electricitymagnetism-8

ECE802 – Lec08 

Note: the red color words or question marks are the words that I cannot recognized!! 

[00:00~10:13]

Electric field can induce dipoles in insolated.  Electrons in insulated are bounded, to the atoms and to the molecules unlike the conductors where they can slowly move.  And when I apply an external filed for instance if the filed is in this direction, that even though the molecule of atoms maybe completely spherical, they will become a little bit elongated in the sense that the electrons will spend a little bit more time there then they used to, and so this part becomes negatively charge and this part becomes positively charge.  And that creates a dipole.  I discussed that with you already during the first lecture, because that is something quite remarkable about this.  But, if you have an insulator, notice that the pluses and minuses indicate the neutral atoms.  And if now I apply the electric field which comes down from the top, then you see a slide shift of the electrons basically spend a little bit more time up and down.  And what you see now is……you see a layer of negative charge been created at the top and a layer of positive charge been created at the bottom.  That is the result of induction.  We called that …… also sometimes “Polarization”.  You polarize in the way the electric charge.  A substance says that do this we called them dielectrics.  And today, we will talk quite a bit about dielectrics.  The first part of my lecture is on the web. If you go 802 web you will see there a document with describes in a great detail what I am going to tell you right now.  Suppose we have a plane capacitor, plan which I charged with certain potential.  And I have on here said a charge plus sigma (+σ), and here I have a charge minus sigma (–σ).  I am going to call this free.  You will see very shortly why I call this free and this is minus free.  So there is a potential difference between the plates.  Charge flows on there.  It has an area A, and sigma free (σ_free) is the charge density – how much charge per unit area.  So we are going to get an electric field, which one is in this direction and I called that E_free.  And the distance between the plates says, is d.  So this is a given.  I now remove the power supply that I used to give the certain potential difference.  I completely take it away.  So that means, this charge here is trapped, can not change.  But now I am moving a dielectric.  I am moving one of those substances.  And what you are going to see here now…… at the top, you are going to see an negative induce layer.  And at the bottom you are going to see a positive induce layer.  I called them plus sigma induced (+σ_ind), and I called this minus sigma induced (–σ_ind).  The only reason why I calling the other free is to distinguish them from the induced charge.  This induced charge which I have in green, will produce an electric field which is in the opposite direction and I called that E-induced (E_ind).  And clearly E_free, is of course the surface charge density (–σ_free) divided by epsilon-zero (ε₀) and E-induced (E_ind), is the induced surface charge density (σ_ind) divided by epsilon-zero (ε₀).  And so the net E field, is the factorial the sum of two.  So E_net, I give the vector, is E_free (E_free vector) plus E_ind (E_ind vector), factorial added. Since I am interesting it, I know the direction already.  Since I am interesting it in magnitudes.  Therefore, the strength of the net E field (E_net) field is going to be the strength of the E-field created by the so called free charge (E_free) minus the strength of the E-field created by the induced charge (E_ind) minus ………… because this effect is down, and this one is in the up direction.  And so, if I now make the assumption that certain fraction of the free charge is induced. So I make the assumption that sigma induced (σ_ind), is some fraction b, times sigma free (σ_free). I just write now an i for induced and an f for free.  b is smaller then one.  If b will 0.1, it means that sigma induced (σ_ind) will be 10% of the sigma free (σ_free).  That is the meaning of b.  So, clearly if this is the case, then also, E of i (E_i), was also meaning b times E of f (E_f).  They are doing it immediately.  They are connected.  And so now I can write down, for E_net.  I can also write down E_free times (1 – b), and that (1 – b) now, we called…… 1 over Kaiba (1/K).  I called it 1 over Kaiba, our book calls it 1 over K.  I am so used to Kaiba that I decided to still hold on to the Kaiba.  And that K, or that Kaiba which ever you want to call it, is called “Dielectric Constant”.  It is a dimension number.  And so I can write down now……in general ……that E (E vector), and I drop the word, net, now.  From now on, when ever I write E, through out this lecture is always the net electric field.  Take both into account.  So you can write down now that E (E vector), equals the free electric field (E_free vector) divided by Kaiba (K).  Because the (1 – b) is (1/K).  And so you see in this experiment that I did in my head……First, bring the charge on the plate, certain potential difference.  Removing the power supply, shaving in the dielectric that E field will go down by the factor Kaiba.  Kaiba for glass is about 5.  That will be an major reduction.  I will show you that later.  If the electric field goes down in this particular experiment.  It is clear that the potential difference between the plates will also go down.  Because the potential difference between the plates, V, is always the electric field (E) between the plates, times d.  And so if this one goes down by a factor of Kaiba, if I just shave the dielectric, not change d, then of course, the potential between the plates is also going down.  None of this is so intuited.  I will demonstrate that later. The question now hold raises …… Does Gauss’s Law still holds?  And the answer is yes.  Of course, Gauss’s Law will still holds.  Gauss’s Law tells me that the closed loop……closed surface I should said, not closed loop.  The closed surface integral of E dot dA, is 1/ε₀, times the sum of all the charges inside my box (Q_ins).  All the charges, the net charges.  That must take to account both induced charge, as well as the free charge.  And so let me write down the “net” to remain you then.  But Q_net is of course Q_free plus Q_ind.  And I want to remain you that this is minus (a minus number for Q_ind), and this was plus (a plus number for Q_free).  A free charge, positive there is plus, and that the same plate, if you have your Gauss’s surface at the top, you have the negative charge to induced.  And so therefore, the Gauss’s Law simply means that you have to take both into account.  And so therefore, you can write down 1/ε₀ times the sum of Q_free.  But now you have to make sure that you take the induced charge into account.  And therefore, you divided the whole thing by Kaiba (K).  Then you have automatically take the induced charge into account.  So you can image a Gauss’s Law very easily by this vector of Kaiba. Dielectric constant is dimension list, I mention it already.  It is one in vacuum by definition.  One atmosphere, gases, typically has dielectric constant, just a hell larger then 1.  We will……most of the time assume that this 1, plus that it has a dielectric constant of 3.  And glass, which is extremely good insulator, as a dielectric constant of 5.  If you have an external field, that can induce dipoles in molecules.  But there is a substances however, which themselves are already dipoles.  Even in the absence of electric field.  If you apply now an external field, these dipoles will start to align along the electric field.  We did the experiment once with some grass seeds.  Perhaps you remember that! 
 

[10:14~20:17]

And as there is a line in the direction of electric field, there were a strength in the electric field.  On the other hand, because the temperature of the substances, these dipoles, these molecules which are now dipoles by themselves, through the chaos motion, will try to dis-align.  Temperature try to dis-align them.  So it is going to be a competition on one hand between the electric field, which was trying to align them, and the temperature was trying to dis-align them.  If the electric field is strong, you can get substantial amount of alignment.  A permanent dipole as a rule, a way stronger then any dipole that you can induced by ordinary means in the laboratory.  And so the substance which are natural dipoles, they have much higher values for Kaiba, much higher dielectric constants then the substances I just discussed, which themselves do not have dipoles.  What is the example is extremely good example.  The electrons spend a little bit more time in the oxygen then near the hydrogen.  And water has a dielectric constant of 80.  That is enormous. And if you go down to the lower temperature.  If you take the ice, –40 degree, it is even higher.  Then the dielectric constant is 100.  I am now going to mesa you through 4 demonstrations, 4 experiments. One of them you have already seen.  And try to follow them as closed as you can because if you miss one small step, then you miss perhaps a lot.  I have two parallel plats which I on that table as you have seen last time.  And I have here a current meter.  I put an A in there.  It means amp meter.  And the plates have a certain separation d.  I am going to charge this capacitor up by connecting these ends to a power supply.  And I am going to connect them to 1500 volts.  I am already going to set my light because that is the way you are going to see very shortly.  I am going to start of with a distance d.  So, this is going to be my experiment-1 with distance of d, 1mm. And the voltage, V, always means the voltage……the…the…potential difference between the plates, is going to be 1,500 volts.  Forgive me for the two Vs.  I can’t help with that!  This means here the potential difference, and this is the unit in volts.  Once I have charge them, I disconnect……this is very important……I disconnect the power supply, which I write the PS.  That’s it!  So the charge is now trapped!  As I charge it, as you saw last time, you will see that the amp meter shows the short surge of current.  Because as I put the charge on the plate, the charge has to go from the power supply to the plate.  And you will see a short surge of current, which will make the handle…the hand of the amp meter as you will see on the wall there, go to the right side, just briefly and come back.  This indicate that you are charging the plates.  And now I am going to open up the gape.  So this is my initial condition.  There is no dielectric.  And now I am going to go d to 7 mm.  And this is what I did last time.  The reason why I do it again because I need this for my next demonstration.  If I make the distance 7mm, then the charge which I called now Q_free, is really the charge on the plate, is now going to change this trap.  So there can be no change when I open up the gape.  That means the amp meter will do nothing.  You will not see any charge flow.  The electric field E is unchanged.  Because E is σ divided by ε₀.  If Q_Free is not changing, σ can not changing.  So, no changes in the electric field.  But the potential V is now going to go up by factor of 7.  Because V equals E times d.  E remains constant, d goes up, V has to go up.  And this is what I want to show you first even though you have already seen this.  And I need the new conditions for my demonstration that comes afterward.  I am going from 1,500 volts to about 10,000 volts.  It goes up by the factor of 7. You are going to see that there.  There you see your amp meter.  I am going to …… you see the …… oh …… this is the porpallor volt meter that we discussed last time.  And here you see the plates.  They are 1mm apart now.  Very close!  Then I am going to charge the plates.  I will count down so you keep your eyes on the amp meter……3……2……1……0……and you saw a current surge.  So I charged the capacitor.  It is charge now.  The volt meter doesn’t show very much 1,500 volts.  Maybe went up a little, but not very much.  But now I am going to increase the gape to 10…oh no…… to 7 mm, and look that amp meter is not doing any thing.  The charge is trapped so there is no charge going into the plate, but look what the volt meter is doing!  It’s increasing the voltage.  It’s now approaching almost 10,000 volts although this is not very quantitated.  And now I have a gape of about 7mm and that is what I wanted!  You see that the plate on the left side here. I now farther apart and there were be 4.  So that is my demonstration number 1.  The repeated of what we did last time.  So now comes number 2.  So now my initial conditions are ……. that V is now 10 kV.  That’s the potential difference between the plates that I have now, and d is now 7mm and I am not going to change that.  At this moment, Kaiba is 1.  But now, I am going to insert the dielectric.  So I take a piece of glass and I just put that into that gape.  Q_free can not go anywhere because I have disconnected the power supply.  So Q_free …… no change.  If there is no change in free charge, the amp meter will do nothing.  So as I plunging this dielectric, you will not see any reading on the amp meter. But, as we discuss that length now, the electric field (E), which is the net of electric field will go down by that factor Kaiba (K).  That’s the whole discussion was all about.  That’s going to be the factor of 5.  And since the potential (V) equals the electric field times d, but I keep d itself 7mm, I am not going to change it.  If E goes down by factor Kaiba, then clearly the potential will also go down by factor Kaiba.  So now you are going to see the second part.  And that is I am going to …… as it is now I am going to plunging this glass, 7mm thick, put it in there.  You expect to see no change on the amp meter, but you expect the voltage difference over the plates to go down by factor of 5.  So you will see that …… that the porpallor volt meter will have a smaller reflection.  Are you ready for this?  There we go!  Now you have smaller potential difference, but there was no current flowing to the plate from the plate. When I take it out again, potential difference comes back to the 10,000 volts.  So that is demonstration number 2.  Now we go to number 3.  But before we go to number 3, I want to ask myself a question …… what actually happened with the capacitance when I bring the dielectric between those plates.  Well, the capacitance (C), is defined as the free charge (Q_free) divided by potential difference over the plates. That’s the definition of capacitance.  And since this experiment as you have seen, the voltage went down by a factor of Kaiba.  The capacitance goes up by the factor of Kaiba because Q_free was not changing.  And so, since the capacitance as we derived last time for plate capacitor, …… I still remember it was the area (A) times ε₀ divided by separation d.  Since we now know there is a glass in plats, that the capacitance is higher by the factor of Kaiba.  This is now the meant we have to make.  To calculate the capacitance we simply have to multiply now by the dielectric constant of the thin layer that separates the two conductors. It’s the layer that has the thickness d that is in between the two plates.  In our case, I brought in the glass. 
 
 

[20:18~30:08]

I could write down a few equations now that you can always hold on to it to your life.  And you can also use them in the two demonstrations that follow.  And one is that E, which is always the net E.  When I write E is always the net one, equals the σ_free divided by ε₀ times Kaiba.  There comes that Kaiba that we discussed today.  Let’s call that equation number one.  The second one, is that the potential difference over the plate, is always the electric field between the plate times d.  Because the integral of E dot dl over a certain path is the potential difference.  That is not going to change.  And then, the third one that make me hand-in is the one I already have there.  C equals Q_free divided by the potential difference (V), which in terms of plate area, is A, times ε₀ divided by d, times Kaiba.  Lets call this equation number three.  Now comes my third experiment.  In the third demonstration, I am not going to disconnect my power supply.  So, now number three, I started out with 1,500 volts, just like we did with number one.  But, the power supply will stay there.  Through out, never take it off.  We start out with d, equals 1mm.  Just like we did in the experiment-1.  No glass.  I am going to charge it up.  Just like what I did with number one.  And of course, I will see that at the amp meter will show this charge. You see the surge of current.  Now I am going to increase d to 7mm.  Now something very different will happen from what we saw in the first experiment.  The reason is that the potential difference (V) is going to be fixed.  Because of the power supply is not disconnected.  Power supply stays its place.  Look now the equation number two.  If that V can not change, and if I increase d by factor of 7.  Now the electric field must come down by the factor of 7.  And so now the electric field (E) will come down by that factor of 7 because I go from 1mm to 7mm.  Now the electric field changes because the d goes up.  In case you will interest in the capacitance, the capacitance will also go down by the factor of 7.  Because if you look at this equation, Kaiba is 1.  If I make d go up by factor of 7, C goes down by factor of 7.  Just look at this as simple as that!  So C must also goes down by factor of 7.  Nothing to do with dielectric!  Nothing!  And so Q_free must now also go down by factor of 7 because if the potential difference doesn’t change.  But if Q_free goes down by factor of 7……oh……if C goes down by factor of 7, Q_free must go down by factor of 7.  This (C) goes down by factor of 7,  this (V) doesn’t change, so the free charge (Q_free) goes down by the factor of 7.  And what does that mean?  That means charge will follow from the plates, away from the plates, and so my amp meter now will tell me the charge flowing from the plate, and so that handle……that hand will go ~~ sphi~~ to the left. And so as I open up, depending upon how fast I can do that, charge will flow from the plates in the other direction.  The charge will flow off the plates, and that current meter will show you every time I open it a little bit…… it will go to this direction.  So, lets do that first.  No dielectric involved.  Simply keeping the power supply connected.  So I have to go back first to 1mm, which is what I am doing now.  I have here this thin sheet to make sure I don’t short them out about 1mm.  And I am going to now connected the 1,500 volts, and keep it on.  And as I charge it, you will see the current meter a surge to the right, right?  That always mean we charge the plates.  So there we go!  Did you see it?  I didn’t see it because I have to concentrate!  Did it go like this?  Good!  So now is charged!  We don’t take this connection off.  It is connected with the power supply all the time.  And now I am going to open up.  And as I am going to open up, the potential remains the same so this volt meter doesn’t give a damn.  It will stay exactly where it is because 1,500 volts remain 1,500 volts.  But now as we open up, we are going to take charge off the plate and so this I expect to go to the left.  Every time that I give a little jerk, I do it now, it went to the left.  I do it now again and I go to 2mm……go to 3mm……go to 4mm……make it 5mm……5mm……6mm……and I finally end up at 7mm.  Every time that I make it larger, you saw the hand goes to the left.  Every time I took some charge off.  So that is the demonstration that the number three.  Why did I go to 7mm?  You guess it!  Now I want to plunging the dielectric. So my experiment number four, I started with 1,500 volts, I started with d equals 7mm, and I am not going to change that.  There is no dielectric in place.  But now I put the dielectric in.  So, Kaiba goes in.  Now is going to happen.  Well, for sure, V is unchanged because it was connected with power supply.  So that cannot change.  What happened with Q_free?  Look at this equation…… When I put in the dielectric, I know that the capacitance goes up by factor of Kaiba.  C will go up by factor of Kaiba.  If C goes up with factor of Kaiba, and if V is not changing, then Q_free must go up by a factor of Kaiba!!  So there is a meaning from equation–3.  So this (Q_free) must go up by the factor of Kaiba.  Well, does that mean that the charge will follow to the plate?  I increase the charge on the plates, and so my amp meter will tell me that.  So my amp meter will say …… a ha! I have to put the charge on the plate and so my amp meter will now do this!  And that is what I want to show you.  The remarkable thing now is that the electric field E, the net electric field E WILL NOT CHANGE!  And you may say…but you put in the dielectric!  Yah…… I put in a dielectric, but I kept the potential difference constant.  And I kept the d constant.  And since the V always E times d, if I kept this 1,500 volts (V = 1,500 volts),  and I kept the 7mm the 7mm (d = 7mm), then the net electric field cannot change.  Exactly what it was before!  That is the reason why Q_free has to change.  Think about that!  Because you do introduce induced charge on the dielectric.  And you have to compensate to that to keep the E field constant.  And the only way that the nature can compensate for that is to increase the charge on the plate, the free charge (Q_free).  So, that is what I want to show you now.  It is the last part.  So I am going now to put in the dielectric.  And what you will see then, is that current will follow on to the plate.  So the porpallor (porpallor volt meter) will do nothing, to sit there.  And you will see this one (amp meter) goes clockwise when I bring in the glass, and then goes back (counterclockwise) of course.  There is only a little charge that comes off, and then it will go back, too.  As I plunging in, you will see charge following on to the plate.  There we go!!  Are you ready for it? ……3, 2, 1, 0!  And you saw charge following on to the plate.  When I remove the glass, of course, then the charge goes off the plate again.  You see that now!  
 

[30:09~40:48]

I have shown you four demonstrations.  None of this is into an intuitive.  Not for you, and not for me!  When ever I do these things, I have to very carefully sit down and think what actually is changing and what is not changing.  I have no got feeling for that.  There is not something in need it says that ……oh, yes, of course, it’s going to happen.  Not at all!  And I don’t expect that from you either.  The only advice I have for you, when you dealing these cases where by dielectric goes in, dielectric goes out, plates separate, plates not separate, power supply connected, power supply no connected, approach in a very cold blood its way, in the real classic MIT way, very cold blooded, think about what is not changing.  And then pick it up from there, and to see what the consequence will be.  How can I build a very large capacitor?  One that has a very large capacitance.  Well, capacitance, C, is the Area (A) times ε₀ divided by d times Kaiba.  In your book always uses K.  So get this K make K large, make A large, and make d smaller as possible as you can.  Ah……but you have the limit for d.  If you make d too small, you make a spark between the conductors.  Because you may exceed the electric field, the break down electric field.  So you must always stay below that break down field, which in air will be 3 million volts per meter.  If you want a very large Kaiba, you will say……well, why don’t you make the layer of water in between.  There is a Kaiba of 80.  Ah…… the problem that the water has a very low break down electric field, so you don’t want the water.  If you take poly-afftulate, I just called it poly here.  Just as a revelation.  Poly-afftulate has a break down electric field of 18 million volts per meter and it has Kaiba I believe is 3.  Many capacitors I made by the layer in between is poly-affulate layer. Although mica will be real superior.  Be as a made, I want to evaluate now with you two capacitors which each have the same capacitance of 100 microfarad (μf).  But one of them, the manufacturer says that you could put maximum potential difference of 4,000 volts over it.  That’s this baby!  And the other, I go to Radioshake, and it says you cannot exceed the potential difference, not more then 40 volts.  Well, if I have the poly-affulate in between the layer of the conductors, then I can calculate what this thickness issue can be before I get break down.  That is very easy.  Because V equals E*d.  And so I put in here 18 million volts per meter, and I go to 4,000 volts, and then I see what I am was d, and it turns out that the minimum value for d.  You cannot go any thinner is than 220 μm.  And so for this one, it is only 2.2 μm.  You can make it much thinner because the potential difference is 100 times lower.  So you can make the layer 100 times thinner before you get the electric break down.  I want the two capacitors to have the same capacitances.  That means, since it have the same Kaiba and they have the same ε₀, it means that A/d has to be the same for both capacitors. So A/d for this one must be the same of A/d, for that one.  But if d here is 100 times larger then this one, then this A must also be 100 times larger.  Because A/d is constant. So if A here is 100, then A here is 1.  But now think about it, what determine the volume of the capacitor?  That’s real the area of the plates times the thickness, and if I ignore for now the thickness of the conducting plates, then the volume of the capacitor clearly is the product between the area and the thickness.  And so, it tells me then the capacitor, which has a 100 times larger area, is 100 times thicker, will have 10,000 times larger volume then this capacitor.  And this baby is 4,000 volts, 100 μf.  It has a length of about 30 cm……10cm like this, 20cm high, that is about 10,000 cm³, 10,000 cm³. You go to Radioshake, and you buy yourself a 40 volt capacitor 100 μf, which will be 10,000 times smaller in volume that will be only 1cm³.  and if I have one of them behind my ear, you wouldn’t even notice that, would you?  Could you tell me what it says here?...……100 μf, how many volts?……… 40 volts.  That is small!  Compare to one, which can handle 4,000 volts, but the capacitance is the same.  So you see now the connection with the area and with thickness by no means trivial.  All this has been very rough on you.  I realized that!  It takes time to digest it and you have to go over your notes.  Therefore, I put the remaining time.  We have quite some time left.  I will try to entertain you with something a little bit easier.  A little nice to digest.  Professor Monsenbrouk in the northland invented………Ya! You can say he invented the capacitor, was an accidental discovery.  You call them a liner jar because he works in liner. And the liner jar is the following.  This is a glass bottle.  So all this is glass.  That’s an insulator.  And he has outside the insulator.  He has two conducting plates.  So, it is a beaker outside, and there is a beaker inside conducting.  That’s the capacitor although he didn’t called it capacitor.  And so you charge this up and so you can have plus charge (+Q) here and minus charge (–Q) on the inside.  And he did the experiment with that.  The energy (u) store in the capacitor, we discussed that last time, equals 1/2, times the free charge (Q_free), times the potential difference (V).  If you prefer (1/2)*CV², that’s the same thing.  I have no problem with that.  Because the C, is Q_free divided by V so it is the same thing.  What I am going to do………I am going to put a certain potential difference over a liner jar.  I will show you the liner jar that we have.  You see there.  And once I had put in……put on some potential difference, put on some charge on the outer surface not on the inner surface.  You can see the outer surface there.  The NO. 1(the projector No. 1) is hard to see. I will show that event later to you. So here you see the glass, and here you see the outer conductor. And there is an No.1 and 2 which you can see very well.  Once I have down that, I will disassemble it.  So I first charge it up.  So there is an energy in there.  This much energy, ( (1/2)CV² ).  Then I will take the glass out.  I will put the outside conductor here, inside conductor here.  I will discharge them completely.  I will hold in my hand.  I will touch it with my face.  I will lick them.  I will do anything to get all the charge off.  And I will reassemble them.  Well, if I get all the charge off, all these Q_free……phi……goes away.  There is no longer potential difference.  When I reassemble that baby then clearly, that couldn’t be an energy left.  And the best way to demonstrate that to you is to take this plunk which I have here, conducting plunk and see weather I can still draw a spark by connecting the inner part with the outer part.  Then you will not expect to see any thing.  So it is something that is not going to be too excited.  But lets do it any how!  So here is the liner jar and I am turning the wheeler to charge it up.  I am going to remove this connection……remove this connection.  Take this out.  Take this out!  Come on!  Believe me, no charge on that anymore.  This one, work out……believe me.  There you go!!  And now lets see what it will happened when I short out the outer conductor with the inner conductor.  Watch it!! ……… Pa!……… (electric spark sound)……That is amazing!  There shouldn’t be any energy on that capacitor.  Nothing!!  And I saw the huge spark.  It is not even a small one.     
 

[40:49~52:35]