Problem Solving Strategies

John Rennie

05:20

@F.Zer Congratulations! :-)

@F.Zer The headings indicate what type of orbital the outermost electrons occupy.

e.g. in groups 1 and 2 the outermost electrons are in the s orbital while in groups 3 to 8 they are in the p orbitals.

@mo-_- Yes, the "flux" is kind of how much electric field flows through the surface.

PinkAura

06:27

@JohnRennie hi sir

John Rennie

Hi :-)

PinkAura

ans given is 2h but im getting h

John Rennie

The shadow is the part of the screen where the light rays cannot reach it because they are blocked by the object.

I've drawn the two light rays at the edges of the shadowed region.

Does this make sense so far?

PinkAura

ohh i considered the second ray to be striking at the foot of the person

i mean for drawing the length of the shadow, i did not draw the last ray

@JohnRennie yes

John Rennie

Yes, that's where the extra ℎ comes from.

PinkAura

06:35

yes thanks

@JohnRennie can you pls explain this question as well

John Rennie

Part (c) is straightforward. I must admit I cannot work out what parts (a) and (b) are supposed to mean.

I would just ignore the question.

For a start it says the disks rotate in different directions but the diagram shows them rotating in the same direction.

PinkAura

yes

John Rennie

It's just a bad question.

PinkAura

maybe

mo-_-

08:14

@JohnRennie Hi !

John Rennie

Hi :-)

mo-_-

Can i ask for help?

@JohnRennie

John Rennie

Yes, go ahead :-)

mo-_-

@JohnRennie What are the characteristics of a hollow conductor?

John Rennie

I'm not sure what you are asking? Is there a specific question that relates to this?

mo-_-

08:21

and what changes between a non-hollow conductor

@JohnRennie yes wait

On a metallic sphere with a radius $R_1 = 5 \, \text{cm}$, a charge $Q_1 = 10^{-6} \, \text{C}$ is deposited. Surrounding it is a metallic spherical shell (hollow metal sphere) with an inner radius of $R_2 = 10 \, \text{cm}$ and an outer radius of $R_3 = 12 \, \text{cm}$. On the spherical shell, a charge $Q_2 = 10 \, Q_1$ is deposited.

Determine the charge density on the inner surface of the spherical shell with a radius $R_2 = 10 \, \text{cm}$ and the potential difference between the two conductors (sphere-spherical shell).

What charge will there be on the
spherical shell surfaces?

John Rennie

Your starting point for questions like this is Gauss's law.

mo-_-

oh okay

John Rennie

Since the system is spherically symmetric we can draw a spherical Gaussian surface and the field will the the same everywhere on that surface. That means there is a simple relation between the field and the charge inside the surface.

And a key point to remember is that the field inside a conductor is always zero.

mo-_-

but shouldn't I first identify where the charges are?

John Rennie

Suppose we take the surface shown by the blue circle.

This is inside the conductor so the field everywhere on the surface must be zero. Yes?

mo-_-

08:34

ok

yes

John Rennie

And that means the flux through the blue surface must be zero, and since φ = q/ε₀ that means the total charge q inside the blue surface must also be zero.

And we know the charge on the red sphere at the centre is Q₁

Yes?

mo-_-

Yes

John Rennie

Let's write the charge on the inner surface at R₂ as Q₂ and the charge on the outermost surface as Q₃

If the total charge inside the blue sphere is zero that must mean:
Q₁ + Q₂ = 0
Yes?

mo-_-

but I didn't understand what is meant by a charge being on the spherical shell

John Rennie

Suppose we start with a spherical shell made up from some metal like copper.

mo-_-

08:40

Q3 would it be external charge?

@JohnRennie ok

John Rennie

This won't be charged because, well, lumps of metal are not charged by default i.e. in some random piece of metal the positive charges of the metal atom nuclei and the negative charges of the electrons on the metal atoms cancel out and sum to zero.

OK so far?

mo-_-

yes

John Rennie

But now suppose I add some extra electrons to the shell. There are various ways we can add electrons to a lump of metal but the way we do it doesn't matter. All that matters is that we now have a negative charge Q = -ne on the metal, where n is the number of electrons we added.

The charge density is always zero inside a conductor, so we know the charge will be on the surfaces of the conductor i.e. we may have some on the inner surface and some on the outer surface.

OK so far?

mo-_-

ok

John Rennie

Now, how much charge is on each surface will depend on what other charged objects are nearby. We know the total charge is just whatever charge Q we added to the shell, but we need to work out how this charge Q is split between the inner and outer surfaces.

And that's what this question is about.

It asks you to calculate how much of the charge we added to the shell goes to the inner surface.

Does this make sense so far?

mo-_-

08:49

yes

John Rennie

The question says a charge 10Q₁ has been added to the shell.

mo-_-

yes

John Rennie

Though in fact we don't need to know that fact to work out what the charge on the inner surface is.

mo-_-

but wasn't it 0?

because we are in a conductor

John Rennie

The field is zero inside the conductor. Yes?

mo-_-

08:52

Yes

Q1 + Q2 = 0

John Rennie

Yes, so Q₂ = -Q₁

If we use Gauss's law as I described above it tells us that the charge on the inner surface must be equal and opposite to the charge on the red sphere at the centre.

And we are told that Q₁ = 10⁻⁶ C

So now we know the charge on the inner surface of the shell must be -10⁻⁶ C.

Yes?

mo-_-

Gauss Law: Q1 + Q2 / ε₀

yes?

John Rennie

Yes

mo-_-

so the flux is 0

Q1 - Q1 / ε₀ = 0

John Rennie

Yes

mo-_-

08:55

right

John Rennie

The question doesn't ask for the charge on the inner surface. It asks for the charge density on the inner surface.

So to complete the answer we divide by the area of the inner surface to get the charge density i.e. charge per unit area.

mo-_-

wait

E = σ/ ε₀

so σ = E ε₀ = 0 * ε₀

the field is zero right?

can't I use this formula?

John Rennie

That equation gives the field in the vacuum just outside a charged conductor.

You can't apply it to the field inside a conductor.

mo-_-

oh

ok

@JohnRennie ok

John Rennie

I don't think E = σ/ε₀ is very useful and you should probably not try and use it to answer problems. In most cases your starting point is Gauss's law.

mo-_-

09:04

I thought it was like Gauss's law, so yesterday I saw you talking about it

John Rennie

It is a particular application of Gauss's law.

mo-_-

so I thought I could use this relation

@JohnRennie oh okay

John Rennie

But Gauss's law is more general.

mo-_-

σ = Q/A = Q/4𝜋R²

John Rennie

Yes

mo-_-

09:07

Q would be -Q1, right?

John Rennie

Yes

mo-_-

nice

now we need to calculate the potential difference between the red sphere and the spherical shell

right?

John Rennie

Yes

@mo-_- Can you suggest how we do this?

mo-_-

but isn't it 0? given that ΔV = - ∫E?

John Rennie

I wonder if you are getting mixed up about what we mean by "the field inside a conductor" ...

When we say "inside a conductor" we mean "in the metal that the conductor is made from".

We don't mean the space inside the inner surface.

So in the diagram above "inside the conductor" means either:
r < R₁ i.e. inside the inner sphere
R₂ < r < R₃ i.e. in between the inner and outer surfaces of the shell

It does not mean the region R₁ < r < R₂

So we know E = 0 if r < R₁ or if R₂ < r < R₃

But the field is not zero for R₁ < r < R₂

OK so far?

mo-_-

09:21

so you mean the part in the smaller blue circle , the white part

John Rennie

Yes

Where I've written E(r)

E(r) ≠ 0

mo-_-

ah ok

John Rennie

Can you calculate E(r) using Gauss's law?

mo-_-

V(R2) - V(R1) = - ∫ E(r) from R1 to R2 ?

John Rennie

Yes :-)

mo-_-

09:25

@JohnRennie mm

wait

im trying

John Rennie

OK :-)

mo-_-

E = Q/4𝜋ε₀R²

John Rennie

Yes! :-)

Well Q₁

E(r) = Q₁/4𝜋ε₀r²

We usually write k = 1/4𝜋ε₀ for convenience, so we get:
E(r) = kQ₁/r²

Yes?

mo-_-

yes

John Rennie

So if we start at r = R₁ and end at R₂ then the potential difference is the integral from R₁ to R₂ of -E(r)dr

mo-_-

09:31

ΔV = (Q1 / (4π * ε0)) * (1/R1 - 1/R2) = 9 * 10^4 V

I have to put the - right?

John Rennie

Yes :-)

mo-_-

but I've seen that some use the formula without the -

John Rennie

The sign depends on whether you go from R₁ to R₂ or whether you go from R₂ to R₁ and the question doesn't say.

mo-_-

ΔV = ∫E

John Rennie

So I think it's just asking for the magnitude of the potential difference.

mo-_-

09:34

ah ok

John Rennie

But you are quite correct that if the direction is specified then the sign matters.

And keeping track of the signs can be surprisingly hard.

mo-_-

yes

the graphic part also seems a little difficult to me

that is, understand the signs well and where I have to draw the circles

John Rennie

It's something you will get used to with practice.

mo-_-

but beyond these requests, what could other exercises ask of me? like calculating the work?

John Rennie

Exercises could ask all sorts of things. The best way to prepare for the JEE is to do as many exercises as you can.

That way you cover everything you could be asked in the exam.

mo-_-

09:39

yes, but I also have to do the theoretical part otherwise I can't do the exercises

that is, at least the theory I need for the exercises

John Rennie

Most students start with the exercises and learn what they need to do that exercise. Then move to the next exercise and learn what they need to that, and so on.

Really the JEE is about learning how to answer JEE questions not learning physics.

mo-_-

ok so you advise me to only do exercises? And learn how to do it?

John Rennie

Are your teachers helping you to prepare for the JEE? Or do you have a tutor?

mo-_-

I had , now I'm studying on my own

John Rennie

OK, I was going to say your teachers/tutor know best how to prepare for the JEE because they have taught students for it so many times. If you are preparing by yourself then you don't have that support.

I would say concentrate on the exercises.

There are lots of books of past JEE questions and practice exams.

Use those and work through them learning how to answer each problem in turn.

mo-_-

09:46

ok I'll try to do it that way then, thanks!

John Rennie

You're welcome :-)

mo-_-

ah thanks also for the help with the exercise

John Rennie

I'm always happy to help.

I'm not always here, but if I am I will reply when you ping me.

mo-_-

Thank you for your availability, I really appreciate it

John Rennie

You're welcome :-)

Pizza

10:46

@JohnRennie Hi!

John Rennie

Hi :-)

Pizza

Are you free now?

John Rennie

Yes :-)

Pizza

Two current-carrying wires are positioned as shown in the figure. The currents are $i_1$ (aligned with the $y$-axis) and $i_2$ (aligned with the $x$-axis).

1. Calculate the resulting magnetic field at point $A$.
2. Identify in which quadrants the resulting field can be zero.
3. Given a charge $q = 1.6 \times 10^{-19} \, \mathrm{C}$ placed at $A$, with a velocity $v$ parallel to and aligned with the $x$-axis, determine the force acting on it.

Data:
- $i_1 = 2 \, \mathrm{A}$
- $i_2 = 3 \, \mathrm{A}$
- $v = 2 \, \mathrm{m/s}$
- $A = (4,1) \, \mathrm{cm}$
- $\mu_0 = 4 \pi \cdot 10^{-7} \, \mathrm{H/m}$

I don't understand... where would point A be located?

I was provided with this image

John Rennie

The point A is at the coordinates (4, 1) i.e. at the point where x = 4 and y = 1
(distances measured in cm)

Pizza

10:54

ah it was hidden in the data...

John Rennie

Yes :-)

Pizza

i have to use biot-savart right?

John Rennie

You don't have to do everything from first principles. We have two infinite wires carrying a current, and the field from an infinite wire carrying a current is a well known result.

It's just B = μ₀I/2𝜋r
Yes?

Pizza

yes

John Rennie

And the fields from the two wires will add together (as vectors)
Yes?

Pizza

10:58

Yes

I have to say whether B is entering or leaving to determine the sign

John Rennie

Yes ... ?

Pizza

yes

John Rennie

Can you take it from here?

Pizza

I get that where i1 is, B is incoming at +x and outgoing at -x

John Rennie

Yes, given where the x and y axes have been drawn it is conventional to define the positive direction for z as coming out of the screen towards us.

Pizza

11:07

for i2 , B is incoming at -y and outcoing at +y

John Rennie

Yes. So the two fields are going to have opposite signs at (4, 1)

B₁ will be negative and B₂ will be positive.

Pizza

but can B1 also be positive and B2 negative?

John Rennie

At the point (-4, -1) B₁ will be positive and B₂ will be negative.

So it depends on what point you choose.

Pizza

oh okay

B at A = -Bi1 + Bi2

John Rennie

Yes :-)

Pizza

11:15

for point 2)

-Bi1 + Bi2 = 0 so when Bi1 = Bi2

John Rennie

(2) just asks in what quarters of the diagram the field could be zero, and that can only happen when the signs of B₁ and B₂ are different. Yes?

Pizza

yes

sorry, I understand

they cancel out when they have opposite directions, therefore I or III quadrant

right?

John Rennie

Yes :-)

Pizza

for the last point we have to consider when B is outgoing / incoming so we have 2 cases and then we calculate the modulus

yes?

John Rennie

(3) is just asking you to calculate the Lorentz force:

F = v × B

Ah, OK, it only says v is aligned with the x axis. It doesn't say if v is positive or negative.

Pizza

11:24

yes

John Rennie

Although I would probably take aligned with to mean v points in the same direction as uₓ i.e. pointing in the pisitive x direction.

Pizza

when the field is incoming, the force is directed upwards while when it is outgoing, it is directed downwards

@JohnRennie yes

John Rennie

Well it says the charge is at the point A, so it depends on the direction of the field at A.

Pizza

cant we do the case of before?

16 mins ago, by John Rennie

At the point (-4, -1) B₁ will be positive and B₂ will be negative.

John Rennie

Yes, you could do the same calculation at this point.

Pizza

11:28

you mean to determine whether B is incoming or outgoing at A?

John Rennie

At (-4, -1) the total field will have the same magnitude as at (+4, +1) but it will be in the opposite direction. Yes?

Pizza

that is B tot that we found before

John Rennie

@Pizza Yes

Pizza

@JohnRennie ok

@JohnRennie yes

Binky

@JohnRennie hello

John Rennie

11:38

@Binky Hi :-)

Pizza

@JohnRennie I didn't understand what you wanted me to understand here

11 mins ago, by John Rennie

At (-4, -1) the total field will have the same magnitude as at (+4, +1) but it will be in the opposite direction. Yes?

hi @Binky

John Rennie

You said:

16 mins ago, by Pizza

when the field is incoming, the force is directed upwards while when it is outgoing, it is directed downwards

So at the point A = (4, 1) the Lorentz force points in the -y direction i.e. downwards.

So you were correct.

But then you seemed to be asking what the general rule was for any point.

Or did I misunderstand you?

Pizza

@JohnRennie I don't understand, but don't we need to find F1 and F2?

@JohnRennie why did you only consider this case

i.e when the field is outgoing

John Rennie

I don't understand what you are asking.

Are saying we should find the force due to B₁ and the force due to B₂ separately?

Pizza

$F_1 = qv B_{Ai_1} \sin \alpha \quad (\text{field is ...})$
$F_2 = qv B_{Ai_2} \sin \alpha \quad (\text{field is ...})$
$F_{\text{tot}} = \sqrt{F_1^2 + F_2^2}$

shouldn't this be done?

John Rennie

11:54

No, just add B₁ and B₂ together to get the total Bₜ

Then the force on the charge is F = qv × Bₜ

Pizza

ok and we had to understand if Bt was incoming or outgoing, right?

John Rennie

Yes

Pizza

I didn't understand this last thing

that is, how can I understand if Bt is incoming or outgoing

but it wouldn't be enough for me to calculate Bt, if it comes out positive, is it outgoing?

John Rennie

We have defined the direction of the z axis to positive z is coming out of the screen towards us. This is the usual convention i.e. if u𝑥, u𝑦 and u𝑧 are the unit vectors we define the z direction using:
u𝑧 = u𝑥 × u𝑦

Yes?

Pizza

yes

John Rennie

12:00

So any field B₁, B₂ or Bₜ is positive if it is coming out of the screen towards us and negative if it is going into the screen away from us.

So all we have to do is calculate B₁ and B₂ making sure we get the sign correct, then when we add them to get Bₜ the sign will tell us the direction of Bₜ

Doe this make sense so far?

Pizza

yes

John Rennie

Does that answer your question?

Pizza

but so I have to go and calculate Bt numerically to see if it comes out positive or negative?

John Rennie

Yes

Pizza

21 mins ago, by John Rennie

So at the point A = (4, 1) the Lorentz force points in the -y direction i.e. downwards.

and how did you understand it?

so have you done the calculations?

John Rennie

12:06

Well let's do the calculation. Start with B₁. We know this is going to be negative because it points into the screen, and the distance from the wire is 4 cm so we get:
B₁ = -μ₀I₁/2𝜋 × 0.04

Yes?

Pizza

yes

John Rennie

I₁ = 2 so that's B₁ = -μ₀/(𝜋 × 0.04)

And if we do the same calculation for B₂ we get:
B₂ = +μ₀3/(2𝜋 × 0.01)

Pizza

ok yes

John Rennie

So we are going to get:
Bₜ = μ₀3/(2𝜋 × 0.01) - μ₀/(𝜋 × 0.04) = μ₀10/(2𝜋 × 0.04)

Pizza

ok so it's outgoing

John Rennie

12:14

Yes, it comes out positive so we know it is outgoing.

Pizza

what I was wondering was whether you had actually done the calculations or had used another way

I mean, I wasn't understanding this

32 mins ago, by John Rennie

So at the point A = (4, 1) the Lorentz force points in the -y direction i.e. downwards.

John Rennie

I hadn't done the calculation but it was obvious that |B₁| < |B₂|

Pizza

when you wrote this, I didn't understand how you came to this conclusion

John Rennie

That's how I knew what the direction was.

Pizza

ah ok, no because I had only written down the formulas :/ so I didn't understand

Anyway, it's clear to me now

John Rennie

12:17

OK :-)

Pizza

A thousand thanks! sorry for the confusion at the end

John Rennie

You're welcome :-)

Pizza

I'm going to lunch now, bye :-)

John Rennie

Bye :-)

F. Zer

21:42

@JohnRennie I didn't want to interrupt the ongoing conversation, but what are outermost electrons ? In group 8, for example, I see the outermost shell has both s and p orbitals. And a second question, it is usual to speak about "s orbitals", "p orbitals" ? I thought orbitals where "rooms" inside each of "s", "p", "d" or "f" subshells.

Transcript for

Problem Solving Strategies