Problem Solving Strategies

PinkAura

05:36

@JohnRennie hi sir

John Rennie

Hi :-)

PinkAura

i was just waiting for you, i have a doubt in semiconductors

John Rennie

I'm free now so go ahead :-)

PinkAura

@JohnRennie why did he put the battery terminals for PN diode like that ? he says that electric field in the diode is from N to P , so P must be at lower potential?

John Rennie

When current passes through a forward biased diode the voltage drops by about 0.7V. This is because the PN junction inside the diode has a potential associated with it called the junction potential.

Do you already know about this?

PinkAura

05:42

i know about the junction potential but N should be at higher potential?

John Rennie

If the current goes though a battery in the usual direction, i.e. from the -ve to the +ve, the potential increases by the EMF of the battery. Yes?

PinkAura

there's emf like that so the current flows?

John Rennie

I'm not sure what you are asking ...

What I'm saying is if you take a simple circuit with a battery and resistor then when the current flows through the battery the potential increases by E (and when it flows through the resistor it decreases by E).

Yes?

PinkAura

@JohnRennie i mean we have battery force pushing the current like that from + to - and so the current flows?

John Rennie

The battery does "push" the current, but all I'm saying is that when current flows through a battery in the forward direction the potential increases. And conversely if current flows through a battery in the reverse direction the potential decreases.

PinkAura

05:49

my network is little slow today, so it might take some time for my messages to be sent

@JohnRennie yes

John Rennie

And when current flows through a forward biased diode the potential decreases by 0.7V.

So a forward biased diode affects the potential in the same way that a reversed 0.7V battery does.

Both reduce the potential by 0.7V.

PinkAura

@JohnRennie umm let me think about it for a moment

John Rennie

Now a diode is not the same as a battery e.g. if you connect a resistor to a diode we don't get a current. But it does affect the potential in the same way as a reversed battery does.

PinkAura

@JohnRennie can you pls explain this?

John Rennie

Are you asking me why the potential decreases by 0.7V when current passes through a diode? i.e. what happens inside a diode to cause it?

PinkAura

05:55

yes

John Rennie

That's a long discussion, and in any case it doesn't matter. It just does!

PinkAura

uhh...okay then

so because potential decreases by 0.7V, it behaves like a battery opposite to current flow?

John Rennie

Yes

PinkAura

okay thanks

John Rennie

We can go into why the potential drop happens if you want, but you don't need to know it for the JEE.

PinkAura

06:08

@JohnRennie we can talk about it if i have time, but i think i can just accept it due to lack of time and also, it isn't in the JEE advanced syllabus so it's just there for mains..

John Rennie

Yes, in your place I would just accept it and move on :-)

NOTE Book

@JohnRennie Hi

John Rennie

Hi :-)

NOTE Book

I read about principle of capacitor

John Rennie

OK ... ?

NOTE Book

06:11

it said that when we bring an uncharged earthed conductor near another conductor its capacitance increases

So is it that capacitance of the system i.e. charged conductor and earthed conductor increases or only charged conductor's capacitance increases?

John Rennie

The capacitance of the two conductors increases, yes, but this is a rather complicated way to look at it. The only capacitors you will encounter in the JEE are parallel plate capacitor and those are much simpler.

@NOTEBook It's the capacitance of the system that increases.

NOTE Book

Um can you tell how? Because electric field between plates will increase because of negatives charges induced on earthed conductor. So dielectric will break down more early and should reduce it capacitance

John Rennie

There are two types of capacitance:
- self capacitance
- mutual capacitance

If you have an isolated conductor then it has a capacitance called the self capacitance.

This depends on the geometry of the conductor. It does not depend on the charge of the conductor.

NOTE Book

Ok

John Rennie

If you have two conductors then the system of the two conductors has a mutual capacitance - this is what we normally mean when we say capacitance as the self capacitance is usually small and not very important.

Again the mutual capacitance depends on the geometry of the system and not on the charges on the conductors.

NOTE Book

06:21

Ok

John Rennie

9 mins ago, by NOTE Book

it said that when we bring an uncharged earthed conductor near another conductor its capacitance increases

What you are doing here is changing the geometry of the system of the two conductors, and this changes the (mutual) capacitance due to the change in geometry.

Is there a question that has caused you to ask about this? If so maybe looking at the question would clarify things.

NOTE Book

It was just a topic that why spherical capacitor alone are poor capacitors where as parallel plate capacitor are very good capacitor

John Rennie

A capacitor works by storing energy in an electric field.

So a good capacitor is able to create a strong electric field from only a small charge.

NOTE Book

Oo

John Rennie

A parallel plate capacitor can create a strong electric field because the field lines can start on the positive charges and end on the negative charges on the other plate so you can have a lot of field lines between the plates.

NOTE Book

06:27

yes

John Rennie

An isolated sphere can have field lines starting on the sphere, but then the field lines have to go off to infinity getting more an more widely spaced as they go.

i.e. there are no opposite charges for the field lines to end on.

That means an isolated sphere cannot create as strong a field (for the same charge) as a parallel plate capacitor can.

NOTE Book

yes

John Rennie

I know this is a bit vague, but it is basically why there is a difference.

NOTE Book

But if a capacitor is able to create stronger fields from small charge then won't dielectric breakdown at smaller charges?

John Rennie

Dielectric breakdown doesn't affect the capacitance. It just affects the maximum charge the capacitor can hold before there is a bang and a lot of smoke :-)

NOTE Book

06:33

Oh right

John Rennie

It's certainly true that parallel plate capacitors always have a maximum voltage rating and if you try to go above that the dielectric will break down and the capacitor will burn out.

@entropy Hi :-) Ping me when you want to discuss this. There is a simple technique for doing these sorts of problems that I can explain.

NOTE Book

I guess i am confused with definition of capacitance. Capacitance shows how much charge it can store for some change in potential difference, right?

John Rennie

Yes

NOTE Book

So if a conductor have a large capacitance it should store more charge for smaller change in potential difference

John Rennie

For mutual capacitance it's the charge divided by the potential difference between the two conductors. For self capacitance it's the potential of the conductor relative to infinity.

@NOTEBook Are you asking about self capacitance now?

NOTE Book

06:45

It should be true for mutual capacitance too right?

John Rennie

Yes, though for mutual capacitance it's the potential difference between the two conductors.

NOTE Book

yes

So how is it that if stronger fields are caused by smaller amount of charge it is a good capacitor? It means more change in potential between the plates for smaller charge

John Rennie

I'd have to think about how to explain that in a simple way. I can't answer right now.

NOTE Book

oh ok

Thanks

John Rennie

Deleted

pi-π

07:06

Hey @JohnRennie

John Rennie

Hi :-)

pi-π

<table class="table-as-div">
<tr>
<td>
<h1 class="text-center" style="margin-top:30px;">some text</h1>
</td>
<td style="text-align:right;">
some text
<br /> <br />
<div style="width: 150px; height: 150px; border: 1px solid #000; float:right;"></div>
<br /><br /><br /><br /><br /><br /><br /><br /><br />
</td>
</tr>
</table>

Here can we not increase the font size of the "some text" in this h1 tag?
<h1 class="text-center" style="margin-top:30px;">some text</h1>

John Rennie

margin-top:30px; doesn't increase the text size ...

pi-π

Shouldn't this work?

<h1 class="text-center" style="margin-top:30px; font-size:240px;">some text</h1>

But it doesn't work when I try?

John Rennie

Where does the text-centre style code come from? An external CSS file?

pi-π

07:14

Yes.

John Rennie

If I do this:

<html>
<head></head>
<body>

<table class="table-as-div">
  <tr>
    <td>
      <h1 class="text-center" style="margin-top:30px;font-size: 100px;">Some h1 text</h1>
    </td>
    <td style="text-align:right;">
      some text
    <br /> <br />
    <div style="width: 150px; height: 150px; border: 1px solid #000; float:right;"></div>
    <br /><br /><br /><br /><br /><br /><br /><br /><br />
    </td>
  </tr>
</table>
</body>
</html>

I get:

So it is working.

pi-π

Okay.

John Rennie

Possibly it's something your CSS file is doing. If you right click the h1 text and choose Inspect you can see exactly what the style info is.

You will see something like this. Note that in my case the style info in the tag is top so it's controlling the size.

pi-π

I see.

Thank You.

John Rennie

You're welcome :-)

Pizza

08:38

@JohnRennie Hi :-)

mo-_-

Hi @JohnRennie

John Rennie

Hi :-)

mo-_-

Are you free?

John Rennie

Yes

Though Pizza asked first ...

Pizza

No problem, he can ask first

John Rennie

08:44

OK :-) @mo-_- go ahead.

mo-_-

ok

A metallic sphere with a volume τ₁, and surface charge density σ, possesses an electric potential V₁.
1. Determine the electric capacity and the electrostatic energy of the sphere.
2. If the sphere is brought into contact with another uncharged metallic sphere of volume τ₂, determine the charge and the final potential of the sphere with the smaller volume.

John Rennie

How far have you got with this?

mo-_-

σ = 11 μC/m², V₁ = 10 kV, τ₂ = 1/8 τ₁, k = 1/(4πε₀) = 8.99 × 10⁹ Nm²/C²

these are the data if they can be useful to us

John Rennie

Do you know how to calculate the potential of a charged sphere?

mo-_-

yes U = 1/2 C ΔV^2

where $C = 4\pi\epsilon_0 R$

John Rennie

08:52

That's the energy of a capacitor, not the potential of a charged sphere.

Yes, you need the capacitance of the sphere C = 4𝜋ε₀R

And the basic equation for a capacitor is Q = CV

where Q is the charge on the capacitor and V is the potential.

OK so far?

mo-_-

@JohnRennie aaaa

V = Q/kR

@JohnRennie yes

John Rennie

So we can find the potential using:
V = Q/C

We just need to calculate Q and C.

And we know C = 4𝜋ε₀R

And we are told the volume it τ₁ and we know τ₁ = ⁴⁄₃𝜋R³

OK so far?

mo-_-

Yes

John Rennie

Now, we are not told Q, but we are told the area charge density σ, and the total charge is σ times the area of the sphere A = 4𝜋R².

Yes?

mo-_-

yes Q = σ A

John Rennie

09:01

So we have:
Q = σ4𝜋R²
C = 4𝜋ε₀R

mo-_-

Yes

John Rennie

V = Q/C = σ4𝜋R²/4𝜋ε₀R = σR/ε₀

Yes?

mo-_-

Yes

John Rennie

The you can use τ₁ = ⁴⁄₃𝜋R³ to find R and substitute it in the equation for V.

mo-_-

$R = \left( \frac{3 \tau_1}{4 \pi} \right)^{1/3}$

$V = \frac{\sigma}{\varepsilon_0} \cdot \left( \frac{3 \tau_1}{4 \pi} \right)^{1/3}$

John Rennie

09:06

OK :-)

mo-_-

and thats point 1. !

John Rennie

Yes :-)

mo-_-

now Q total = Q1 + Q2 but Q2 = 0

so Q total = Q1

John Rennie

For part 2 we use the fact that if two conductors are connected they will have the same potential.

mo-_-

ah so V1 = V2

John Rennie

09:10

Yes. We know the total charge is Q, and this gets divided between the two spheres so we have Q₁ on one sphere and Q₂ on the other.

And for both spheres V = Q/C where Q is the charge on that sphere and C is the capacitance of that sphere.

Yes?

mo-_-

but it is said that the other sphere is uncharged

isnt Q2 = 0?

John Rennie

Initially uncharged. But when you touch the spheres together charge will flow off the first sphere onto the second sphere.

mo-_-

ok

John Rennie

So we end up with a charge Q₂ on the second sphere and a charge Q - Q₂ on the first sphere.
Yes?

mo-_-

Yes

John Rennie

09:16

So we have a potential V₁ = (Q - Q₂)/C₁ on the first sphere and V₂ = Q₂/C₂ on the second sphere.
Yes?

mo-_-

Yes

John Rennie

And the potentials have to be the same so V₁ = V₂

Can you take it from here?

mo-_-

$\frac{Q - Q_2}{C_1} = \frac{Q_2}{C_2}$

John Rennie

Yes

mo-_-

so i have to find Q2

John Rennie

09:19

Yes, you can solve for Q₂

mo-_-

$Q_2 = \frac{Q}{1 + \frac{C_1}{C_2}}$

John Rennie

Yes. And C₁/C₂ = 4𝜋ε₀R₁/4𝜋ε₀R₂

And the question tells you that τ₂ = 1/8 τ₁
Yes?

mo-_-

@JohnRennie yes

@JohnRennie yes so we can find R2

John Rennie

Well you can find the ratio R₁/R₂. So what is R₁/R₂ ?

mo-_-

2

do I also have to find C1/C2?

John Rennie

09:27

C₁/C₂ = 4𝜋ε₀R₁/4𝜋ε₀R₂ = R₁/R₂

Yes?

mo-_-

right

C1/C2 = 2

This ratio indicates that the larger sphere has twice the capacitance of the smaller sphere

John Rennie

Correct :-)

And since the potentials are equal that will mean Q₁ = 2Q₂

mo-_-

I got lost for a moment, but so how much is the total charge Q ?

However, I didn't understand the fact of the uncharger sphere, why only initially? it is not specified

John Rennie

We know Q = Q₁ + Q₂, so if Q₁ = 2Q₂ that means Q = 3Q₂

mo-_-

oh okay

so Q2 = Q/3

John Rennie

09:41

Yes

mo-_-

do i also have to find Q1 = ?

Q1 = 2Q/3

$V_f = Q_1/C_1 = Q_2/C_2$

John Rennie

Part 2 just asks for Q₂ and the potential of the small sphere.

mo-_-

ah ok so just Q2 = Q/3 and V2 = Q2/C2

John Rennie

Yes :-)

cOnnectOrTR 12

Hi @JohnRennie

mo-_-

09:48

@JohnRennie Thanks ! :-)

John Rennie

@mo-_- You're welcome :-)

@cOnnectOrTR12 Hi :-)

cOnnectOrTR 12

If a body moves from a to b on a line in n sec with variable velocity then its average velocity = b-a/n. Is average velocity average of all the instantaneous velocities ?

John Rennie

Yes

mo-_-

@JohnRennie I'm sorry but why did we find the potential before (1. point)? Shouldn't we have found electrostatic energy?

cOnnectOrTR 12

@JohnRennie But how does displacement/ time gives average of all the velocities of motion ?

John Rennie

09:55

@mo-_- It asks for both the potential and the energy,

For the energy you can just use U = ¹⁄₂Q²/C

mo-_-

ah ok, sorry for the interruption

cOnnectOrTR 12

10:09

average of a and b is a+b/2. Here average velocity in this sense is v1+v2/2. So Both definitions of average velocity should be equal. But how ?

Pizza

@JohnRennie Hi :-)

John Rennie

Hi :-)

Pizza

Can I ask or do you have to answer connector first?

cOnnectOrTR 12

@JohnRennie Is there any answer to this silly question? 😅

Pizza

Anyway I would like to ask this

Two flat metallic plates, each with a surface area $\Sigma = 0.8 \, \text{m}^2$, are placed parallel to each other at a distance $h = 4 \, \text{mm}$, forming a parallel-plate capacitor. The two plates are connected to a generator providing a potential difference $V$.

1. Calculate the capacitance of the parallel-plate capacitor.
2. The lower plate $A$ is fixed, while the upper plate $B$ is kept in mechanical equilibrium by a mass $M = 0.8 \, \text{kg}$, as shown in the figure.
Assuming the masses of the plates, the string, and the pulley are negligible, calculate the voltage $V$ at which t…

the green circles would be the pulleys

John Rennie

10:28

@Pizza Part 1 is easy. Yes?

Pizza

C = Q/V

but this is the capacitance of the capacitor

John Rennie

C = ε₀A/d
Yes?

See here for example

Pizza

wait we are in this case

right?

$C_0 = \frac{\varepsilon_0 \Sigma}{h}$

$E_0 = \frac{\Delta V_0}{h} = \frac{\sigma_0}{\varepsilon_0}$

$u_e = \frac{1}{2} C_0 \Delta V_0^2$

$u_e = \frac{1}{2} \varepsilon_0 E_0^2$

we have to use these formulas in our case

John Rennie

I don't understand what you are asking. If you have a parallel plate capacitor with plate area A and plate separation d then the capacitance is C = ε₀A/d

$C_0 = \frac{\varepsilon_0 \Sigma}{h}$
Yes, this if Σ is the arega.

Pizza

$C = \varepsilon_r C_0$

$E = E_0 = \frac{\Delta V_0}{h}$

$U'_e = \frac{1}{2} C \Delta V_0^2 = \varepsilon_r U_e$

$u'_e = \frac{1}{2} \varepsilon E_0^2 = \varepsilon_r u_e$

@JohnRennie these formulas are if a plate were inserted inside the armor, right?

so not our case

@JohnRennie ok

NOTE Book

10:37

@cOnnectOrTR12 average velocity is that velocity by which if you travelled for the same time interval you will reach the exact point that you did when you have variable velocity at different time. So avg. V = total displacement/ total time. V1+V2/2 is only a special case.

John Rennie

So part 1 is just that equation.

Pizza

yes ! :)

John Rennie

Can you see how to do part 2?

Pizza

@JohnRennie right?

John Rennie

Yes.

There is a force between the two plates pulling them together, so the string tension needs to balance that force.

So you have to find a way to calculate that force.

Pizza

10:46

$F = P \Sigma$

where P would be the electrostatic pressure

John Rennie

I would do it a different way. Suppose you pull the plates apart a distance dx, then you have done work W = Fdx. Yes?

Pizza

Yes

John Rennie

And that work has to go into the energy stored in the capacitor.

And we know the energy stored in a capacitor is U = ¹⁄₂CV²

Yes?

Pizza

yes

John Rennie

So you can calculate dU/dx, and from that get the force

Pizza

10:52

mm okay so : $U = \frac{1}{2} \left( \frac{\varepsilon_0 A}{x} \right) V^2$

@JohnRennie here is F = dU/dx or -dU / dx?

$F = \frac{\varepsilon_0 A V^2}{2x^2}$

John Rennie

I always get mixed up with signs. I usually ignore the sign and figure out what it should be the the end.

Pizza

i got this

NOTE Book

@JohnRennie sorry for interrupting we can use ∫dq*σ/2ε = F net on each plate too right?

John Rennie

@NOTEBook Yes.

NOTE Book

ok

John Rennie

10:57

There are several different ways to do this.

NOTE Book

yes

Pizza

now $V = Eh$ right?

John Rennie

Yes

Pizza

ok so now

t = Mg and t = F

John Rennie

Yes

Pizza

11:00

so F = mg

$\frac{\varepsilon_0 A V^2}{2x^2} = Mg$

John Rennie

I have to do something for a few minutes so I'll have to leave this with you.

Pizza

oh okay

John Rennie

There is a good article about it here

Pizza

$V = \sqrt{\frac{2x^2 Mg}{\varepsilon_0 A}}$

I think so, but maybe x = h?

NOTE Book

yes i got the same answer

Pizza

11:09

:-)

nice

Thanks!

PinkAura

@JohnRennie hi sir

John Rennie

Hi :-)

PinkAura

for ideal diodes, forward bias, resistance=0, reverse bias , resistance = infinity

John Rennie

Yes

PinkAura

but for non-ideal diodes, forward bias, resistance depends upon cutoff voltage (we can use that in the circuit to find current and all), what about reverse bias which is non ideal?

John Rennie

11:18

Reverse bias is actually pretty close to ideal.

But in reverse bias the current is very small but not zero.

That's where the non-ideality comes from.

PinkAura

what if we had to find that current (current through it), how would we do that?

John Rennie

Use an ammeter I guess. I think for a silicon diode the reverse current is of order 1μA so it is very, very small.

PinkAura

for example in this question, i neglected current through reverse bias and i could easily workout the ans, but what if it asked, current through reverse bias (D2)

John Rennie

The reverse bias current depends on the design of the diode so unless the question tells you what the reverse bias current is you wouldn't be able to answer.

PinkAura

ohh okay, thanks

and can you just tell what is cut in voltage of diodes?

John Rennie

11:24

I doubt a JEE question would ever need to you to take into account the reverse bias current unless it specifically told you what the reverse bias current was.

@PinkAura The current voltage curve of a diode looks like this.

PinkAura

yes

wait, it doesn't go on till infinity? the forward bias one?

John Rennie

There isn't a sharp cutoff voltage as it's a gradual transition, so we kind of mark Vd somewhere in the region where the current starts to rise.

For silicon diodes it's about 0.6 to 0.7V

PinkAura

@JohnRennie yes

John Rennie

@PinkAura Do you mean why doesn't the voltage on the x axis go off to infinity?

PinkAura

@JohnRennie yes

John Rennie

11:28

The current rises approximately exponentially with the voltage.

PinkAura

ohk yes, i am aware of that

@JohnRennie that's cut off voltage, ryt?

John Rennie

In principle the voltage could be taken to infinity but the current rises so quickly with increasing voltage that in practice the diode overheats and burns out before the voltage rises much above Vd.

PinkAura

@JohnRennie so it cannot rise further?

John Rennie

@PinkAura Yes, the cutoff voltage (Vd in this diagram) is somewhere around the "knee" of the curve.

PinkAura

@JohnRennie but what is "cut-in" voltage?

John Rennie

11:31

"Cut-in" voltage? I don't think I have ever heard that term used.

Did you see it in a question?

PinkAura

@JohnRennie umm yes

John Rennie

I think that's the same as the "cut-off" voltage i.e. two different terms for the same thing.

PinkAura

this is the ans

@JohnRennie but if it were the same as cut off then actually there wouldnt be any voltage, below 0.6V mark?

John Rennie

Do you want me to explain the answer or do you get it?

PinkAura

@JohnRennie i get this that, voltage cannot rise above this cut-in voltage,

but i have not heard of it either...

John Rennie

11:34

I think it just means "the voltage above which current starts flowing"

While the cut-off voltage means "the voltage below which current stops flowing"

So they mean the same thing just stated in different ways.

PinkAura

@JohnRennie im not sure how this definition justifies the answer?

John Rennie

Suppose we start the PSU at zero volts then gradually increase it i.e. V becomes positive.

If V < Vd (cut-off ot cut-in) no current flows. Yes?

PinkAura

yes

ohk i got it

John Rennie

And if no current flows the voltage drop across the resistor is zero because the voltage drop is ΔV = IR.

PinkAura

thank you so much

John Rennie

11:38

OK :-)

PinkAura

@JohnRennie yes , yes!

NOTE Book

@JohnRennie it would be better to say C=dQ/dV then just C=q/Δv right?

John Rennie

I guess so, though C is normally a constant so either way is fine.

NOTE Book

Ok

Thank you

John Rennie

You're welcome :-)

PinkAura

11:49

@JohnRennie sir, there's one more thing

in zener diode if there's no breakdown, it behaves as a simple reverse bias right?

ohh i guess, he left

mo-_-

12:13

@JohnRennie sorry but we have to assume $\tau_1 = 1 m^3$?

σ = 11 μC/m², V₁ = 10 kV, τ₂ = 1/8 τ₁, k = 1/(4πε₀) = 8.99 × 10⁹ Nm²/C²

then in the data it was written how much V1 was

can i write $R_1 = V_1 \cdot \epsilon_0 / \sigma$ ?

$R_1 = \left( \frac{3 \tau_1}{4 \pi} \right)^{1/3}$

$R_1 = \sqrt{\frac{{Q}}{{4\pi \sigma}}}$

since I don't know $Q$ and $\tau_1$ , then I should use the first way ?

Shashaank

15:25

Heyo @JohnRennie

Can you explain how the cat is both alive and dead in Schrodinger's cat thought experiment

Pizza

21:47

A square loop with side length $d$ is hinged along one of its sides to the $z$-axis, while the sides perpendicular to this axis form an angle $\theta$ with the positive direction of the $y$-axis. This loop carries a current $I_0$ in the direction indicated in the figure. A magnetic field is present in the space, parallel to and aligned with the $z$-axis, with a magnitude of $B_0$.
Calculate:
a) The force acting on each of the free sides of the loop.
b) The magnitude, direction, and orientation of the resulting torque with respect to the $z$-axis.

can we discuss this tomorrow morning? @JohnRennie

Transcript for

Problem Solving Strategies