Problem Solving Strategies

John Rennie

06:57

@F.Zer The outermost electrons are the ones with the highest energy. So for Neon the outermost electrons are the 2p because the energy of the 2p is higher than the energy of the 2s. If you ionise Neon to remove an electron you'd be left with 2s²2p⁵.

By "orbital" we mean 1s or 2s or 2p or 3d, and so on.

Maybe strictly speaking "orbital" should mean 2pₓ, etc the 2p contains three "orbitals".

But I think the meaning is not that precisely defined, and th word can be used either for the 2pₓ or the whole 2p.

mo-_-

08:55

@JohnRennie Hi

John Rennie

Hi :-)

mo-_-

can i ask for help?

John Rennie

Yes, I'm free :-)

mo-_-

The electrostatic field in a certain region of space is given by:
E = (5x uₓ - 4y uᵧ + 3z u𝓏) · 10⁵ V/m
Calculate: a) The flux Φ(E) through the closed surface shown in the figure with sides $a = 10 \, \text{cm}, \, b = 15 \, \text{cm}, \, c = 20 \, \text{cm}$.
b) The charge $q$ inside the surface.
c) The charge density $\rho$ assuming it is constant within the surface itself.

im sending the diagram

the main problem is how to find $\cos(\theta)$

John Rennie

How do you want to do this? Do you want to say what you've done so far? Or shall I start explaining it?

mo-_-

09:02

$\cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \|\vec{v}\|}$

i was thinking about this

John Rennie

In fact you don't need cosθ

mo-_-

but I don't think it should be used, I think there is a shortcut

John Rennie

There is an easy way to do the question

mo-_-

then it's better if you explain

John Rennie

OK :-)

Let's start with the top surface of the box. On this surface we have some field E with x, y and z components:
Ex = 5x
Ey = -4y
Ez = 3z

OK so far?

mo-_-

09:05

isnt Ey = -4y ?

John Rennie

Oops

mo-_-

@JohnRennie ok

John Rennie

Now to find the flux through the top surface we need to find the dot product E.A, where A is the vector area of the top surface.

And the direction of the vector area is along the z axis because the direction of a vector rea is always normal to that area.

mo-_-

A = b * a so E * a * b uz

John Rennie

Does this make sense so far?

mo-_-

09:08

yes

John Rennie

Now, Ex and Ey are parallel to the top face of the box, so their angle with the vector area is 90°

That means that the flux through the top of the box due to Ex and Ey is zero.

Only Ez creates some flux through the vector area.

mo-_-

yes uz * ux and uy * uz = 0

John Rennie

OK so far?

mo-_-

yes

John Rennie

@mo-_- Yes, exactly :-)

mo-_-

09:10

so (3z * 10⁵) * b * a

John Rennie

@mo-_- That's the x component of the field ...

mo-_-

oh right

ops

John Rennie

:-)

So the flux through the top of the box is just φ = 3z × 10⁵ × a × b

mo-_-

yes

but so that red arrow was just to confuse

John Rennie

Yes :-)

Well, I suppose not really.

Look at it this way

Suppose we write:
E = 5x u𝑥 - 4y u𝑦 + 3z u𝑧
A = ab u𝑧

mo-_-

09:14

ok

John Rennie

Where I've written the area A as a vector.

mo-_-

yes

John Rennie

Then:
φ = E.A = (5x u𝑥 - 4y u𝑦 + 3z u𝑧).(ab u𝑧)
= 5xab u𝑥.u𝑧 - 4yab u𝑦.u𝑧 + 3zab u𝑧.u𝑧

So I'm writing out the dot product in full using the field and area vectors.

Does this make sense so far?

mo-_-

yes

John Rennie

And as you said u𝑥.u𝑧 = u𝑦.u𝑧 = 0 and u𝑧.u𝑧 = 1

So this immediately simplifies to:
φ = 3zab

mo-_-

09:18

yes, it's clear now, that's all I didn't understand

John Rennie

And I didn't use any arguments about what vectors were parallel or normal, I just calculated the dot product.

And you can do this for all six faces. Yes?

mo-_-

so the bottom part is zero, because z = 0?

John Rennie

@mo-_- Correct :-)

And that's going to be true for all three faces that pass through the origin.

mo-_-

ah ok, it should be easy now

John Rennie

OK :-)

mo-_-

09:21

Can I ask something else?

John Rennie

Post the question, but I need to take a couple of mins to make a coffee. I'll read the question as soon as I'm back at my desk.

mo-_-

ok :-)

A charge $q = 1.5 \cdot 10^{-8} \, \text{C}$ is located in the median plane of a charge uniformly distributed with a density $\rho = 10^{-8} \, \text{C/m}^3$ between two infinite parallel planes separated by $d = 2 \, \text{cm}$. Calculate:
a) the electric field $E(x)$ due to the distributed charge;
b) the work $W$ done by the electrostatic forces to transport $q$ to a point $P$, located outside the charged region and at a distance $h = 3 \, \text{cm}$ from the closest plane.

I don't understand the text

John Rennie

What it is saying is that you have an infinite charged slab between the two planes. So the thickness of the charged slab is 2cm.

And the charge density inside the slab is 10⁻⁸ C/m³

So the total charge per unit area of the slab is:
σ = 10⁻⁸ × 0.02 C/m²

OK so far?

mo-_-

so there are 3 plans?

John Rennie

No. There is just a single infinite slab of the charged insulator. The two planes shown are the front face and the back face of the slab.

The slab is infinite but the drawing shows only a small part of it because obviously it's hard to draw an infinite slab!

The dashed lines are supposed to show the slab going out to infinity in all directions.

Is this clearer?

mo-_-

09:38

yes

so now we have a charge that is located in the median plane

John Rennie

The question refers to the slab as the "distributed charge"

@mo-_- Yes, the charge q is at the centre of the slab

mo-_-

and what is around the charge what is it

John Rennie

Do you mean what do the + symbols around q mean?

mo-_-

yes

John Rennie

The slab is an insulator with a constant charge density of 10⁻⁸ C/m³ everywhere inside the slab.

The + symbols are showing this charge density inside the slab.

mo-_-

09:45

ah ok

Can we go trough the exercise?

John Rennie

Yes, shall I start explaining or do you want to start?

mo-_-

explain pls

John Rennie

OK, the first part of the question is asking us to calculate the electric field due to the slab. It says "due to the distributed charge" but by "distributed charge" it means the slab.

And as usual to find the electric field we are going to use Gauss's law.

mo-_-

ah ok

can we use this diagram?

John Rennie

Yes, that is exactly how we are going to do this.

We will start by finding the field outside the slab i.e. to the left side and right side of the slab.

mo-_-

09:50

$\Phi(\vec{E}) = \iint_{\text{CIL}} \vec{E} \cdot \hat{n} \, dS =
\iint_{\text{LAT}} \vec{E} \cdot \hat{n} \, dS +
\iint_{\text{B}_1} \vec{E} \cdot \hat{n} \, dS +
\iint_{\text{B}_2} \vec{E} \cdot \hat{n} \, dS$

John Rennie

You seem to immediately reach for complicated equations, when there is a far simpler way to do this ...

mo-_-

so we do the basis (B1 - B2) first

@JohnRennie ah ok

John Rennie

Since the slab is symmetric the fields E on the two sides have the same magnitude. The only difference is one points left and one points right.

mo-_-

Yes

John Rennie

So the flux out of the cylinder on the left side is φₗ = EA and the flux out of the cylinder on the right side is also φᵣ = EA.

So the total flux flowing out of the cylinder is φₜ = 2EA.
Yes?

mo-_-

09:54

@JohnRennie why is not 0?

the vectors are opposite

John Rennie

On both sides the field flows out of the cylinder. Yes?

mo-_-

yes

John Rennie

The signs can be a little confusing here. Remember that the flux is E.A and on both sides E and A are in the same direction. Yes?

mo-_-

yes

John Rennie

So the product E.A is always positive even though the field E is in a different direction on both sides.

That's why the fluxes add instead of cancelling.

So we get EA + EA not EA - EA.

mo-_-

10:00

ok

John Rennie

Think out it this way: when we talk about the flux through a closed surface we mean the total flux flowing out of the surface. It doesn't matter about the exact direction - all that matters is whether the field is flowing outwards or inwards.

mo-_-

the one lateral should be 0 but I don't know why

John Rennie

And in this case both Es flow outwards so they are both positive. Yes?

mo-_-

yes

John Rennie

So the total flux flowing outwards is 2EA.

Are you happy with this?

mo-_-

10:04

yes

John Rennie

OK. Now we have to find the flux through the curved sides of the cylinder. Can you see what this is (it doesn't need a calculation)?

mo-_-

in theory it should be 0, but I don't know why

I don't see both an incoming and outgoing flux

Isn't it directed on the lateral surface like a ray?

John Rennie

It's zero because the electric field is normal to the slab, so it's parallel to the curved sides of the cylinder.

mo-_-

oh right

I was getting confused with the infinite thread case

John Rennie

mo-_-

10:11

yes

John Rennie

If we consider some small patch of the curved surface then the vector area points radially outwards from the axis of the cylinder. So everywhere on the curved surface E.dA = 0.

We only get a non-zero flux at the left and right ends of the cylinder.

mo-_-

yes

John Rennie

So that means the total flux flowing out of the cylinder is φ = 2EA

Yes?

mo-_-

yes

John Rennie

now to use Gauss's law we need to find the charge Q inside the cylinder. Then we can write φ = Q/ε₀.

So can you find the charge inside the cylinder?

mo-_-

10:15

$Q=\phi\cdot\epsilon_0$

$Q=2EA\cdot\epsilon_0$

John Rennie

Well yes, but we need to relate Q to the properties of the slab.

suppose the slab has a charge density of σ coulombs per unit area.

The area of the slab that the cylinder encloses is A

So the charge inside the cylinder is Q = σA

Yes?

mo-_-

but then does the area of the slab approximate that of the cylinder?

John Rennie

Thje part of the slab that is inside the cylinder is shown by the red line. Yes?

mo-_-

yes

John Rennie

And that red circle has the same radius as the two circular ends of the cylinder, so it has the same area A. Yes?

mo-_-

10:23

yes

John Rennie

So the area of the slab that is inside the cylinder (the red line) is also A.

And if the charge per unit area is σ then the charge Q inside the cylinder is the charge density σ times the area inside the cylinder A.

So Q = σA

Yes?

mo-_-

yes

John Rennie

So now we know φ = 2EA

And we know Q = σA

So we can write:
2EA = σA/ε₀

Yes?

mo-_-

yes

John Rennie

Then rearrange to find E and we get:
E = σ/2ε₀

Yes?

mo-_-

10:28

yes

John Rennie

This is a result that is worth memorising. The field outside an infinite charged slab is given by:
E = σ/2ε₀

Anyhow, to find the field outside the slab in our question we just need to calculate the area charge density σ. Yes?

mo-_-

yes

John Rennie

Can you see how to do this? The question tells us that the charge per unit volume of the slab is 10⁻⁸ coulombs per cubic metre.

And the thickness of the slab is d = 2cm.

mo-_-

i dont know

John Rennie

Suppose we take a 1m by 1m area of the slab. The slab has a thickness of d, so we get a box of size 1m by 1m by d metres. Yes?

Would a diagram help?

mo-_-

10:36

yes pls

John Rennie

OK give me a minute ...

@mo-_- We have a slab that is infinite, so I've drawn its edges as dashed lines, and the slab is d = 2cm thick.

And I've selected an area within the slab of 1m by 1m. So this part of the slab forms a box with sides 1m by 1m by 𝑑.

Is this clear?

mo-_-

oh sorry

John Rennie

It's only infinite in two dimensions i.e. it is infinitely wide and infinitely tall, but its thickness is finite and equal to 2cm.

mo-_-

the thickness does not change

John Rennie

Yes.

Like an infinite plane, but make the plane 2cm thick.

mo-_-

10:45

yes

John Rennie

Have you got it now?

mo-_-

yes

John Rennie

OK :-)

So I've selected an area of the slab 1m by 1m i.e. an area of 1m².

So this forms a box of size 1m by 1m by d.

mo-_-

yes

John Rennie

Then the charge per unit area is the total charge contained inside this box divided by the 1 square metre area.

Yes?

mo-_-

10:48

yes

John Rennie

And the charge inside the box is the volume V times the volume charge density ρ, and the question tells us ρ = 10⁻⁸ C/m³.

And the volume is V = 1 × 1 × d = d.

mo-_-

yes

John Rennie

So the charge inside the box is Q = ρd.

And since the area we chose was 1 m² the area charge density is:
σ = Q/A = ρd.

Ok so far?

mo-_-

clear

John Rennie

And we found earlier that outside the slab E = σ/2ε₀ so finally we can write:
E = ρd/2ε₀

And that's the answer for the field outside the slab.

Does this make sense so far?

mo-_-

10:57

I'm lost as to why he's the one outside

John Rennie

1 hour ago, by John Rennie

We used this diagram to find E, and the ends of the cylinder are outside the slab.

So we are finding E at the ends of the cylinder outside the slab.

mo-_-

so h cylinder >d

John Rennie

Yes. The slab is shown as the orange plane.

It's actually a sheet 2cm thick not a plane.

But the cylinder is longer than 2cm so it extends outside the slab on both sides.

mo-_-

now should we instead consider the case in which the cylinder is inside the slabs?

John Rennie

Yes, and in fact there's a shortcut we can use to do this easily.

mo-_-

11:02

:)

John Rennie

Give me a moment to draw a diagram ...

I've drawn an edge on view of the slab.

And I've drawn a dashed line through the middle of the slab.

And I'm going to find the field at the dashed line.

Does this make sense so far?

mo-_-

Yes

John Rennie

Now suppose I cut the slab in half along the dashed line.

So instead of a single slab of width d we have two slabs of width ¹⁄₂d

Yes?

mo-_-

Yes

John Rennie

And the fields outside the two half slabs are given by the equation we got earlier but now the thickness is ¹⁄₂d so the fields are:
E = ρ(¹⁄₂d)/2ε₀

Yes?

mo-_-

11:13

Yes

John Rennie

Now I've drawn in the fields from the two inner sides of the slabs as well.

I've coloured them to show what side is producing what field.

mo-_-

ok

John Rennie

And since electric fields add, the total field is the sum of all the fields I've drawn.

OK so far?

mo-_-

yes

John Rennie

So on the right side we have the black ¹⁄₂ρd/2ε₀ from the right side of the right slab plus the red ¹⁄₂ρd/2ε₀ from the right side of the left slab.

So the total field on the right side is ρd/2ε₀

And that's the same as we had before.

OK so far?

mo-_-

11:18

yes

John Rennie

And likewise on the left side when we add up the fields we get the same E = ρd/2ε₀ that we had before.

Now, what do we get when we add up the fields in the space in between the two slabs?

mo-_-

0?

they are opposite

John Rennie

Yes :-)

The field in between the two slabs is zero. And it stays zero no matter how big or how small the spacing between the two slabs is.

Yes?

mo-_-

Yes

John Rennie

So let's bring the slabs back together to form a single slab again, i.e. reduce the spacing between them to 0. Then the field where the two slabs join will still be zero.

And we've shown that the field at the centre of the slab is zero!

Yes?

mo-_-

11:27

Yes

John Rennie

So we have found the field at the middle of the slab, but suppose we want to find the field at some distance x from the middle, where 0 < x < ¹⁄₂d.

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

I've left the dotted line in to show the middle of the slab, but now I've drawn a dashed line a distance x from the middle of the slab, and we are going to find the field at the dashed line.

Can you suggest how we do this?

mo-_-

E = ρx/2ε₀

by intuition

John Rennie

Kind of. What we are going to do is split the slab like this:

Now the left slab has thickness ¹⁄₂d + x so the fields it creates are E = ρ(¹⁄₂d + x)/2ε₀

And the right slab has thickness ¹⁄₂d - x so te fields it creates are E = ρ(¹⁄₂d - x)/2ε₀

Yes?

mo-_-

Yes

John Rennie

Now are the fields outside still the same i.e. is the total field outside still
E = ρd/2ε₀

mo-_-

11:43

Yes

John Rennie

Yes :-) But now, what is the field in between the slabs?
Take right to be positive so the red field is E = +ρ(¹⁄₂d + x)/2ε₀ and the blue field is E = -ρ(¹⁄₂d - x)/2ε₀

mo-_-

px/ε₀

John Rennie

Yes :-)

And since the fields stay the same as we bring the slabs back together again we have now found the equation for the field at a distance x from the centre line!

mo-_-

nice!

John Rennie

OK :-)

Now if we set x = ¹⁄₂d that's back to the surface of the slab, so it should be the same as we got earlier for the field outside the slab.

And we can use this to check we got the correct equation.

So if we set x = ¹⁄₂d do we get the same as the field outside the slab?

mo-_-

11:51

$\frac{\rho d}{2\epsilon_0}$

yes

John Rennie

Yes :-)

So everything looks good.

Weg now have equations for the field both outside and inside the slab.

Yes?

mo-_-

Yes

John Rennie

So we have the complete answer to part (a)

mo-_-

yes

John Rennie

Now for part (b) we start with the charge q at the centre of the slab (where the field is zero)

mo-_-

11:53

ok

John Rennie

The we move q 1cm from the centre to the surface of the slab, then we move it 3cm more to a distance of 3cm from the surface of the slab.

mo-_-

ok

$\Delta V_1 = \int_0^{d/2} E(x) \, dx = \int_0^{d/2} \frac{\rho}{\epsilon_0} \cdot x \, dx = \frac{\rho}{2 \epsilon_0} \cdot \left(\frac{d}{2}\right)^2$

John Rennie

mo-_-

$W_1 = q \cdot \Delta V_1$

John Rennie

Yes, exactly :-)

mo-_-

11:57

@JohnRennie we break point b) in two steps

right?

John Rennie

Yes

Outside the field is constant so that's easy, it's just E₂ × 3cm

And inside we integrate

mo-_-

its - or + E2?

John Rennie

And that's the answer to part (b)!

mo-_-

$\Delta V_2 = E_{\text{2}} \cdot h$

?

John Rennie

If we start at the middle and move right then both E₁ and E₂ point right so they will have the same sign. Yes?

mo-_-

12:00

right

so $+$

John Rennie

Yes, the two parts will add.

mo-_-

$W_2 = q \Delta V_2$

$W = W_1 + W_2$

right?

John Rennie

Yes

mo-_-

the solution does this

John Rennie

I guess the solution added the two ΔVs first to get a total ΔV then multiplied that by q.

It's exactly the same as finding the two works separately and adding them.

mo-_-

12:06

ah ok

but so the whole slab has moved?

John Rennie

No, the slab stayed still and we moved the charge from the centre of the slab to 3cm outside the slab.

So the force on the charge was q times the field due to the slab.

mo-_-

So the charge is point-like?

John Rennie

Yes. We have a point charge q at the centre of the charged slab.

Pizza

Hi

John Rennie

Hi :-)

mo-_-

12:16

@JohnRennie and why did we before calculate Ex only of the slab and not both (the charge and the slab)?

John Rennie

Because the charge q is not affected by its own field. It is only affected by field of the slab. So we only needed to calculate the field of the slab.

Pizza

@JohnRennie Hi :-), is there time to ask something or is it too late and i come tomorrow morning?

John Rennie

@Pizza Is it quick?

Pizza

I would need a review of what I did because I don't have the solutions and I don't know who to ask for confirmation :(

That is, it would be an exercise

mo-_-

@JohnRennie ah ok

John Rennie

12:18

Post the exercise and I'll have a look.

If I can do it quickly I'll do it now.

mo-_-

A thousand thanks! This exercise was something absurd

John Rennie

@mo-_- It covered a lot of useful concepts!

The sort of calculations we did to solve it are techniques that you need to know.

Pizza

Consider four long straight parallel wires $f_n$, with $n = 1, 2, 3, 4$, which geometrically form the vertices of a square cross-section prism with side $L$. These wires carry steady currents of the same intensity $I$ and in the directions indicated in the figure.
Determine:
1. The total magnetic field at the center $C$ of the square.
2. The intensity of the force per unit length $F/l$ acting on wires 1 and 3.
Data: $L = 6 \, \text{cm}; I = 120 \, \text{A}; \mu_0 = 4 \pi \cdot 10^{-7} \, \text{H/m}$

ok so

John Rennie

I don't want to do it now.

I'll try to find time to go through it later today and I'll post the solution.

Pizza

ah ok, I had done it, but if you want I can write tomorrow

@JohnRennie ok, if you don't have time don't worry anyway

John Rennie

12:25

Ok

mo-_-

@JohnRennie Good to know

@JohnRennie anyway sometimes when I do the exercises I also compare myself with chatgpt to see how it does them. In your opinion I do wrong since it's a computer can it make mistakes?

John Rennie

ChatGPT can sometimes be helpful, but it often gets thongs wrong so I wouldn't rely on it.

There are lots of sites that have solutions to JEE questions, so I think it's better to Google and see if you can find the solution on a web site.

mo-_-

Yes, I do this too

Anyway I'm going to do some more exercises, bye and thanks for the help again

John Rennie

You're welcome :-)

John Rennie

13:26

@Pizza I get the field at the centre is Bₜ = μ₀I/3𝜋

For part 2 the field of the other three wires at wire 1 is Bₜ = μ₀I/4𝜋√2, and the force on wire 1 is just F = BₜI𝓁 where set 𝓁 = 1 to get the force per unit length.

I did this quickly so I can't promise there are no calculation errors.

Pizza

$F = \frac{\mu_0 i^2}{2\pi d}$

I found $F_{41y}, F_{42y}, F_{43x}, F_{12x}$

Maybe I'm completely wrong

$B_t$ should be $\frac{2\sqrt{2} \mu_0 i}{\pi L}$

@JohnRennie I don't know... I have to look at it better

John Rennie

13:44

We can discuss it now if you want ...

Pizza

Yes, that's fine for me

John Rennie

OK, start with I₁

The field lines from I₁ are anticlockwise circles centred at I₁ and with magnitude:
B = μ₀I/2𝜋R
where R is the distance from I₁ to C.

@Pizza OK so far?

Pizza

Yes

John Rennie

And the distance from I₁ to C is 3√2 cm
Yes?

Pizza

Yes

John Rennie

13:50

So B₁ = μ₀I/6𝜋√2

The magnitudes of all the fields will be the same since the currents and distances are all the same.

And B₁ and B₃ are in the same direction while B₂ and B₄ are in the same direction.

So B₁ + B₃ = μ₀I/3𝜋√2

Pizza

right

John Rennie

And B₂ + B₄ has the same magnitude but pointing top right.

Now when we add the two vectors the resultant is going to point straight upwards. Yes?

Pizza

Yes

B₂ + B₄ = μ₀I/3𝜋√2 ?

John Rennie

That resultant is the y component of B₁ + B₃ plus the y component of B₂ + B₄, so it is:
Bₜ = (B₁ + B₃)cos45 + (B₂ + B₄)cos45

@Pizza Yes

And cos45 = 1/√2

Pizza

yes from how we did it here it comes

Bₜ = μ₀I/3𝜋

John Rennie

13:56

Yes :-)

Pizza

then I thought that whoever wrote on the paper made a mistake

can i quick share?

John Rennie

Yes ...

Pizza

he just multiplied *4 and rewrote the radius another way

John Rennie

Their radius is the same as ours. They wrote r = L/√2 = 6/√2 = 3√2

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