8.02x - Lect 8 - Polarization, Dielectrics, Van de Graaff Generator, Capacitors

[@lecturesbywalterlewin.they9259]

This website contains all my 94 course lectures (8.01, 8.02 and 8.03) with improved resolution. They also include all my homework problem sets, my exams and the solutions. Also included are lecture notes and 143 short videos in which I discuss basic problems. ENJOY!

[@aditya-lr2in]

Lectures by Walter Lewin. They will make you ♥ Physics. Hi sir how are you

[@srcsohan382]

Which website?

[@Jan-vz8kj]

Sir, Thank you so much for these Lectures, I really enjoy them!

[@andredelacerdasantos4439]

Please, professor, please make sure your videos will continue to be freely accessible after you leave us, we will always need you!

[@egghog3342]

@@andredelacerdasantos4439 when did he said he'll remove?

[@positivegradient]

A hundred years from now, young minds will see the beauty and feel the irresistibe excitement of physics because of these lectures. They are no less than the pyramids of Giza or the masterpieces of Picasso.

[@theunknownscientist3249]

I'm having so much fun just by watching these, I love all Mr. Lewin's spirit, he's the best teacher I never had.

[@rahuljaiswal-hb3ex]

sir , really you are god for billions of students, your teaching is absolutely great and unbelievable good .

[@lecturesbywalterlewin.they9259]

:)

[@varunahlawat9013]

@%F0%9F%98%9E/UCiEHVhv0SBMpP75JbzJShqw Sir, thanks lecture is amazing!! But I still don't get that, why needle goes to left in demonstration 3. please anyone?

[@ДҜ74Џ]

Billions is a stretch but he is amazing

[@jonathanseagraves8140]

The demonstration included in this video is mind blowing and simple. This is one of the gems that offers a bridge to connect you to a higher intuition.

[@hdheuejhzbsnnaj]

When I saw that girl sleeping on camera, I had to laugh because she paid $40k to sleep and I paid $0 to be on the edge of my seat.

[@rasmusturkka480]

U r such a clever little boy

[@syoudipta]

0:54 Dipole formation 15:21 Demo 1 18:20 Demo 2 20:22 Equation set 24:52 Demo 3 29:15 Demo 4

[@homosapien5684]

You sire, are a true hero

[@mewsicman9541]

Thanks, appreciate it

[@lochinvar00465]

I've had half a century of experience in the electronics field and I have never before had such a clear description of the way capacitors work.

[@lecturesbywalterlewin.they9259]

:)

[@anubhaagarwal4224]

Sir I’m enthralled and I just can’t thank u enough for ur contributions as a soul to the education how I wish u be blessed forver and we can take ur legacy forward

[@lecturesbywalterlewin.they9259]

So nice of you

[@owl6218]

Prof. Lewin has put in SO much work into the demonstrations.....hope the students make good use of the demonstrations. Because, in my experience as a teacher I have faced frustration in the matter. No matter how you put it, science has a high 'intrinsic cognetive load', so, no matter how it is put across, the student still needs to walk all the way to grasp it.....

[@nawalsingh666]

Brilliant Lectures with amazing experiments. I hated Electromagnetic theory before this, now I can't stop watching your lectures. You are resolving a new mystery in every experiment. ♥❣

[@lwhite4763]

You have a wonderful way of explaining information that is often very diffifcult to comprehend; Your Sharing this knowledge with any one who wants to learn and doing so both effectively and efficiantly. Your work is appreciated. Thank you

[@ryanarendt6196]

Thank you so much for keeping these on the web! Physics has always been something I've loved....a looooong time ago I watched many of these on open courseware, but that was years ago, much of which went over my head but it was a great introduction. Now taking university level physics it's so great to come back with a conceptual and mathematical foundation and soak in as much as I can. Thank you!

[@lecturesbywalterlewin.they9259]

:)

[@rga8895]

Your lectures are truly a gift!

[@lecturesbywalterlewin.they9259]

:)

[@liangchen3605]

Hello, Prof.Lewin, at 44:26(you say we do positive work), if there is no E field inside the Van de Graaff sphere, what force is repelling us to bring the charge inside the sphere? Thanks.

[@lecturesbywalterlewin.they9259]

there are 2 issues here. 1. energy is needed to place charge on the belt 2. energy is needed to rotate the belt. 3 There is E-field between the belt and the metal "comb" *inside the dome* which removes the charge from the belt

[@liangchen3605]

So does this mean the mechanical energy used to rotate the belt is partially converted to the electric potential energy of the charge near the metal comb inside the dome? what I am thinking is something like lifting up a ball, where the work done by us against the gravity is converted to potential energy of the ball. I don't know if there is any similarity between these examples. On the Wikipedia page, it says "The larger the sphere and the farther it is from ground, the higher will be its peak potential". So I think the gravitational PE is related the electrical PE for the Van de Graaff generator. Is this correct? Sorry about this long message. And thanks again for the quick reply.

[@asdfbaseqsfawefbaef7738]

Thank you professor for the most excellent lectures! You're the reason I am taking this course!

[@lecturesbywalterlewin.they9259]

kzbin.infoesyw1xihCWk

[@NicolasSchmidMusic]

I understand the Liden Jar like this: The strong E field between the two conducting bottels creates some corona distrages, so charges go on the dielectric "bottel". Then wenn you discharges the conducting bottels and put everything back together, the charge on the dieletric bottel (different sign inside and outside) creates a potential difference between the two conducting bottels and thus, wenn you shortcut them, it creates a spark :)

[@lecturesbywalterlewin.they9259]

:)

[@cricworld6797]

Indeed it will take me considerable time to digest those.Sir you are the hero of my world.❤❤

[@Praveensingh-vj3fg]

Thanks so much sir ,these lectures with full of experiement is great work in the history of world and Who knows how many physicists will emerge with the help of these lectures

[@berserkslayer8638]

29:58 this is what I love about these lectures.

[@miranaangel9622]

You are the best professor I have ever seen. I got an A+ in my class by only watching your videos, didn't even do homework. LOL

[@mustaphaalkhafaaf5512]

this is what I call lecturing , not like the rubbish universities which just talk like the parrots and do it for the sake of their salary .... walter U bent geweldig dank U , Ik heb 20 geleden in nederland gewoont en ik kan nog steeds nederlands spreeken ,,, you are a legend Sir.

[@montserratzetina9950]

So if you, a student, had to establish a question from the minute 36:19 - 41:15, what would be the question if you’re expected to answer that same question based on the information given to you from minute 32:5 - 50:09 ?

[@aakashsrivastava6486]

I love u sir for anything and everything u do and speak. I am learning a lot from u😊 Wish u a busy, healthy and happy life😊

[@johnbalvin5401]

That was a wonderful demostration , I would have wanted to be in that class

[@lecturesbywalterlewin.they9259]

what was your explanation?

[@johnbalvin5401]

The glass its polarized, I think, or maybe the base has some charge that its tranfer into the plates

[@lecturesbywalterlewin.they9259]

You will have to be wayyyyyy more specific for a proper explanation

[@johnbalvin5401]

There are two cylinders(that previusly I called plates) made of copper , 1) when the cylinderes are been charged, some of the charge go to the base (I dont know what material it is) so after discharging the cylinders , the charge that was in the base is returned to the cylinders, so can be a potencial diference between the cilinders. or maybe 2) When the cylinders are been charged, it creates a magnetic field inside the glass, so the atoms in the glass line up with the magnetid field, and when the cylinders are without charge, the magnetid field that is in the glass makes in any way I dont know charged the cylinders again. or maybe. 3)Everything is discharged ; but when putting back the second cylinder there is friction, some of the charge go to the glass and the other charged its in the the cylinder inside, so when putting near the "promp" the charge the was in the cylinder inside the glass it travels to the "promp" so it discharge it. (I dont know what it name is of the "promp" (I´m spanish speaker))

[@lecturesbywalterlewin.they9259]

you are not even close. I do give the explanation + demo in a later lecture. I suggest you watch it.

[@王溯]

Hello Dr. Lewin, at 18:30 you put the 7mm glass between the two plates and then take it out, there is expected to see no current through the amp meter, but I see the amp meter moved a little bit when the edge of the glass is going in and out of the area between the two plates. In addition, the amp meter goes towards two different ways when the glass is put in and taken out. Would you please explain this? THANK YOU.

[@lecturesbywalterlewin.they9259]

The amp meter is doing exactly what it is supposed to do. Please figure it out!

[@王溯]

I think the little move of the amp meter is caused by the little bit current which is flowing towards the propeller volt meter and back. Am I getting it right？

[@dihan6130]

Lectures by Walter Lewin. They will make you ♥ Physics. I do see a tiny bit movement about the amp meter too. I think it may caused by the creation of the induced electric field which has opposite direction and creates a temporary weak current going through the amp. May I have your opinion sir?

[@TheGREDATV]

Very nice lecture like always Dr. Lewin, today playing whit the factor K is pushing integrated electronics beyond his limits, in the same circuit we can have high-K materials for high capacitance in transistor gate that ultimate can translate in more speed, and low-K materials for reduce the capacitance between interconnections and have lower noise in the system, all that in the nanometer scale at will, how amazing. And like always physics works

[@lecturesbywalterlewin.they9259]

:)

[@amalsrivastava6853]

During this lockdown i was really worried how will i do physics when i dont understand a bit of electrostatics. Thx to Dr Walter Lewin not only do i understand it but i also experienced it. Greetings from India

[@praveenkumarsingh8178]

Real definition of teacher

[@sanatanisher4639]

Brilliant lecture sir....love❤️ from India 🇮🇳🇮🇳🇮🇳

[@afifakimih8823]

"physics is so beautiful...."

[@InventTwig]

At 18:43 when Mr lewin inserts the glass plate a small deflection is seen on the ammeter and when he removed it there was negative deflection , even when he said that no current would flow , how did it happen ???? I'm terribly confused please help

[@lecturesbywalterlewin.they9259]

InventTwig I looked at the demo. There is a HUGE change in the potential difference (factor kappa), there is a minute current when the glass enters and when I remove it. It is so small that I had never even noticed it and no one in the class commented on it. Good for you that you noticed it. Compare the current with the current when I keep the power supply connected. The difference is HUGE. Notice that the minute current is larger when I bring the glass in than when I removed the glass. That's because I moved the class in much faster than I removed it (look again at the demo). I conclude that if I had moved the glass in very slowly and removed it also very slowly that there would have been NO current. That should be a clue to what caused these minute currents. I suggest we both think about why this happened and why it is related to the speed at which I move the glass.

[@lecturesbywalterlewin.they9259]

Lectures by Walter Lewin. They will make you ♥ Physics. I think I found the explanation. Key is that the minute current that we see depends on the speed with which I move the glass (that's an observational fact). Moving the glass changes the E field. The faster I move the glass the larger will be dE/dt. Maxwell's equations will tell me that this causes a B field. But since my dE/dt is not constant, there will be a dB/dt thus a change in magnetic flux and that, according to Faraday's law will cause a current. What do you think of this?

[@InventTwig]

Lectures by Walter Lewin. They will make you ♥ Physics. Yes Mr lewin I think you are right , I too was that thinking about this , I did too notice that while removing the glass plate the back current in the opp direction was low , I thought that it might be happening because of the changing potential ,that when you shove in the dielectric really fast the potential decreases real fast and the changed pd causes the charge to move and when the glass plate is removed slowly the current is low and in the opposite direction because the pd changes slowly because the glass plate is removed slowly. Mr lewin what do you think about this. I think that your explanation explains the deflection more correctly. Please tell me your thoughts on this. Thank you Mr lewin :)

[@InventTwig]

Lectures by Walter Lewin. They will make you ♥ Physics. Ps:Mr lewin can you please tell me how the connections to of the galvanometer and the plates and the power supply are made ?

[@lecturesbywalterlewin.they9259]

InventTwig the /dt is the key. I ONLY looked at Maxwell's eqs. There is a dphi/dt and a dE/dt. Thus I ONLY used those. They are the most basic.

[@amisharawal3967]

Hello sir that was a very wonderful lecture and I am very thankful to you for making the concepts comprehensible...... And also, Greetings from India..... I have a possible explanation for the results seen in the demonstration of the Leyden jar, ( time stamp 40:25 )...... My explanation is as follows.... It could be possible that when you charged the conducting cylinders (thereby making a charged capacitor), the molecules in the glass beaker became induced dipoles...... And when you disassembled the jar setup, the dipoles held on to the charge somehow (I still can't make a clear explanation as to how that occurred, since it was induce)..… And after discharging the metallic conducting cylinders, and putting them back around the glass (which was holding on to the charges it aquired through the induced dipole), when you connected the two metal cylinders, the dipole charges got a pathway to discharge themselves, going through the metal cylinders and through that device you used to connect the 2 cylinders, which completes the circuit pathway, thus causing the spark, which was actually the induced dipoles getting discharged...... Can this be a possible explanation? And once again, thank you for such an amazing lecture and a lot of greetings from India.....(I m in 11th standard, by the way, and I still understand mostly all that you teach in the first time........)

[@veerenpawar9868]

Wonderful lecture sir 😃 Love from India

[@thomasrad6296]

So irritating when it zooms onto a students face when he is trying to speak about something on the chalk board.

[@peterpace3379]

Thomas rad Nah you get to see the beauty of MIT xD

[@CrazyGamer-xi8rf]

Those are moments where the professor accidentally spoke something incorrect... They edit those moments out..

[@nikhilnagaria2672]

@@CrazyGamer-xi8rf you hear the voice man

[@achyutsingh5298]

50:38 sir correct me if I'm wrong- you just touched a 30,000 V source, in India we receive household voltage of sockets equal to 220v which even though a lot smaller than 30kv is large enough to fry people but you just touched a WHOPPING 30Kv SOURCE! wasn't that experiment a little too life threatening?

[@obayev]

That was amazing in many ways! Thank you, Professor!

[@lecturesbywalterlewin.they9259]

You're very welcome!

[@krishvasa7644]

Sir what is the answer to the question you posed with respect to the capacitor still having a potential difference after you clearly removed the charge?

[@stargazer7644]

He explains it in a later video

[@shivamkakkar4027]

spark between two conductors is due to alignment of dipole in the glass hence +ve side of insulator attract negative charge. nd -ve side of insulator attracts positive charge of conductor. Hence two conductors charged again by the property of insulator i.e dielectric.

[@bhanu8391]

At 42:51 why doesn't all the charge transfer on to vandfegraph and on the second time itself both would become equipotential?

[@maneeshkumar1725]

prof., in your fourth experiment, you said that in order to compensate for the E field induced in the dielectric, we must add more charge on the plates - so when the dielectric is placed more charge flows onto the plates. but doesn't more charge on the plates mean that the dielectric is polarized more and the opposing electric field is stronger? which in turn requires us to put more charge on the plates? I hope you understand my question.

[@banajadandasena4142]

Now I am started to love physics... thanks MIT.. thanks for these videos.i love this teaching style.thank you sir..

[@NormaNsNs]

Thank you Dr.Lewin for sharing this jewel with us !

[@Lostverseplays]

The lyden jar one i think because maybe of the induced electric field inside the glass??

[@DanielKRui]

@18:14 the amp meter did twitch both putting the glass plate in and taking it out

[@maaan8494]

I think maybe charging the plates in the Leiden jar induces a dipole in the glass that stays in the glass when you remove the plates. When you put the plates back they are themselves polarized and because the top of the jar is open, the electric fields don't cancel and a voltage is generated

[@giplochon]

(@ 15:45) Prof, what would happen if you just kept separating the the two plates? How far can you go until something breaks down? In theory, voltage would just keep increasing I guess, but to what point?

[@madakagnaneswari4973]

Love for walter lewins teaching from India 😍😍❣❣

[@360wheelz5]

3:44 why isn't the induced electric field surface charge density divided by *the* *permittivity* *of* *the* *medium* *of* *dielectric* ? After all the field *E_i* is induced *inside* the dielectric medium? Help me.

[@lecturesbywalterlewin.they9259]

The E-field is the resut of induced charges on the surface of the dielectric - for more info use google

[@360wheelz5]

@@lecturesbywalterlewin.they9259 but the E field *pass through* the dielectric so shouldn't the permittivity of dielectric be taken into consideration in some way or the other? I posted a question on stack exchange physics.stackexchange.com/q/562671/238425 But have no satisfactory answers yet. And also, I searched my book and google but couldn't find any relevant query. Your help is needed!

[@lecturesbywalterlewin.they9259]

@@360wheelz5 permittivity is the dielectric constant and, ofcoz, that is taken into account.

[@InventTwig]

Professor why isnt there an electric field from +sigma induced to -sigma free? at 3:17

[@lecturesbywalterlewin.they9259]

question unclear there are free charges and induced charges as I explained. The net E-field is the sum of the free charges and the induced charges.

[@canmumcu9804]

@@lecturesbywalterlewin.they9259 Professor, I quess the question was about the small gap between the dielectric and the plate. Should be an electric field between them right ?

[@MuhammadWaqas-fc3db]

The spark between the glass beaker and the copper cylinders experiment may be due polarization of glass. Since glass is an insulator, electric dipoles remain somewhat alligned even after the removal of external electric field. When you reassemble the whole thing, the dipoles induce charges on the metallic beaker which causes spark when shorted.

[@lecturesbywalterlewin.they9259]

Not correct. All capacitors have dielectrics. When you discharge a capacitor INSTANTANEOUSLY all induced charge on the dielectric disappears. Something else is at work here.

[@Aryan_Playz]

Can I get to know in which lecture has sir given reason for the sparks even after disassembly of the capacitor ??

[@lecturesbywalterlewin.they9259]

I don't rememebr which lecture - but I did explain it and did a demo to demostrate the explanation

[@emanuelefeola4116]

I can't understand one thing about the Van de Graaf explanation. Reading my notes about previous lectures, inside the sphere E = 0, so I do no work and there is no potential difference between the inside and the surface of the sphere. So how can charge move to the sphere if the potential difference is 0? Can someone please tell me where my reasoning is wrong? Thank you in advance By the way, I really enjoyed this particular lesson, mind-blowing

[@abdullaalmosalami]

This is a bit late to reply but I think your confusion is E is 0 but that definitely does not mean that the electric potential is zero. There was an E field on the outside of the speaker that you have to go against first, which means you do work, and you keep going until you get inside, and all that work you did is still there as potential energy, and that is why the electric potential inside on the smaller sphere is higher than the bigger sphere one the small charged sphere is inside. I would also think of it as follows: there is an electric field still coming out of the smaller sphere, and they go from the smaller sphere to the surface of the larger sphere (because the inside of the larger sphere has no electric field since it is a conductor). This nonzero field between the smaller sphere and the larger sphere means there is a potential difference between their surfaces, hence the possibility of charge exchange.

[@gulzarali6370]

@@abdullaalmosalami great

[@The_Green_Man_OAP]

6:50 Dielectric constant κ values: Free space: 1 Air: ~1 Glass: ~5 15:33 v=1500V, d=1→7mm 15:45 d=1mm, i≥50µA 15:55 v>1500V ("went up a little") 15:59 Increase d to 7mm 16:06 i=0μA 16:11 v changes, increasing? 16:15 v=10kV 16:21 d~7mm 18:30 κ: 1→5 18:32 v decreases to...2kV? But again i=0μA. (~30°=⅙π rads swept = 8kV change => 12*8=96kV max?) 18:41 Back to v=10kV 21:18 Summary of eqns: (1) Enet's: Eg=Eo/κ, (2) Supply volts 1.5kV initial, but then disconnect when cap charged, vary dist:V=Ed, (3) Cg=κCo "o" meaning "original value, in air" "g" meaning "value in glass" 21:43 Keep supply connected 26:03 κ~1, d~1mm, v=1500V (fixed), i~-50μA as charge goes to ammeter from plates. 26:11 d~4mm , i~ -5μA 26:13 d~5mm, i~ -2μA 28:50 Keep supply volts, d~7mm & put in κ => E cannot change, free Q increases. 29:10 Charge must increase by ∆Q=(κ-1)Q. ...yes? 29:21

[@Leonardo-el6sq]

At 3:50 (regarding the induced charge on a plate by a dielectric) why shouldnt It be used the permittivity of the dielectric instead of the permittivity of free space (Epsilon zero)?

[@lecturesbywalterlewin.they9259]

ask Mr Maxwell!

[@mguieb1988]

He used the dielectric constant, kappa or k, instead of using an epsilon relative. It's another way of notating the same concept. Personally, I use epsilon relative because kappa can be confused with coulomb's constant k.

[@yuwang6549]

I have a guess on your demonstration of Leyden Jar and it might have something to do with the metal fork tool you have on your hand: When you remove the inner plates you attach the metal fork to it -> some charges go to the metal fork; Then during the disassembling you again attach the metal fork to the outer plate, it might neutralize some charges on the metal plates but it gains some opposite charges. When you assemble the jar, you used metal fork to attach the inner plate therefore charging the inner plate -> now the jar gains energy again! The reason why I think so is that I noticed that your metal fork has a glass handle therefore cannot be discharged.

[@lecturesbywalterlewin.they9259]

no

[@agrawal.akash9702]

At 33:46: The maximum potential difference one can put across the first capacitor is 4000 V and for the second, it's 40 V. So, if someone puts 4000V across the first, then the minimum distance is 220 microns, and if someone puts 40V across the second, then the minimum distance is 2.2 microns. But, if someone puts 40V across the first one, won't the minimum distance also be 2.2 microns?

[@lecturesbywalterlewin.they9259]

question unclear how many minutes into the video?

[@agrawal.akash9702]

32:40 to 33:50 Sorry for not being clear. My question was: The first capacitor had a breakdown voltage of 4000V, which means that if it's connected across a voltage of 4000V, then the minimum distance between the parallel plates would be 220 microns (with polyethlene in between) or else breakdown would occur. But if we reduce the voltage across the capacitor, won't we be able to bring the plates even closer, thus, in a way, reducing the minimum distance required to avoid breakdown?

[@lecturesbywalterlewin.they9259]

If you buy a capacitor you do not have an option to make changes. In principle you could take the 4000 V cap apart and you can replace the 220 microns dielectric to to 2.2 microns and the breakdown would then occur at 40 V. In doing so you have increased the capacitance by a factor of 100.

[@agrawal.akash9702]

Lectures by Walter Lewin. They will make you ♥ Physics. Thank you. I understand it now.

[@owl6218]

I was making small videos demonstrating electrostatic charges produced by rubbing different materials. Glass rubbed with silk is supposed to produce an opposite charge on glass than that produced on ebonite/PVC when rubbed by fur/wool. I used two pieces of glass - one made of borosilicate glass, another made of ordinary soda lime glass. The ordinary glass when rubbed with silk cloth was getting attracted to both PVC rubbed with fur, and ALSO to boro glass rubbed with silk....I was scratching my head for a while. Then I realised that the rubbed piece of plain glass was not getting any significant charge (as shown by a pith ball electroscope). It was not charged, it was just showing the dielectric response of getting attracted to both the +ve and the -ve charged object nearby... it was the first time I got to see the dielectric effect of glass, practically. Seeing fine bits of paper flying towards a charged ebonite rod is one thing, seeing a 6 inch long (uncharged) glass strip rotating to follow the ebonite rod is another thing......tributes to all the scientists of 400 years back who had to patiently tease apart the confounding factors to distill Gauss' law out of their experiences in the lab 😊

[@ukartikey]

in leyden jar experiment i think the energy produced afterwards even after taking off all the charge from the metal plates is due to the charge induced on the dielectric(glass) which then acts as a source for recharging the two metal sheets and produce energy.

[@lecturesbywalterlewin.they9259]

If you remove the free charge of a capacitor, the induced charge on the dielectric goes away INSTANTANEOUSLY.

[@dylanhaymore608]

The Leyden jar experiment is particularly interesting in the sense that the parts that you think should retain the charge actually do not once it is dissected. The glass jar is prone to holding water molecules from the air on the surface and thus the charge from the capacitor is transferred to the surface of the glass jar via coronal discharge when the system is pulled apart. The charge is then able to be discharged through the conductors when the parts are put back together.

[@lecturesbywalterlewin.they9259]

your explanation is incorrect

[@dylanhaymore608]

Lectures by Walter Lewin. They will make you ♥ Physics. Awe shucks. I guess I will have to finish the rest of your lecture series to find out the answer! Still an intriguing phenomenon nonetheless. I do enjoy your lectures! Thank you for taking the time and effort to make them available to the general public.

[@lecturesbywalterlewin.they9259]

:)

[@dylanhaymore608]

Lectures by Walter Lewin. They will make you ♥ Physics. I just watched the lecture in which you explain the Leyden jar. I was close in how the charge was conserved yet wrong in the physics behind why it works. For the full explanation behind the physics I recommend continuing to watch Prof. Lewin's lectures.

[@markschmidt9852]

20:11 are these true for any configuration? aren't they only valid for parallel plate capacitors?

[@AnonymousAnonymous-dn1rk]

They are always true because electric field always falls by a factor of kappa

[@jakofo]

hello Teacher, thx for this video. I want to ask you, how do you measure the potential difference in the experiment 46:42, how do you connect your probe and also I can´t see how you charge te small sphere. Thank you very much.

[@lecturesbywalterlewin.they9259]

I suggest you google power supplies.

[@jakofo]

Thank you , I have done the experiment successfully, but I still can´t measure the differential potencial.

[@servingthroughyoutube4205]

44:31 Why you had to do work ? Isn't inside the shell charge zero , so that E inside the shell is zero ?

[@lecturesbywalterlewin.they9259]

yes E-filed inside is zero. Thus if you move a charge inside the area where E=0 you do not have to do work. However, to get to that area you may have to do work depending on existing E-fieds

[@servingthroughyoutube4205]

@@lecturesbywalterlewin.they9259 Sir, you are a superhero teacher . May god give you the longest life .

[@nagashreey.g7696]

namasthe sir lectures are so inspirational, thank you so much

[@lecturesbywalterlewin.they9259]

:)

[@ZodiacZubeda21]

At 28:30, you mention that the electric field does not change when you introduce a dielectric because the voltage and distance stay constant. So by the relationship E=σ/ε0κ, does this mean the free charge density increases by a factor of κ?

[@lecturesbywalterlewin.they9259]

*what I said is correct. If V does not change and d does not change, then E can not change. Watch it again and try to follow my arguments (remember that Q_free will increase if I keep V constant and place a dielectric between the plates) or use google.*

[@ZodiacZubeda21]

Thank you!

[@safanazim9745]

22:15 at constant potential as distance increases q must increase but why is it decreasing

[@lecturesbywalterlewin.they9259]

increase distance => C decreases. V=C/Q V does not change. C decreases, thus Q must decrease.

[@mykulpierce]

Just putting my prediction here while I continue the course. The Leyden jar stores the charge on the glass. In a demonstration with Franklin's Bells you can induce a charge THROUGH glass likely for the same reason. Charge is provided to one side, attracts charge from the other. This is why Q is increased with a dielectric, its stored on the surface of it.

[@lecturesbywalterlewin.they9259]

you have not properly explained my bizarre demo. I give the correct explanation in a later lecture.

[@mykulpierce]

@@lecturesbywalterlewin.they9259 I look forward to it! Hopeful not before another sleepless night.

[@xiaomanguo9199]

Still watching it in 2020-5-17!! Is it possible to calculate how many charges the sphere conductor has after connecting with the power supply?

[@lecturesbywalterlewin.they9259]

yes, Q=VC, C is the capactance

[@xiaomanguo9199]

Thanks. But if I want to know the amount of charge on the conductor (appears in 47:17) how to calculate the capacitance of the sphere conductor? The equation C = A*epsilon/d is for the parallel capacitor. Is there an equation for sphere or cylindric conductor?

[@abdullaalmosalami]

@@xiaomanguo9199 Watch the previous lecture #7.

[@sharudeva]

Sir, at about 43:50 you said that work is done to move the small sphere into the van de graff sphere. But when u demonstrated did u not do work to move the small sphere to the outside of the van de graff sphere ?

[@lecturesbywalterlewin.they9259]

when I bring a charged sphere inside the vdG dome I have to do positive work. Then I touch the inside of the dome with my sphere, the potential of the dome and the sphere become equal and there is no longer an E-field inside the dome. I now move the sphere outside the dome in a zero E-field, thus no work has to be done, no negative work and no positive work.

[@VickysTuition]

I got very confused @ 31:24 .. so i m writing this comment hoping it helps others !! If you switch off the power supply on the capacitor.. then electric field is CONSTANT and doesn't increase when you decrease d !!! But if the power supply is switched on and if you decrease d... then the capacitance goes up ... And to achieve the same potential difference ... the battery now charges the plates more.. increasing electric field between them !!

[@ahmadeldesokey9844]

Sir , at 44:38 , will all the charge on the inner sphere transfer to the outer shell or will some charge still on the inner sphere after contact with the outer shell ?

[@lecturesbywalterlewin.they9259]

depends on geometry. Most will go to the outside.

[@cosmologicalvision2250]

Hey Sir.... In your second demonstration....Why you not consider the Net Q ......instead of Free Q......

[@kargamelpeyo6784]

Hi Sir Lewin, thanks for you amazing and inspiring work. Would you please answer my question ? There's a tiny problem i can't clearly understand in this video. It may be simple but not for me ;-) After 3,45"", you write down : E free = sigma free divided by epsilon zero (It's ok for me) then you write E induced = sigma induced divided by epsilon zero. Here is my problem. In that case, the dielectric is now inside the capacitor, so why are you allowed to keep episilon zero in this last expression as it is not "vacuum" anymore but a material inside (dielectric) ? Thanks a lot for your reply. Take care. Chris (Belgium)

[@lecturesbywalterlewin.they9259]

I watched at 3:45. It's x-tal clear, I cannot add to the clarity. watch it again.

[@kargamelpeyo6784]

​@@lecturesbywalterlewin.they9259 Thanks. Is then this point of view correct : The fact that the dielectric is polarized on its external surface (as you showed with you sliding slides) is equivalent to replace it by putting the induced charge on the 2 plates of the capacitor ? If it's true then i would understand. Just tell me. ;-)

[@bhanuranjan7702]

Your doubt seems relevant here. Please explain sir!🙂🙂

[@melonboi6318]

@@kargamelpeyo6784 hello there, I am having the same issue as you did 8 months ago :I did u come up with a good explanation? Mine is that the epsilon zero is correct because epsilon R is to calculate the total field = a material doesn't actually have a different epsilon for its own electric field but we use epsilon R to calculate the total field because it's faster. K (or epsilon r) is just a number that shows how many oriented dipoles u get in a material

[@رهامسرور-م9ه]

What is the difference between electric potential and the electric flux? Both physically and mathematically

[@KaviPriyan-qt6vc]

The dielectric (glass) is polarized so on assembling the plates , both plates become oppositely charged and on short circuiting they produce spark am i right sir?

[@lecturesbywalterlewin.they9259]

not right

[@abdullaalmosalami]

@@lecturesbywalterlewin.they9259 Work was done to polarize the dielectric material - the electric field torque that tends to align the material's molecules in the direction of the electric field does work on those molecules. I'm going to guess that this alignment does not completely go away once the electric field is removed (i.e., the charge on the plates is removed) and so some of that energy is still retained in the dielectric, or one can just say the material is still polarized even if the field is zero. Upon reassembling the plates, this energy can be partially released. The polarized dielectric induces charge separation in each of the plates separately, and with the shorting instrument you used touching the outer plate, some of those free charges from the outer plate due to the induced charge separation move to the other tip of the instrument, creating an electric field in the area around it. Once you brought this tip close enough to the inner plate, you reach the electric breakdown voltage difference and hence the spark.

[@samanthacipher8456]

I'm not gonna sleep tonight for this, really. :D I need an explanation for that capacitor with glass as dieletric! Greetings from Italy mr Lewin, It's my pleasure to watch your old lectures.

[@lecturesbywalterlewin.they9259]

I give the explanation in a later lecture.

[@fredsmhmd8105]

@@lecturesbywalterlewin.they9259 the dielectric stays polarized after removing two plates then after putting them back in the dielectric induces charge,thats the reason for the E field maybe?

[@wus9472]

Sir did you given an explaination for this phenomenon. Which lecture can I watch to see that explanation ? Thank you

[@virenderbhardwaj3137]

is it necessary that capacitors should have equal and opposite charges always when connected to a battery? if yes, why? its hard to believe for me because here the electrons coming on one plate will repel the equal no. of electrons on the opposite plate, is it just a fact or is there a reasonable explanation for it? at first I used to think that if the charges are not equal and opposite than there would be a field in the capacitor plates but it is not necessary that they should be equal and oppositely charged to get the electric field inside the plates to be 0.

[@erikmjelde4428]

17:26 he says nothing will happen on the amp meter but clearly the meter moves when he plunges in the plate. Why?

[@lecturesbywalterlewin.they9259]

there was indeed a very very very small current when I put the glass in. I did not see any current when I removed the glass. I suspect that when I moved the glass in that I accidentally removed some free charge which is on the glass side of the plates. Compare the minute current with the factor of 5 increase in V.

[@erikmjelde4428]

Thank's! I love your videos.

[@csandoval82]

Awsome lesson. Where do you get your lab equipment like the amp meter, the plate capacitor and the voltage source? I teach and need some lab equipment. Also thankyou for doing this. Your a great teacher. 👍

[@aikasei1693]

I have a question. Why the potential of sphere would not change if the power supply is active according sphere????. Maybe because of the space that is full. I dk.

[@iriscaballeroperez5115]

My guess for the Leiden jar is that the dielectric is kept polarized and when the conductors are placed again, a surface charge will be induced on them. The top part of the top jar will be +, and bottom part of bottom jar -. Then, joining these to faces with the "wire" will cause a discharge.

[@lecturesbywalterlewin.they9259]

incorrect

[@iriscaballeroperez5115]

@@lecturesbywalterlewin.they9259 I saw the solution in a further video. Very interesting Walter! I thought that maybe the dielectric could be permanently polarized - seems not so. Big hug from Spain.

[@wus9472]

Which lecture did you see the explanation?

[@mathieuhansz2210]

Bonjour Pr Lewin, je suis français Je pense que les charges que l’on retrouve dans le condensateur (Leyden) reconstitué sont les charges induites dans le diélectrique. The charges in the capacitor at the end are coming from the inducted in the dielectric glass.

[@mechminds50]

dear prof. at 50:38 it is 30000 volts and it's written 300000volts

[@lecturesbywalterlewin.they9259]

I am not responslible for the cc. *I do say 30 thousand.*

[@guilhemescudero9114]

14:40 I think I may have an other unenlightenment about why the potential difference ΔV between the two plate is growing as we are growing up the gap between them : As you say at 14:29, E = σ/𝛆₀ , so E is a constant at each point between the two plates, if we grow up the gap, E must remains the same, and since E⃗ = - ∇⃗V and ΔV= ||- ∇⃗V||·d , ΔV between the plate 1 and the plate 2 is getting bigger as d is growing.

[@ayoubtarhouchi1433]

Sir , your lectors are very helpfull for me so thank you so much i'am a moroccan student and we learned that we use in place of KAPPA we use EPSILON r which depend on the material my question Sir , is it the same thing ??? HAve a good day Sir

[@lecturesbywalterlewin.they9259]

www.britannica.com/science/dielectric-constant

[@ayoubtarhouchi1433]

thank you so much Sir , it really helps Thanks again :)

[@devesharya5184]

When capacitor was charged, the dipole is created inside glass, while disassembling the device the induced charge on plates got fade away but the free charge dipole of glass was as it was as before so when the device is reassembled the created dipole again induce charge but less in quantity compared to before in the device, now the plates are charged again, hence you can still use the capacitor but with less energy than before. ( maybe i am wrng but i like the demos) love from INDIA ... :)

[@mohammad3691]

. What will happen if you touched the two conductors with the fork before getting them back . i thought that will make us sure that they are completely discharged . i thought the cause of the spark after replacing them back is that the glass surfaces having charges induced from the electric field between the two conductors and this charges transferred to the conductors. Am i right?

[@stephenwise9551]

When the Liden jar is veing reassembled... will it get induced charge since glass is Dielectric?

[@lifelyrics5659]

Leydan*

[@mr_gerber]

@@lifelyrics5659 Leiden or Leyden *

[@positivegradient]

How does the voltmeter measure the potential difference when there is no current flowing? I am aware only of voltmeters based on moving coil galvanometers that measure P.D. by measuring current.

[@positivegradient]

I am talking about a situation where the power supply is removed after the capacitor is charged. If the power supply is still connected, the voltmeter will draw a current and will be able to measure the P.D.

[@lecturesbywalterlewin.they9259]

>>>How does the voltmeter measure the potential difference when there is no current flowing? >>>> A Voltmeter is an Amp meter with a HUGE internal resistance 100 MΩ and more

[@KeithandBridget]

Very nice demonstrations in the last third of the lecture.

[@lecturesbywalterlewin.they9259]

Glad you liked it

[@thienthanhtranoan6723]

Is All whole equations in 19:23 right for all situation, prof.?

[@thalesagricola2827]

Happy birthday, professor Lewin!

[@lecturesbywalterlewin.they9259]

+Thales Agricola Thank you.

[@sohamrane5362]

Hello Professor. Lewin, I was wondering what would be the mathematical equation for capacitance. If in case the distance between the plates d, was greater than the thickness of the dielectric. As if the dielectric was just placed between the plates but did not touch either of the plates. The dielectric would then be a layer of air, then let us say, a layer of glass, and then again a layer of air. Would the mathematical a direct sum of Ctotal= Cair+ Cglass+ Cair? or would the change in Kappa affect the capacitance in a different way? PS: I love your lectures, they are truly eye-opening and have increased my pre-existing love for physics. I am looking forward to becoming a physicist someday in the future, and perhaps a physics professor, even half as good as you. Thank you so much for these lectures. I am eagerly awaiting the answer to my question.

[@lecturesbywalterlewin.they9259]

if d is greater than tickness of the dieletric the Capicitance can easliy been calculated. I probably do that in one if my 8.02 Help Sessions.

[@sohamrane5362]

@@lecturesbywalterlewin.they9259 Okay Professor, I will be completing the entire course of 8.02, I will search for it as I go along. Thank you :)

[@abdullaalmosalami]

I believe you would use the definition C = Q/V, and recalling that V is potential difference, which is the same as saying how much work you need to do to carry a 1 coulomb charge across the plates, you can go ahead and compute it out. For example, say the distance between the plates with some charge Q (disconnected plates so Q is stuck) is d and the thickness of the dielectric is t and it is placed exactly center (just as an example). Then on either side of the dielectric, the distance it and the plate on its side is (d-t)/2, which let's just call r. We will note the electric field in the gaps will be the electric field as it is in a vacuum - let's called it E_vac - and the electric field in the dielectric is what has been shown in the video (E_vac divided by the dielectric constant) - let's call it E_d. So, the potential difference between the plates of the capacitor with a dielectric in it as described will be V = E_vac*r + E_d*t + E_vac*r. E_d will be E_vac/k (kappa), so overall you have V = E_vac * ( 2r + t/k). Notice that d (the distance between the plates) is 2r + t, and without the dielectric, the voltage potential would be E_vac*(2r + t). The ratio between V in these cases is (2r+t/k)/(2r+t), and since the charge is unchanged in either case, and C = Q/V, C changes by a factor of (2r+t)/(2r+t/k). Observe that since k>1, this factor is also >1, so capacitance still goes up as expected. Also observe if t = d (the thickness of the dielectric is the same as the distance between the plates), r = 0 and we get the same case as was shown in the lecture.

[@marketmail49]

Greetings from India, sir! Im a keen follower of your lectures. Always helpful to me. I have a small confusion in 3.55 when you said the induced field E_ind = σ_ind/ε0... I will be more than happy if you may kindly explain why have we used ε0 here rather than the permittivity of the dielectric (may be ε say). Thank you in advance.

[@lecturesbywalterlewin.they9259]

I explain in great detail in one of my 8.02 lectures why anywhere at the surface of a conductor that E= σ/εo

[@marketmail49]

@@lecturesbywalterlewin.they9259 Thank you sir.

[@ummi_zuhair1626]

Why E=Enot/K formula is not valid for the case when dielectric doesn't span equipotential surface Plz do reply I am stuck at this and unable to find solution Your help will mean a lot to me

[@gabrielfacendabueno143]

you are the best professor!

[@lecturesbywalterlewin.they9259]

Glad you think so!

[@chaitanyagiri76]

at 3:57 shouldnt E induced be charge densuty divided by permitiviitty of that MEDIUM

[@lecturesbywalterlewin.they9259]

no

[@chaitanyagiri76]

It will be great if u explain me why so ? 👏

[@lecturesbywalterlewin.they9259]

use google.

[@abhayshankar8762]

Permittivity of that medium is kappa times epsilon zero. He already corrected for that.

[@rezamaram2439]

Thanks, Prof. Walter Lewin for the great lecture. Could you please explain about the probing you used for the Van de Graaff voltage measurement? One side of the probe is grounded and the other side is on a distance from the Can you are adding the charge on? Thanks

[@lecturesbywalterlewin.they9259]

your question is not clear. Refer to how many minutes into the lecture and then rephrase your question.

[@rezamaram2439]

@@lecturesbywalterlewin.they9259 Thank you very much for responding and sorry for the unclear question I had. Actually, what I was referring to is the following: you examined your own Van de Graaff with a paint can and measured its voltage using a voltmeter as you were accumulating charges on the paint can. To measure the electric potential difference of two points, I think we need to connect both probes of the voltmeter to the two points. But what I was seeing in the lecture video was that one of the probes of the voltmeter was hanging on the air, close to the paint can, assuming the other probe is grounded (Maybe I don't see it well, though). Is that how should it be?

[@lecturesbywalterlewin.they9259]

@@rezamaram2439 if you want to know the details of how we measured the potential contact Andy Neel [e-mail]

[@rezamaram2439]

@@lecturesbywalterlewin.they9259 Thanks a lot. I certainly will.

[@manuferre7186]

Best explanation about electricity that I have ever seen (And I am electronic engineer!!!!!! hahahah). Physics works, You are telling us!!!!!!

[@lecturesbywalterlewin.they9259]

thank you

[@generalginger7804]

18:40 There is small deflection in Ammeter.

[@generalginger7804]

Due to charging of the glass slab.