if there's a way I can make use of the fact that point A is hinged

John Rennie

@entropy I'm going to make a coffee. I'll be 5 mins.

entropy

Okay, sure.

John Rennie

@entropy Hi

entropy

Yes, sir

John Rennie

08:08

You'll see lots of these problems where you have to relate the velocities of different parts of the system, and some will be quit complicated.

But there is a simple approach to solving them that always works so I strongly recommend you use it.

Give me a moment and I'll draw a diagram ...

In this case we know what vb is and we need to find va. Yes?

entropy

Yes

John Rennie

The wat we do this is to find a relationship between a and b.

entropy

Yes. I guess we could pythagoras theorem for that

John Rennie

Then we can differentiate d/dt on both sides and that gives us a relationship between da/dt and db/dt.

entropy

Okay.. yes got it

John Rennie

08:14

And da/dt and db/dt are the two velocities we need.

@entropy OK :-)

This may seem a bit long, but it always works even in complicated questions.

entropy

Thank you for the help

Yes..

John Rennie

Do you want to go through it or can you take it from here?

entropy

I will give it a try... I understand now

John Rennie

OK :-)

I'll be here for several hours more so you can ping me if you want to discuss anything.

entropy

Thank you

John Rennie

08:16

You're welcome :-)

Pizza

08:27

@JohnRennie If you want I'm here now

John Rennie

Hi :-)

Pizza

Hi :-)

John Rennie

There's something a bit odd about part (c)

If the field is parallel to the z axis the flux through the loop will always be zero so part (c) seems an odd question.

Are you sure the direction of B is correct?

Pizza

Strange, because it also asks us for the angle where the flux is max

@JohnRennie Yes, the text is definitely this one

$\Phi_B = B_0 \cdot A \cdot \cos(\theta)$

John Rennie

OK. That means we cannot answer (c) since φ = 0 for all angles.

But we can do parts (a) and (b) if you want.

Pizza

08:37

Isnt $\theta = 25^\circ$ the angle between the normal vector to the loop and the field $B_0$. ?

John Rennie

The vector area of the loop is horizontal i.e. parallel to the xy plane. Yes?

Pizza

Yes

John Rennie

And B is vertical so B and A are always at right angles to each other regardless of what value θ has.

Yes?

Pizza

Yes

John Rennie

So B.A is always zero.

Pizza

08:43

Right

We can do the other 2 points

John Rennie

How do you want to do this? Do you want to start or shall I?

Pizza

Can you start?

John Rennie

Ok, so (a) asks us to find the forces on the sides.

Now we know the force on a conductor of a length 𝓁 carrying a current I in a field B is:
F = BI𝓁

Yes?

Pizza

F = I • B x L

@JohnRennie yes

John Rennie

Let me draw in some letters to identify the sides ...

Pizza

08:47

👍

John Rennie

The sides AB and CD are parallel to B and the cross product of parallel vectors is zero. So the force on those two sides is zero.

Yes?

Pizza

Yes

John Rennie

The sides BC and DA are normal to the field so the force on both of those is:
F = B₀I₀𝓁

But we need to get the directions of the forces.

The way I like to do this is just use the Lorentz force equation F = qv × B

The velocity v of the charges in the wire is in the same direction as the current, so using the right hand rule we can find the direction of the force.

Can you do that?

Pizza

Yes

Wait

John Rennie

...on phone

Pizza

08:56

Side BC:
- Direction: $+x$(outward).

Side DA: Direction: $-x$ (inward)

These two forces (on BC and DA) create a torque about the z-axis.

John Rennie

I'm back!

@Pizza Yes, although they're not really +x and -x because their direction will change as the loop rotates.

The BC force is clockwise and the CD force is anticlockwise if we are looking down the z axis at the loop.

Yes?

Pizza

09:15

@JohnRennie Sorry my internet was gone

John Rennie

No problem, I was on the phone anyway :-)

Pizza

@JohnRennie yes

@JohnRennie 👍

John Rennie

But then that makes part (b) a bit weird because the two equal and opposite forces create equal and opposite torques about the z axis so the total torque is zero.

pi-π

Hey @JohnRennie

John Rennie

Hi :-)

Pizza

09:23

But however, there is still a net torque about an axis within the plane of the loop, which causes the loop to rotate. If the axis of the hinge allows for free movement, this torque will cause continuous rotation.

I crashed again :(

John Rennie

There is indeed a torque about the x axis, but the question says the loop is hinged about the z axis not the x axis.

So the loop cannot rotate about the x axis.

pi-π

Why does event binding occur automatically when a row is selected in a DevExpress Grid? @JohnRennie Any Ideas?

John Rennie

This question would make more sense if the field was parallel to the x or y axes.

@pi-π I've never use WinForms so I don't know. Sorry :-(

pi-π

Okay. Thanks!

Pizza

@JohnRennie Oh ok, thanks

PinkAura

10:15

@JohnRennie hi sir

John Rennie

Hi :-)

PinkAura

can you just tell how avg( sin^2 x+ cos^2 x) = avg sin^2x + avg cos^2x

i mean without actually finding average ?
what i mean to ask is , when can we write avg ( fx + gx) = avg fx + avg gx?

John Rennie

Suppose we do a discrete average. That is take the interval x = 0 to x = X and take N values of x in that range and evaluate f(x) and g(x) at all N points.

Then the average of (f(x) + g(x)) is:
1/N Σᵢ (f(xᵢ) + g(xᵢ))

OK so far?

PinkAura

yes

John Rennie

And we can rewrite that as:
1/N Σᵢ f(xᵢ) + 1/N Σᵢ g(xᵢ)

And that's just the discrete average of f(x) + the discrete average of g(x)

Yes?

PinkAura

10:21

yes

John Rennie

Now, we'd actually use an integral not a discrete sum for a continuous function, but an integral is just our discrete sum with N ⟶ ∞

PinkAura

so is it always true?

John Rennie

Yes

Actually we could have just used an integral directly couldn't we?

PinkAura

okay yes

thanks

John Rennie

If we are taking the average over a range Δx i.e. from x = some value to x = some value + Δx then the average is:

avg(f(x) + g(x)) = 1/Δx ∫(f(x) + g(x)) dx

and the integral splits up into two integrals just like our sum did.

So again we see it's always true.

mo-_-

10:26

Hi @JohnRennie :-)

John Rennie

Hi :-)

mo-_-

@JohnRennie are you free?

John Rennie

Only if it's quick.

I have to leave in 15 mins.

I will be back in about an hour if you want to chat then.

Shall I ping you when I'm back?

mo-_-

Ok

Then I just post the problem

A spherical conductor, centered at O with radius R₁, carries a charge Qₐ. It is concentric with a hollow spherical conductor with inner radius R₂, outer radius R₃, and total charge Qᴮ. The potential difference between the two conductors is Vᴮ - Vₐ = 500 kV. A test charge q, placed at a distance R from the center O of the system, is subjected to a repulsive electrostatic force of intensity F = 1.5 × 10⁻⁶ N.

Assuming that the presence of q does not affect the charge distributions on the two conductors, determine the total charges Qₐ and Qᴮ present on the two conductors.

PinkAura

11:04

@JohnRennie sir , i have a quick question, why two antiparallel progressive waves superposition isn't a progressive but two parallel progressive waves superposition is a progressive wave?

NOTE Book

@mo-_- you found answer?

mo-_-

11:21

@NOTEBook Mm no

Have you tried doing it?

NOTE Book

yes

is it 1/3 milliC

Qa+Qb?

mo-_-

11:38

@NOTEBook Have you calculated the potentials (V(Ri))?

NOTE Book

R1?

mo-_-

It Is like ∆V = V(R3) - V(R1)?

NOTE Book

yes

this is 500kV

From here i tried to calculate Qa and then Q_a +Q_b

mo-_-

V(R3) = kqB/R3?

NOTE Book

no

potential due to inner charge as well

mo-_-

11:41

I assumed V(infinity) = 0

NOTE Book

@mo-_- Ok

mo-_-

So for r>R3, ∆V=V(R3) - V(infinity)

NOTE Book

No also due to Q_a

mo-_-

So V(R3) = - integral (E(r>R3)) from infinity to R3

@NOTEBook why? Isn't there just Qb on the external conductor?

NOTE Book

yes but it will cause induction

mo-_-

11:45

Don't I have to find V(R1) separately?

Like ∆V = V(R1) - V(R2)

NOTE Book

Like potential due to inner sphere at R3 +potential due to outer sphere at R3

mo-_-

So V(R1) = V(R2) - integral (E(R1<r<R2)) from R2 to R1

NOTE Book

You have done induction of concentric sphere as a topic right?

Wait

Electrostatic Potential and Capacitance 05 : Potential in Concentric Shells JEE MAINS/NEET

►

check this lec or any other similar or you will create many doubts. Or your teacher will teach you after some topic so you can ask him

mo-_-

@NOTEBook But how did you use q?

The text says his contributions can be overlooked

NOTE Book

oh

that meant that q is small and won't cause induction

if q was very large it will attract shells electrons

mo-_-

11:57

But the value was given in the data, so I think it is useful for something.

NOTE Book

yes

see force between spheres and q is given

Practically q will cause induction on spheres but it will be hard to calculate so we just neglect its induction on spheres

So q is very small

mo-_-

Ah, I didn't consider that force...

Can you by any chance show me how you did it?

NOTE Book

Oh i see

mo-_-

If you wrote it on paper

NOTE Book

wait

you know how charges will be induced because without it we can't solve the problem?

mo-_-

12:02

No maybe it's better if I watch the video above first

NOTE Book

yes

mo-_-

Anyway, thanks for the help! Then maybe we'll see if @JohnRennie agree with you

NOTE Book

I will be active till 6 then at around 8 so i can help if you want

@mo-_- ok

John Rennie

@mo-_- Hi, sorry I was away so long ...

mo-_-

@JohnRennie Hi

John Rennie

12:10

I see you and Notebook have been discussing the problem. Are you done or is there more to discuss?

mo-_-

The problem was that I wrote V(R3) = kQb/R3

And that's wrong

And I didn't know why

That is, I basically thought that the charge q was going from infinity to R3 , but from what I understand this q should not be considered

So I wrote

V(R3) = - integral (E(r>R3)) from infinity to R3

Where E(r>R3) = kQb/r²

V(R3) = kQb/R3

V(R2) = V(R3) , since E(R2<r<R3) = 0, so it's constant

V(R1) = V(R2) - integral (E(R1<r<R2)) from R2 to R1

Where (E(R1<r<R2)) = kQa/r²

John Rennie

Outside a spherical charge distribution the field is the same as if we had a single point charge with same charge as the system. Yes?

mo-_-

So V(R1) = kQb/R3 + kQa/R1 - KQa/R2

@JohnRennie yes , E(r>R3) = kQb/r²

The charge is not Qb? I don't understand

John Rennie

So the force on the charge q is:
F = kq(Qa + Qb)/R²

And this is one of two equations we will need to find Qa and Qb.

OK so far?

mo-_-

Ok

John Rennie

12:21

The second equation comes from the potential difference between the sphere and the shell.

Inside the shell the field depends only on Qa and it's given by:
E = kQa/r²
where r is the distance from the centre of the sphere.

Yes?

("inside the shell" means r < R₂)

I'm rushing a bit because I need to go soon.

mo-_-

@JohnRennie Yes

John Rennie

I've done this diagram showing what the fields are outside the shell, and in the space in between the sphere and the shell.

Are you confident you understand why I have written those equations for the fields?

mo-_-

Ok this Is clear

@JohnRennie yes

You used Gauss's law

John Rennie

Yes :-)

So for the charge outside the shell E = k(Qa + Qb)/R² and F = qE so F = kq(Qa + Qb)/R²

mo-_-

Yes

John Rennie

12:38

And we are told R = 1m, F = 1.5 × 10⁻⁶ N and q = 0.5 pC so we get:
FR²/kq = Qa + Qb = 3.33 × 10⁻⁴C

Yes?

mo-_-

Yes

John Rennie

Now the field in between R₁ and R₂ is E(r) = kQa/r²

mo-_-

Yes

John Rennie

And to get the potential difference between R₁ and R₂ we need to integrate E(r) from r = R₁ to r = R₂

Yes?

mo-_-

Yes

John Rennie

12:43

So we are going to get:
ΔV = kQa/R₂ - kQa/R₁
(I'm not sure about the sign but we can work out the sign later)

mo-_-

Ok yes

John Rennie

So that is ... ΔV = 5 × 10⁵ = Qa × 8.99 × 10⁹ × (1/0.4 - 1/0.5)

mo-_-

Ok

John Rennie

Qa = 1.11 × 10⁻⁴ C

(assuming I did the arithmetic correctly)

NOTE Book

yes

mo-_-

12:46

Nice !

John Rennie

OK :-)

I feel like I didn't give you the chance to do it for yourself because I'm in a rush :-(

mo-_-

8 mins ago, by John Rennie

And we are told R = 1m, F = 1.5 × 10⁻⁶ N and q = 0.5 pC so we get:
FR²/kq = Qa + Qb = 3.33 × 10⁻⁴C

So we sub Qa here?

John Rennie

But has that helped? Do you understand it now?

@mo-_- Yes

mo-_-

@JohnRennie Yes, then I'll reread everything calmly

NOTE Book

Total charge is asked right? So i guess it is Qa+Qb

John Rennie

12:48

Qa + Qb = 3.33 × 10⁻⁴C
Qa = 1.11 × 10⁻⁴C

mo-_-

@JohnRennie I was wrong, at least I know now that it's correct like this

John Rennie

@NOTEBook Hmm, that's true, which seems a bit odd. Maybe it meant for us to find both charges.

@mo-_- OK :-)

NOTE Book

you are going right?

mo-_-

11 mins ago, by John Rennie

And we are told R = 1m, F = 1.5 × 10⁻⁶ N and q = 0.5 pC so we get:
FR²/kq = Qa + Qb = 3.33 × 10⁻⁴C

John Rennie

@NOTEBook I have a few more minutes if you have a question ...

mo-_-

12:50

@NOTEBook But do you mean to stop here?

@JohnRennie thanks for the help !

NOTE Book

i was calculating equivalent capacitance in series... we took charge on each Capacitor to be Q

So why is it that equivalent capacitance also have Q charge on it?

is it because battery can suck only that much charge from each plate or something?

John Rennie

If you connect a battery to your two capacitors the a charge Q flows out of the battery.

NOTE Book

yes

John Rennie

And that charge must flow onto the equivalent capacitor because there is nowhere else for the charge to go.

NOTE Book

oh

John Rennie

12:54

So as far as the battery is concerned it supplied a charge Q, and the battery EMF is E, so the equivalent capacitance must be:
Cequiv = Q/E

NOTE Book

yes

John Rennie

Now we know that charge Q actually flowed onto the top plate of the top capacitor, and then there were other charges flowing between the two capacitors.

But it doesn't matter.

NOTE Book

yes

John Rennie

The equivalent capacitance is just the charge supplied by the battery divided by the battery EMF.

NOTE Book

oh

John Rennie

12:56

The battery doesn't care what happened to that charge after it left the battery :-)

NOTE Book

ok

Yes i got it thank you

John Rennie

OK :-)

NOTE Book

@mo-_- i will go off for like 1.5 hr then will come online at around 8 to do lec you can ping

Pizza

16:12

@JohnRennie I'm sorry but using the left hand rule, I get different directions of F from the ones you drew

for BC , F pointpoints towards +z, while for BA I think -z

but isn't there a general rule I can use? , because I always get confused which one I should use for each case

Or am I doing it wrong, I don't know...

ah no, I think I was wrong, I find the same, sorry for the ping

mo-_-

18:16

@JohnRennie but shouldn't it be V(R₁) = V(R₂) -∫ kQₐ/r² dr (limits from R₂ to R₁) ?

@NOTEBook

F. Zer

19:35

@JohnRennie Thank you John !

Transcript for

Problem Solving Strategies