Problem Solving Strategies

Abcd

5:51 AM

@JohnRennie Hello.

John Rennie

@Abcd Morning :-)

Abcd

@JohnRennie Aren't 1 and 2 in series?

samjoe

Nope!!!

John Rennie

Start at the battery cathode. Is there a route that takes you round the circuit and back to the anode that passes through C1 without going through C2? If so then they are not in series.

Abcd

Okay, got it.

Abcd

6:30 AM

@JohnRennie Today can you answer that dielectric question?

Farcher

Abcd

Abcd I am sorry not to arrived here before but I am not really conversant with chat.

Abcd

@Farcher Okay, are you free right now to discuss about it?

Farcher

For about 20 minutes.

Abcd

Okay

@Farcher I have marked the imaginary metal plate with red. How can we say that the lower plate of the capacitor has the same magnitude of charge as the red plate?

Farcher

The red metal plate has a net zero charge on it.

Abcd

6:43 AM

yes

Farcher

If the top plate of the capacitor is has positive charge on it the top of the red plate has an equal amount of negative charge on it.

Abcd

why?

Farcher

That charge is

Abcd

What about the induced charges of the dielectrics? Don't they affect it?

Farcher

Those induced charges on the dielectric induce the charges on the metal plate.

Abcd

6:46 AM

but that magnitude is less

magintude of induced charges is given by $Q(1- 1/k)$ if I am not wrong

Farcher

If there were no induced charges on the red plate then there would be an electric field inside the red plate.

Abcd

my point is that the magnitude of the induced charges is less than the magnitude of charge on the capacitor plate

@Sid Do you have any idea

Farcher

Forget about your arrangement and consider a capacitor with dielectric inside it. One plate of the capacitor is earthed. To the other plate I add a charge +Q. How much charge is there on the bottom plate of the capacitor?

Abcd

@Farcher -Q/2 ?

Farcher

So when you charge a capacitor the magnitude of the charges on each of the plates is different?

Abcd

6:56 AM

inner surfaces of the plates would have same magnitude

Farcher

So that is what happens to the top of the red plate of your capacitor.

I wiill have to go in a minute or so as I have a dental appointment.

John Rennie

8:07 AM

@Abcd I think I've come up with a way of showing this ...

Alex K Chen

Can anybody tell me where do I read stuff to understand how electricity flows in some object ? I was reading chapter 29 in HRK, but it mentioned stuff like "lattice crystals" and "valence electrons" and "polarisation" which I didn't understand a bit.
Or alternatively, you can also tell where I can read to have an understanding of "lattice crystals" and "valence electrons" and "polarisation" (I have these in my chemsitry book too but my book is very very very very bad).
Thanks

Madhuchhanda Mandal

9:56 AM

Any wbjeeian here?

Madhuchhanda Mandal

12:33 PM

Which is more preferable among IIEST and JU if I want to go for higher studies? Any ideas? @Blue

Sid

12:45 PM

@MadhuchhandaMandal Not of WB, but JU trumps IIEST in all respects, I think. NIRF rankings say so too

Madhuchhanda Mandal

@Sid Triumphs you mean?

Sid

@MadhuchhandaMandal See the verb form

Madhuchhanda Mandal

@Sid okay :-)

Abcd

4:21 PM

@JohnRennie when will you be telling it?

John Rennie

@Abcd now if you want ...

Abcd

@JohnRennie Okay.

John Rennie

OK. Let's start with this diagram:

Abcd

@JohnRennie by $\epsilon $ you mean dielectric constant $k$ ?

John Rennie

I'm not going to insert any imaginary metal plates. What I'm going to do is calculate the voltage $V$ by minimising the electrostatic energy.

No, I mean $\epsilon$ where $\epsilon = k\epsilon_0$

Abcd

4:23 PM

@JohnRennie how and why will you minimise it

@Sid can you please see this problem^

John Rennie

To start with you have to take on trust the equation for the energy density of an electric field i.e. the energy of the field in joules per cubic metre. The equation is $$ \rho = \tfrac{1}{2} \epsilon E^2 $$

Abcd

@JohnRennie yes, I have derived it too

John Rennie

where $E$ is the field strength in volts per metre

Abcd

yes

John Rennie

OK, for the upper part of my diagram the potential difference is $V_0 - V$ and the distance is $d_1$ so the field strength is $$ E_1 = (V_0 - V)/d_1 $$

Abcd

4:26 PM

yes'

John Rennie

And likewise for the lower part the field strength is $$ E_2 = V/d_2 $$

Abcd

yes''

John Rennie

If $A$ is the area of the plates then the volume of the upper part is $Ad_1$

Abcd

yes

John Rennie

And the total energy is the energy density times the volume so: $$ U_1 = \tfrac{1}{2} \epsilon_1 \left(\frac{V_0 - V}{d_1}\right)^2 A d_1 $$

Abcd

4:29 PM

yes'

John Rennie

And likewise the total energy of the bottom part is: $$ U_2 = \tfrac{1}{2} \epsilon_2 \left(\frac{V}{d_2}\right)^2 A d_2 $$

Nehal Samee

@Abcd ... Is it $C= \frac{4}{3}$ ... ? I might be wrong ...

Abcd

@NehalSamee no

John Rennie

And we add the two to get the energy of both parts:

Abcd

@JohnRennie yes''

John Rennie

4:31 PM

$$ U = \tfrac{1}{2} \epsilon_1 \left(\frac{V_0 - V}{d_1}\right)^2 A d_1 + \tfrac{1}{2} \epsilon_2 \left(\frac{V}{d_2}\right)^2 A d_2 $$

Abcd

yes'''

John Rennie

Which I'll rewrite to simplify it:

$$ U = \frac{\epsilon A}{2d_1} (V_0^2 - 2V_0V + V^2) + \frac{\epsilon_2 A}{2d_2} V^2 $$

Abcd

yes

Sid

C is just 4, I think.

Abcd

@Sid why?

John Rennie

4:34 PM

Now I want to find out the value of $V$ that minimises the total energy $U$, so I just differentiate to get $dU/dV$ then set that expression equal to zero. OK so far?

Abcd

@JohnRennie why do you want to minimise it?

Sid

Let JR finish. Then, we can proceed with this...

John Rennie

Any system will settle into the lowest energy possible for it. That means the value of the voltage $V$ will be the one that minimises the total electrostatic energy.

3

Abcd

K

John Rennie

OK, so if we take my equation for $U$ and differentiate it then set it equal to zero we get:

$$ 0 = \frac{\epsilon A}{d_1} (V - V_0) + \frac{\epsilon_2 A}{d_2} V $$

Or:

$$ \frac{\epsilon_1 A}{d_1} (V_0 - V) = \frac{\epsilon_2 A}{d_2} V $$

Abcd

4:40 PM

then?

John Rennie

Now this should be starting to look familar ...

Abcd

its C

C1V0 - C1V = C2V

John Rennie

Because $\epsilon_1 A/d_1$ is the capacitance of the upper part if you treat it as a separate capacitor. And likewise for the bottom part.

And CV = Q, so this is just the condition for two capacitors in series: $$ Q_1 = Q_2 $$

And that's why the upper and lower parts behave just as if they were two separate capacitors in series.

Abcd

Oh wow Thanks @JohnRennie, got it.

John Rennie

@Abcd sorry it took a while, but I was trying to think of a simple way to explain it.

Abcd

4:44 PM

Okay.

@JohnRennie Do you have any idea about the previous problem?

John Rennie

@Abcd tomorrow ...

Abcd

My teacher did it in 2 seconds saying that "it's hit and trial and obvious."

@JohnRennie Okay. @Sid will explain then I guess.

@Sid Are you there

Sid

Yes. First tell me, what is your first idea on seeing that problem.?

Abcd

@Sid my first idea isnt good... And I now know its wrong ...so leave it.

Sid

Being wrong isn't the point. I want to know where to start. So, Your first idea will be helpful.

Abcd

4:49 PM

I thought that maybe inifnite capacitors would maximise it

so we need C to be greater than that

which is obviously wrong

And we know how to calculate the equivalent capacitance of infinite ladder

but not of finite ladder having n sections.

Sid

Good. Your idea isn't far off. And, this is actually an infinite ladder..

We know, the value of capacitance is independent of "n". So, even if there's one ladder or infinite ladder, the capacitance will remain same.

Do you agree?

Nehal Samee

@Sid ... Can we say that the equivalent capacitance of C , 4 and 2 will again give C ...?

Sid

@NehalSamee Exactly!

@Abcd Is it clear or should I draw you a diagram?

Abcd

@Sid Why are you considering it to be infinite

Sid

@Abcd Okay. Let it be finite. For the value of C, I can make the value of "n" as large as I want, but the equivalent capacitance won't change. Yes?

Abcd

5:01 PM

@Sid yes

Sid

@Abcd Okay. So, as large as I want, means I can keep adding the ladder and the equivalent capacitance won't change => This is an infinite ladder.

Abcd

@Sid he has clearly mentioned its a finite ladder

Sid

@Abcd No. He mentions that the ladder ends at C. What if, C is also a network of many capacitance?

I agree this might not be very intuitive though..

Abcd

@Sid Please read the first three words of the question

Sid

@Abcd So? If you read the whole question, it only says that the ladder terminates at C. It doesn't say what C could be.

Anyway, let's assume it's not infinite.

At least, we know that the equivalent capacitance doesn't depend on "n".

Jasmine

5:07 PM

He has also mentioned that the equivalent capicitance is independent of sections in between so it doesn't matter if it is 1,2,3 or infinity

Sid

(Be back in 15 minutes. I have to go take supper)

Jasmine

@JohnRennie awesome explanation sir. This was taught to us in a way I didn't get but the way you explain it's perfect.

Sid

@Abcd Okay back. Let's look at it this way.

Abcd

ok

Sid

Let the equivalent capacitance be $C_eq$

if I remove the elements to the left of the Yellow line, Then, still the equivalent capacitance will remain same because I just removed one ladder and equivalent capacitance is independent of "n".

So, basically everything right to the Yellow line has equivalent Capacitance same as the original. Do you agree? @Abcd

Abcd

5:25 PM

yes

Sid

So, we can replace the entire thing in the right as $C_e$

Meaning, we have only a simple Ckt. to solve now. @Abcd

Do you see why this is similar to the infinite network problem?

Abcd

yes

Sid

There's another small thing that you need to take care of. That is, the equivalent capacitance between A and B is actually C!

To prove that, since equivalent Capacitance is independent of "n" where "n" is the number of ladders of 2 and 4, let's put n=0. What do you get? @Abcd

Abcd

C

Sid

And since, equivalent capacitance is independent of "n", C is actually the equivalent Capacitance between A and B!

Abcd

5:29 PM

yes

then @Sid

How to calculate the value of C