8.02x - Lect 7 - Capacitance, Electric Field Energy

[@GabiMartin97]

So many years later your lectures still help students like me. Thank you very much! Greetings from Argentina.

[@lecturesbywalterlewin.they9259]

:)

[@shuvashishsharma1299]

@@lecturesbywalterlewin.they9259 Sir can you please explain 3:22 why we have taken avg of E(inside) and E( outside)? Why not E(inside)+E (outside) from superposition principle?

[@lecturesbywalterlewin.they9259]

@@shuvashishsharma1299 factor of 2 Let the charge on each plate of the capacitor be Q=100q (+Q on one plate and --Q on the other). Let the area of each plate be A. The E-field between the plates is then Q/(A*eps_o). Let the distance between the 2 plates be d. Now release q from the + plate to the negative one. The energy that is released is force*distance which is very close to q*d*Q/(A*eps_o). But after this transfer the E-field is reduced, it’s now only 0.99*Q/(A*eps_o). I now release another q thus the energy that is released is now 1% less than with the first q. By the time I have released 10 times one q, the E-field is only 0.9*Q/(A*eps_o) and thus the force on the charge q is only 0.9*q*Q/(A*eps_o). The force keeps going down as I transfer more charge. By the time I have released the last q (of 100) the E-field has become 1% of the original E-field thus the force on the last q is about 1% of the force during the first transfer. Thus by the time that I have released 100q=Q the average E-field was only 0.5*Q/(A*eps_o), NOT Q/(A*eps_o). Thus the average force on a charge q during transfer was NOT q*Q/(A*eps_o) but only half that. Thus, by spooning off the chaInfamousrge from one plate to the other the energy release is NOT Q*d*Q/(A*eps_o) but only half of that. The energy that is released this way (also called potential energy) is identical to the work that YOU would have to do to separate the 2 plates over a distance d.

[@intothephysics1262]

@@lecturesbywalterlewin.they9259 Sir, I am from India, I am preparing for iit jee. Sir, how can I contact you plzzz sirr

[@kalpanaghartimagar2301]

Me too 😊 from NEPAL

[@ianmichael5768]

Watch this. Then watch it again and listen. He is explaining concepts! The math in physics exists to help you understand the concepts! What he is explain is a reality in nature. Ask yourself, why is this possible. If you plan to study grad physics(or higher), then do not lose sight of what is happening behind the math. Nature is beautiful! Never stop asking why things are the way they are. Just my opinion. Excellent lecture!

[@marxiewasalittlegirl]

Thanks

[@mewsicman9541]

Is it okay? I'm not taking the assignments associated in each lecture. Because I'm only here for understanding the concepts

[@hkl103]

​@@mewsicman9541Sakurai says in his quantum mechanics book, someone who has read the book but cannot calculate stuff has learned nothing. My theoretical physics professor has always said exercises are more important that the lecture. I think those statements are worth considering.

[@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!

[@sharudeva]

Thank You Sir! You are seriously a Steve Jobs for physics. This is really a place to experience Physics.

[@lecturesbywalterlewin.they9259]

thanks for your kind words. You can write me a personal msg on my channel just as you wrote your last 2 msgs

[@MikaelNutsos]

Dear Sir. I totally agree that you not only have a fantastic knowledge in physics, but you are blessed with a multi-talent in teaching. It was exiting to see you teaching about these things. I am the inventor of the patent The family that comprises The Following patents: European patent EP1638666 Japanese patent JP4597969 , United States patent US7594959 and US8323385 I would like to ask your help me to explain what happens from a physicist's point of view. A lot of big-sized companies have made third-party testing and fantastic results have been demonstrated. Very thin synthetic fiber filter layer have the capability to filtering the air as god as HEPA filters but without the enormous pressure drop these HEPA filters cause. Please send an email to my address to discuss more about the posibilyty for a collaboration. My email is [e-mail] Best regards Michael

[@MikaelNutsos]

This is the first time for me using You-tube sending message to an other , isn't public for all? I prefer to emailing you more private .

[@knowledgeunlimited]

Lectures by Walter Lewin. They will make you ♥ Physics. Sir.. If the plates are rolled in the capacitor then will the distribution of E field will remain same as explained in the lecture??

[@ahhhhhhhh6947]

Watched these lectures as a 19 years old preparing for JEE, now I am 22 with degree in EE. Watching again for fun. Lectures hits different now as a graduate. You are a true Guru to me Walter

[@larcomj]

Yeeeeeessssssss, same here. Ive been working in industry as an EE for over 5 years and im brushing up on my Physics. These videos are such a gift. Hope your doing well as an EE! Keep on studying, you will be amazed at how quick you loose it if you dont use it.

[@yurilsaps]

You are the teacher of the world!! In some years, everyone will watch thesee classes because no teacher can do what YOU DO!!!

[@lecturesbywalterlewin.they9259]

+Yuri Aps That would be very cool!

[@ITVishal]

Yeh I am seeing

[@armankhajuria]

@@lecturesbywalterlewin.they9259 That's right. Love from India sir...

[@bogee647]

on behalf of every student who has ever been taught by you, my thanks. your enthusiasm and genuine interest in your field has been key in the lives of your learners!

[@lecturesbywalterlewin.they9259]

+Mike Odie Thanks Mike

[@jerjsmit6806]

I am a former student who graduated 4 years ago and, after many years of working in a different field, have almost entirely forgotten most of what I have learnt. Thanks to your lectures, I have renewed fascination for physics. Electrodynamics was always a weak area for me and so having such a great amount of detail taught in such an interesting manner is beyond priceless. I have been inspired to pursue my masters. Thank you for inspiring me and so many others, and for continuing to post inspirational content during these dark times :)

[@srishtiparihar960]

sir I am a class 12 student & I have seen your lectures they all are awesome .. your lectures are very interesting really sir you are the best physics teacher . . hats off to you sir

[@lecturesbywalterlewin.they9259]

:)

[@priyabratadas4864]

AS SOON AS A PROBLEM ARISES WHILE STUDYING PHYSICS, WALTER LEWIN SIR IS ALWAYS THERE.....💯💯💯❤️❤️

[@MrAyangan]

The combination between the amount of math and demonstrations are just right to keep you extremely hooked to these lectures, its like I'm binge watching them, in fact I just finished with 8.01 a week days ago. It is truly, as you once said "art". A very big thank you again professor!

[@lecturesbywalterlewin.they9259]

+Ayan Gangopadhyay That's great Ayan. Are you now going to watch my Electricity & Magnetism (8.02) lectures?

[@MrAyangan]

+Lectures by Walter Lewin. They will make you ♥ Physics. yes I'm already at lecture 7 thank you again!

[@lecturesbywalterlewin.they9259]

+Ayan Gangopadhyay superrrrrr!

[@MrAyangan]

+Lectures by Walter Lewin. They will make you ♥ Physics. Do you explain uniqueness theorems in any of these lectures, Professor?

[@lecturesbywalterlewin.they9259]

+Ayan Gangopadhyay no I don't

[@veronicanova7036]

Thank you, Professor Levin, for this class.

[@lecturesbywalterlewin.they9259]

Veronica Melo You are very welcome Veronica

[@rabikumarch1173]

Dear Sir, for me, at the age of 52, your lectures arouse great interest in physics .Common day problems cannot deterr my attention to your lectures. I haven't come accross such a great teacher.

[@lecturesbywalterlewin.they9259]

so kind of you

[@mandarkulkarni9999]

I'm really loving physics sir. U r best sir. No tough calculus to afraid students but just showing practically everything to clear the concept by which we can solve any problem... Hats off sir....

[@lecturesbywalterlewin.they9259]

:)

[@TBoy205]

Mandar Kulkarni when you’ve taken through differential equations and the calculus in this class still scares you

[@stevenm3914]

Thank you so much for having these videos available for those who want to learn. Currently taking this Physics 2 for engineering and wouldn't have a chance at passing because of corona virus if it wasn't for your content. Truly appreciate it from the bottom of my heart.

[@shantanu6892]

I have already studied all these topics in my school but I watch these lectures again and again because they are just too interesting.

[@janitajacob7066]

I loved your lecture sir.I was not able to understand the concepts taught in my class maybe because I was not interested but watching your videos every time makes me fall in love with physics. Thank you professor Lewin for changing my perspective towards physics.Greeting from India...

[@Miguel-tr2ev]

often I laugh when you make "sacrifices" for the sake of science but then It also made me truly appreciate physics. nothing beats watching physics works than simply reading books. You rock sir!

[@achyutsingh5298]

29:47 that experiment was just beautiful!

[@terrymulvaney3046]

i just wished i had a math teacher that had interested me in physics like this, finally learning physics at 66yo

[@nickyraja3793]

Go and check Professor Leonard, he is the Walter Lewin of Maths.. you won't regret it

[@varunahlawat9013]

DAMNNN I LITERALLY CAN SAY THAT THIS WAS THE BEST LECTURE OF MY LIFE TILL NOW!

[@manuferre7186]

I have studied Engineering and it's the first time I see a capacitor in real action (not only as a secondary character). I also did my final project on super capacitors, it would have been amazing to do this demonstrations with one, but maybe firefighters should have been called first XDD. Thanks Walter, as always ;)

[@fredericchopin4821]

YOU ARE THE BEST TEACHER EVER MR. LEWIN!

[@sarangpahwa619]

How can a person sleeps in his lectures.. he is taking this precious gift of taking his classes live granted...not gd.

[@javiersuarez385]

Muchas Gracias profesor Walter, sus clases son lo máximo! Saludos desde Perú

[@lecturesbywalterlewin.they9259]

javier suarez Gracias por sur amables palabras. \\/\////////@lter

[@MrAbhinavgour]

Every time I watch these lectures I gain a new perspective. Kudos to you sir.

[@digitalMathMechanic]

I'm really thankful for you professor, I used to extremely love physics but now by looking your lectures I'm better understanding my love. thank you.

[@MrMEldegwy]

I really love these lectures, Thank you very much

[@lecturesbywalterlewin.they9259]

+Mohamed Eldegwy Thank you and have a great 2016

[@VitoOporto3D-WORLD]

Dear Professor Lewin. Besides being a great physicist and teacher you are great at drawing. Specially on perspectives. (in case no one else noticed)

[@piyushbehera1987]

Wish you were my teacher at school... The concepts are so simple when you explain them.. Thanks sir.. You are my hero now.

[@physicsbymr.halderstudentl8561]

I 'm seeing your conceptual physics lecture from India. It's creating amazing feelings in my mind.I want to read in MIT for M.Sc in physics but i think i haven't option.Thank you.

[@PauloConstantino167]

You will be forever my hero!! You peel off all my difficulties.

[@lecturesbywalterlewin.they9259]

:)

[@WR8YsDunceCapLabs-e6d]

Oh, how I wish I could have been your student in person. Thanks for making all of this public and free.

[@lecturesbywalterlewin.they9259]

you are very welcome

[@marcoantonioalcazarperedo8846]

WOOOOOOW!! THAT'S SO AMAZING!! thank you Walter Lewin for this amazing lectures!!

[@manavshetty4449]

Sir, I think I have a better way of arriving at the same result as in 3:33 . To move upwards, the upper plate has to do work against the electric field provided individually by the bottom plate whose electric field is sigma/2*epsilon. The charge of the upper plate is Q and distance covered is x, Therefore force is = F = QE = Q*sigma/2*epsilon....... but sigma/epsilon = Electric field btw the plates, therefore required force will be F = QE/2... Therefore required work to be done will be W = QEx/2

[@mirunion]

Has this been confirmed as sustainable proof?

[@hurricanelamp6413]

@@kxb6098 He's talking about relative electric field(so 2 speak) strength btwn d plates,he nvr mentioned it as d electric field pertaining to one plate only,then taking d average btwn d two(both being d field strengths btwn d plates) sounds fine as well. When talking bout dipoles, d work being done to move 1 charge away has definitely got to do wd d electric Field strength of not jss 1 charge but both. Lastly,quit judging others!

[@raghulsankar1153]

@@kxb6098 you dumbo . he asks the question cause prof lewin didn't reply to the explanation and acknowledge it , which he usally does btw. so he wants to be sure if it's right . conviction is good for every physicist .. don't look down on people like this

[@eamon_concannon]

Thanks a lot. Saved me time trying to figure this out. I'm not sure that his explanation makes sense and I'm a bit surprised he did not give your explanation.

[@eamon_concannon]

Thanks a lot. Your explanation makes complete sense and you saved me time trying to work it out. I'm not sure that his explanation makes sense and I'm a bit surprised that he didn't give your explanation.

[@adityabhatia2866]

Really cool experiments.. physics never fails to fascinate!!! Huge fan professor...

[@pratapanurag757]

I'm loving the practical experiments from your lectures too.. love from india 🥰

[@kaushiksenthilkumar682]

I'm a high school kid prof.lewin you're videos are really beautiful and engaging. Inshorts "You made me love Physics!!!"

[@satyamsanodiya6908]

✌️✌️✌️

[@matteop700]

Damn, 2 years study Physics in one of the best AeroSpace Engineering University in Italy and he needed only 15' it gave me as much as needed more reality to understand Capacity and Capacitors. DAMN.

[@lecturesbywalterlewin.they9259]

:)

[@AnthonyFrancisJones]

Complete mastery of the topic and lecturing in general. Helps greatly with my physics teaching.

[@digitalMathMechanic]

love you professor. really you are great, you are the storehouse of knowledge and logics. I'm really greatful to get lecture from you.

[@digitalMathMechanic]

You are always welcome professor. Professor do you have visited Nepal?

[@bird9]

< I go to infinity, I put +q in my pocket, I approach B, and the work I have to do per unit charge is the potential of B > :)

[@pragati6218]

Thank you for this educational videos. I am very grateful. I sincerely hope that professor and MIT would continue to share this invaluable and amazing educational resources with the world.

[@lecturesbywalterlewin.they9259]

You're very welcome!

[@srichakraraj2338]

Sir you are my favorite lecturer. Truly inspired by you sir....i love your lectures

[@lecturesbywalterlewin.they9259]

:)

[@crescendowook1733]

This is really incredible. Love from South Korea 👏🏻

[@cricworld6797]

Schools are still closed. We find tremendous obstacles to studying here. Every moment when I deal with physics, I remind your word, "Physics is not difficult, physic is there to make a difficult thing easier. I suppose that the English language is incomplete. Because there is no specific word for me to describe your contribution to my studies.Thank you a lot.Greetings from Sri Lanka🇱🇰🇱🇰

[@shubysheikh83]

i have no words to express my gratitude towards u sir.. i have been looking for the real experience of physics wich i finally found out here.. u hv been amazing so far n i hope u continue with this extra ordinary work of urs.. ur contribution to physics is beyond anythng.. thanku so very much professr 😁

[@lecturesbywalterlewin.they9259]

Thank you Shuby for your kind words

[@Aryan_Playz]

The only person who could tell me what a capacitor actually is🙏🙏

[@PauloConstantino167]

These students were so lucky. So lucky. I bet they didn't even realize this. Being at the best university in the world and in front of the best teacher in the world. This is beyond words. I wish I was that lucky.

[@lecturesbywalterlewin.they9259]

:)

[@phyindia001]

Really, You made me to love physics more & more

[@archithrd2651]

Very well explained sir .Completely soothing to here 😌 This lecture helped me a lottt.❤

[@obayev]

Thank you very much Professor. Your lectures are a pure joy!

[@lecturesbywalterlewin.they9259]

You are very welcome

[@simplyenigma3]

I'm here watching ur videos sir after 5 years for my neet exam😊 and yes its helping me to understand better🙂. Thank you sir❣ love from🇮🇳 . - Alhamdulillah🤗

[@lecturesbywalterlewin.they9259]

Keep it up

[@walpurgoffnacht]

sir if i may ask, why do you take the average of the two fields back in minute 3:31 ? is there any mathematical explanation about it?

[@lecturesbywalterlewin.they9259]

+Usamah Jundi When I move a charge Q in a field E over a distance x, then I have to do work QEx. However, the field through which I move the charge Q is not everywhere E. It's E on one side and zero on the other. Thus the work I have to do is 0.5*QEx.

[@jessiegoodman9620]

@@lecturesbywalterlewin.they9259 for some reason it is difficult to wrap my head around

[@guilhemescudero9114]

​@@jessiegoodman9620 Me too, and I have an alternative explanation of why the result seems true : Suppose you ignore the second charged plate, then the first charged plate which we move experiences no electric forces when we move it : if you move a charge in an empty space, there is no electrical forces which act on this charge. So it is only the charged plate below which has an influence on the charges of the upper plate when we move it. Since the electric field generate by the (infinite) second plate is E₂ = σ/2𝛆₀ (you can find this with Gauss's Law) then to move the first plate for along a x distance we have to do the work of W₁= Q₁· E₂ · x = Q₁· σ/2𝛆₀ · x =1/2 (σ·A· E · x) because between the to plate, the magnitude of the electric field is E = E₁ + E₂ = 2 · E₂ E₁ = E₂ so E₂ = E/2 Hence W₁= 1/2 (σ· 𝛆₀/𝛆₀· A· E · x ) = W₁= 1/2 (E·A· E · x /𝛆₀) =1/2 (E²A·x· 𝛆₀) = 1/2 (E²∆V· 𝛆₀) with ∆V the volume crossed throw the displacement of the first plate.

[@arqampatel5122]

The electric field is due to the charge on both plates (2 *sigma/2(epsilon)) but the force on one of the plates is only due to field created by the other plate so we would ignore its own contribution to the field in calculating the force. Is this right?

[@lecturesbywalterlewin.they9259]

@@arqampatel5122 yes that is right. Charge one plate only, it does not create a force on itself.

[@sanidhyasrivastava2315]

That really made me fall in love with Physics... Greetings from India Sir ❤️❤️

[@lecturesbywalterlewin.they9259]

Glad to hear that!

[@SMahanta2001]

Sir, you are a great teacher in my mind.

[@growwithgaurav3225]

Every teacher should be like you🙏 Love from india

[@murtazaabbas2166]

I think Professor Walter Lewin should give a lecture on teaching too. He's a living Legend

[@jatinsingh8423]

Than you, sir the way you present the concepts with demonstrations is fun to learn

[@markfar4837]

What a great lecture prof!!! Those experiments were crazy. I have a request Sir Walter Lewin :- Please upload demonstration of the questions in which a solid sphere is taken (having uniform volume density) and a small spherical cavity is removed from random part of sphere and it is asked for the electric field at any point inside spherical cavity.

[@lecturesbywalterlewin.they9259]

if the sphere is a conductor all the charge will be uniformly at its outer surface also when you cut out any piece inside the solid sphere. Thus if you are outside the sphere and you measure the E-field at and near the outer surfaceyou will never have any idea of what is inside the sphere.

[@JaiPrakash-bk3uv]

sir I will definitely become a quantum physicist. sir I like you. I find me very happy when chat with you

[@absalommax]

no teacher can do what YOU DO!!!

[@almohaimen]

Greetings from Bangladesh 🇧🇩 Favourite one 💙💙

[@learner2278]

Can anyone further explain the knowledge at 3:37? I understand the electric field above the Q is 0, and the electric field below the Q is E. However, I do not know why we need to calculate the average electric field and why the average electric field is calculated by dividing by 2. Thanks a lot!

[@learner2278]

Dear Professor Walter, I saw your explanation about moving positive charges and reducing E. However, I wonder if charges are being moved when we are moving up the plate. Therefore, E may not be reduced. However, things change if these two plates are connected by an ideal wire and a battery. If these two plates are connected by an ideal wire and a battery, the electric field will be reduced. Thus, we can get the average E = 1/2*E. This is my guess and assumption, but I don't know how to explain by using the layer you mentioned. Hope you can solve my problem. Thanks a lot.

[@learner2278]

I found that my solution is wrong since E will not reduce to 0. Thus, I do not know how to solve this problem

[@lecturesbywalterlewin.they9259]

factor of 2 Let the charge on each plate of the capacitor be Q=100q (+Q on one plate and --Q on the other). Let the area of each plate be A. The E-field between the plates is then Q/(A*eps_o). Let the distance between the 2 plates be d. Now release q from the + plate to the negative one. The energy that is released is force*distance which is very close to q*d*Q/(A*eps_o). But after this transfer the E-field is reduced, it’s now only 0.99*Q/(A*eps_o). I now release another q thus the energy that is released is now 1% less than with the first q. By the time I have released 10 times one q, the E-field is only 0.9*Q/(A*eps_o) and thus the force on the charge q is only 0.9*q*Q/(A*eps_o). The force keeps going down as I transfer more charge. By the time I have released the last q (of 100) the E-field has become 1% of the original E-field thus the force on the last q is about 1% of the force during the first transfer. Thus by the time that I have released 100q=Q the average E-field was only 0.5*Q/(A*eps_o), NOT Q/(A*eps_o). Thus the average force on a charge q during transfer was NOT q*Q/(A*eps_o) but only half that. Thus, by spooning off the charge from one plate to the other the energy release is NOT Q*d*Q/(A*eps_o) but only half of that. The energy that is released this way (also called potential energy) is identical to the work that YOU would have to do to separate the 2 plates over a distance d.

[@ketanthegoodreddy8485]

I'm an eighth grade student and I can understand everything you teach

[@vedantvaideswar5386]

I don't understand how anyone can sleep through your lectures

[@noob_chess_player]

fall in love with physics. love from india .

[@kiandanesh081]

I love your teaching dear teacher ❤❤

[@mdselimhabib5209]

Great lecture Prof. Lewin.

[@sangeetayadav6665]

Loved your lectures ☺️

[@lecturesbywalterlewin.they9259]

Glad you like them!

[@deepat3294]

You are a really good teacher Love from india

[@JohnSmith5555]

At 19:14, I tried calculating the plates' capacitance with the video paused, just for practice, and got ~1.1M farads. I put 1.25 sq. metres for the area, multiplied it by epsilon 0, then divided by their separation, 0.00001 metres. What was my mistake?

[@joshuacharlery5826]

Thank you! Greetings from the Virgin Islands.

[@Zhaneka-ow4sb]

the best professor!!!

[@kiandanesh081]

From Afghanistan I inspire my student like you

[@VickysTuition]

@14:37 In the presence of A... shouldn't capacitance of B reduce due to the electric breakdown between them ?? In last lecture when the ball was brought near Van De Graff, sparks started between them... so shouldn't capacity to hold charge reduce for B due to discharge with A ??

[@VickysTuition]

I think the paper they place in between is a better dielectric than air.. therefore resists the breakdown better than air

[@yash29210]

the reason why there is a factor of half at 3:33 is that when we consider force on the upper plate then that force will be by the lower plate in downward direction................... and the magnitude of that force will be equal to (charge on upper plate)*(E-field due to lower plate) = Q*[(sigma/2)*(epsilon_0)] (just as we used to in case of a chrge "q" in the E-field of "Q")............................

[@lecturesbywalterlewin.they9259]

we discussed this in depth long ago.

[@yash29210]

What I am trying to say is that a positively charged plate would feel a force of Q*E in the E-field of a negatively charged plate and that E=(sigma*epsilon_0)/2

[@JohnDoe-tr1cx]

I recall you saying earlier that electric potential is energy per unit charge. So why is it that the energy stored in a capacitor is (1/2)QV? Why not simply QV? Also, I've already taken physics II at UT Austin, but I am using your lectures to refresh my memory as I'm now tutoring the subject. I just want to say that I really appreciate you putting up your lectures, and the way you teach is very inspiring. I'm considering a career in academia because of it.

[@lecturesbywalterlewin.they9259]

I derive in my lectures that the energy in a capacitor is 0.5*C*V^2. Since C=Q/V the energy is also 0.5*QV,

[@haupham5086]

Professor, I still do not understand why the force is 1/2QE instead of QE? Why you used the average electric field ?

[@lecturesbywalterlewin.they9259]

The infamous factor of 2 Let the charge on each plate of the capacitor be Q=100q (+Q on one plate and --Q on the other). Let the area of each plate be A. The E-field between the plates is then Q/(A*eps_o). Let the distance between the 2 plates be d. Now release q from the + plate to the negative one. The energy that is released is force*distance which is very close to q*d*Q/(A*eps_o). But after this transfer the E-field is reduced, it’s now only 0.99*Q/(A*eps_o). I now release another q thus the energy that is released is now 1% less than with the first q. By the time I have released 10 times one q, the E-field is only 0.9*Q/(A*eps_o) and thus the force on the charge q is only 0.9*q*Q/(A*eps_o). The force keeps going down as I transfer more charge. By the time I have released the last q (of 100) the E-field has become 1% of the original E-field thus the force on the last q is about 1% of the force during the first transfer. Thus by the time that I have released 100q=Q the average E-field was only 0.5*Q/(A*eps_o), NOT Q/(A*eps_o). Thus the average force on a charge q during transfer was NOT q*Q/(A*eps_o) but only half that. Thus, by spooning off the charge from one plate to the other the energy release is NOT Q*d*Q/(A*eps_o) but only half of that. The energy that is released this way (also called potential energy) is identical to the work that YOU would have to do to separate the 2 plates over a distance d.

[@haupham5086]

Lectures by Walter Lewin. They will make you ♥ Physics. Thank you very much, professor. Your explanation is very clear and easy to understand. That 1/2 was the hardest thing in this lecture.

[@Paczification]

You're the boss, Lewin!

[@hugomuller1609]

Professor Lewin, at 2:55, why does the charge should be inside the plate? I thought that as the plate is a conductor, there can not be charge inside...

[@lecturesbywalterlewin.they9259]

+Hugo Muller With the plate capacitor, as I have drawn it, the charge has to go to the upper surface of the lower plate and to the lower surface of the upper plate (Gauss' Law). If you put charge on one plate of a plate capacitor an equal but opposite charge is automatically created (induction) in the other plate. That charge will move to the inside surface. You can draw the thin layer of charge AT the surface if you prefer that. But in reality "at the surface" is a thin layer and I have drawn that thin layer. Apply Gauss Law and you will see that inside each of the plates the E-field is zero. The field starts at the surfaces.

[@solomiaz3697]

Lectures by Walter Lewin. They will make you ♥ Physics. The opposite charge on the other plate is created only when the plate is grounded, is not it?

[@arielfranco8251]

Amazing classes. Thank you very much for your lectures.

[@lecturesbywalterlewin.they9259]

Glad you like them!

[@IamLegend573]

Sir still i can't understand the reason why the force due to electric field becomes half there ? At 3:35 🙁.. please sir explain

[@lecturesbywalterlewin.they9259]

factor of 2 Let the charge on each plate of the capacitor be Q=100q (+Q on one plate and --Q on the other). Let the area of each plate be A. The E-field between the plates is then Q/(A*eps_o). Let the distance between the 2 plates be d. Now release q from the + plate to the negative one. The energy that is released is force*distance which is very close to q*d*Q/(A*eps_o). But after this transfer the E-field is reduced, it’s now only 0.99*Q/(A*eps_o). I now release another q thus the energy that is released is now 1% less than with the first q. By the time I have released 10 times one q, the E-field is only 0.9*Q/(A*eps_o) and thus the force on the charge q is only 0.9*q*Q/(A*eps_o). The force keeps going down as I transfer more charge. By the time I have released the last q (of 100) the E-field has become 1% of the original E-field thus the force on the last q is about 1% of the force during the first transfer. Thus by the time that I have released 100q=Q the average E-field was only 0.5*Q/(A*eps_o), NOT Q/(A*eps_o). Thus the average force on a charge q during transfer was NOT q*Q/(A*eps_o) but only half that. Thus, by spooning off the chaInfamousrge from one plate to the other the energy release is NOT Q*d*Q/(A*eps_o) but only half of that. The energy that is released this way (also called potential energy) is identical to the work that YOU would have to do to separate the 2 plates over a distance d.

[@venkatakishore7276]

At 45:50 it was mentioned that when a high current runs through a fuse it will get melted. Consider a circuit in which we have Power supply , switch , fuse , Bulb connected in series. Now i want to run a current of 1 amp with a voltage 100v. Power = 100watts. The minute i close the switch the fuse got melted. Consider the same circuit with a new fuse. Now i am going to run a current of 0.1 amp with a voltage of 1000v. Power 100watts. That means i am going to keep the power constant. The minute i close the switch will the fuse still melts ? The material and dimensions of the both fuses are same.

[@lecturesbywalterlewin.they9259]

fuses melt due to I^2*R in the fuse

[@venkatakishore7276]

It means the fuse got melted due to large number of electrons flowing through it. Because high current means large number of electrons right ?

[@gourangzalke6291]

Love from India ❤️❤️

[@banwarichoudhary2428]

Hello sir, I think at time around 2:58min you went wrong (may be I'm), according to me(as I understood physics from your lectures) "the electric force on a charged body is exerted by the electric field created by OTHER charged bodies". But you had taken electric field (sigma/epsilon zero) which is the sum of both the plates. But according to upper statement it should have taken electric field (sigma/2epsilon zero) of only one plate for force on other plate. By the way answer is correct but it is confusing me to take average.

[@lecturesbywalterlewin.they9259]

what I have is correct! it's a bit naive of you to think that i would make a mistake in a lecture at 1 first year college level.

[@banwarichoudhary2428]

@@lecturesbywalterlewin.they9259 yes sir you are right. I just wanted to know that I'm right or not. Sorry sir

[@mattheoswho1010]

Can you please suggest a resource for better understanding the 1/2 factor in the expression for the electric field at 3:30?

[@lecturesbywalterlewin.they9259]

Infamous factor of 2 Let the charge on each plate of the capacitor be Q=100q (+Q on one plate and --Q on the other). Let the area of each plate be A. The E-field between the plates is then Q/(A*eps_o). Let the distance between the 2 plates be d. Now release q from the + plate to the negative one. The energy that is released is force*distance which is very close to q*d*Q/(A*eps_o). But after this transfer the E-field is reduced, it’s now only 0.99*Q/(A*eps_o). I now release another q thus the energy that is released is now 1% less than with the first q. By the time I have released 10 times one q, the E-field is only 0.9*Q/(A*eps_o) and thus the force on the charge q is only 0.9*q*Q/(A*eps_o). The force keeps going down as I transfer more charge. By the time I have released the last q (of 100) the E-field has become 1% of the original E-field thus the force on the last q is about 1% of the force during the first transfer. Thus by the time that I have released 100q=Q the average E-field was only 0.5*Q/(A*eps_o), NOT Q/(A*eps_o). Thus the average force on a charge q during transfer was NOT q*Q/(A*eps_o) but only half that. Thus, by spooning off the charge from one plate to the other the energy release is NOT Q*d*Q/(A*eps_o) but only half of that. The energy that is released this way (also called potential energy) is identical to the work that YOU would have to do to separate the 2 plates over a distance d.

[@luizalmeida1209]

See you friday, professor! :)

[@chessematics]

3:21 is that really the reason why the force is QE/2? Isn't it due to the fact that when considering work and force ON that plate we only consider the electric field acting ON the plate, which is σ/2 ?

[@lecturesbywalterlewin.they9259]

an easy way to see the factor of 2 is the following. E-fields due to charges on the upper plate do not act on these charges - E-fields due to charges in the lower plate act on the charges in the upper plate and E-fields due to the charges in the upper plate act on the charges in the bottom plate.

[@chessematics]

@@lecturesbywalterlewin.they9259 yeah that's exactly what I meant

[@lecturesbywalterlewin.they9259]

@@chessematics also view this video - solution to problem 189 kzbin.info/www/bejne/nF7EfJiYjZ16jNE

[@lecturesbywalterlewin.they9259]

an easy way to see the factor fo 2 is that the E-field due to some charges do not act on these charges. Thus the E-field due to the charges at the lower plate act on the charges of the upper plate and the E field due to charges on the upper plate act on the charges on the lower plate.

[@lecturesbywalterlewin.they9259]

see also kzbin.info/www/bejne/nF7EfJiYjZ16jNE

[@guilhemescudero9114]

14:32 Does it depends of the way we approaches the sphere B with our test charge? I think of this because the configuration of the electric fields will change when we add the A sphere, and E⃗ = -∇ (V) so V(r) (r the distance from the center of the B sphere) changes?

[@lecturesbywalterlewin.they9259]

The Capacitance of B does not depend on *the way* we move A. It only depends on the final position of A relative to B.

[@guilhemescudero9114]

@@lecturesbywalterlewin.they9259 I probably expressed myself badly (English is not my mother language), I wanted to say that if we take sphere B alone, with a certain net charge QB, its electrical potential is the same everywhere on this sphere, and the equipotential areas are concentric around this sphere. But when we add the sphere A with an opposite net charge -Qb, Qb remains identical on the sphere B, however the equipotential areas will no longer be spherical around the sphere B because the lines of electric fields leaving B will be reoriented towards A, so when we approach a test charge of B, depending on which side we approach it (I said approach not touch), a test charge will not undergo the same potential gradient for the same distance traveled, is this not? So the work necessary to bring a charge close to B (at ∆r from the surface of B) will not be the same depending on which side we approach B (I said approach, not touch), is not it? My question had nothing to do with capacitance, it was referring to what you said at 14:10 I think I got the timing of the video wrong in my first message. Thanks a lot

[@michealbingham3686]

Hi Professor Lewin, At time 3:33, how can the charge reside on the surface inside the plate if the electric field inside is 0? Shouldn't the charge reside outside since by Gauss's law inside the plate should be 0 charge if the electric field is 0?

[@lecturesbywalterlewin.they9259]

the charge is at the surface of the plate but the charge has a finite thickness which is my pink layer. There is NO charge inside the conducting plate. I have a better way to explain the factor 2. Maybe that helps you: factor of 2 Let the charge on each plate of the capacitor be Q=100q (+Q on one plate and --Q on the other). Let the area of each plate be A. The E-field between the plates is then Q/(A*eps_o). Let the distance between the 2 plates be d. Now release q from the + plate to the negative one. The energy that is released is force*distance which is very close to q*d*Q/(A*eps_o). But after this transfer the E-field is reduced, it’s now only 0.99*Q/(A*eps_o). I now release another q thus the energy that is released is now 1% less than with the first q. By the time I have released 10 times one q, the E-field is only 0.9*Q/(A*eps_o) and thus the force on the charge q is only 0.9*q*Q/(A*eps_o). The force keeps going down as I transfer more charge. By the time I have released the last q (of 100) the E-field has become 1% of the original E-field thus the force on the last q is about 1% of the force during the first transfer. Thus by the time that I have released 100q=Q the average E-field was only 0.5*Q/(A*eps_o), NOT Q/(A*eps_o). Thus the average force on a charge q during transfer was NOT q*Q/(A*eps_o) but only half that. Thus, by spooning off the chaInfamousrge from one plate to the other the energy release is NOT Q*d*Q/(A*eps_o) but only half of that. The energy that is released this way (also called potential energy) is identical to the work that YOU would have to do to separate the 2 plates over a distance d.

[@michealbingham3686]

Thanks so much! Makes perfect sense @@lecturesbywalterlewin.they9259

[@shahinulshakib2009]

Greetings sir, In 3:26, Why you took the electric field as the average of 0 and (sigma/epsilon) ? Yes, the electric field inside the conductor is zero. But why it is mandatory to take the average and not the non zero one?

[@lecturesbywalterlewin.they9259]

factor of 2 Let the charge on each plate of the capacitor be Q=100q (+Q on one plate and --Q on the other). Let the area of each plate be A. The E-field between the plates is then Q/(A*eps_o). Let the distance between the 2 plates be d. Now release q from the + plate to the negative one. The energy that is released is force*distance which is very close to q*d*Q/(A*eps_o). But after this transfer the E-field is reduced, it’s now only 0.99*Q/(A*eps_o). I now release another q thus the energy that is released is now 1% less than with the first q. By the time I have released 10 times one q, the E-field is only 0.9*Q/(A*eps_o) and thus the force on the charge q is only 0.9*q*Q/(A*eps_o). The force keeps going down as I transfer more charge. By the time I have released the last q (of 100) the E-field has become 1% of the original E-field thus the force on the last q is about 1% of the force during the first transfer. Thus by the time that I have released 100q=Q the average E-field was only 0.5*Q/(A*eps_o), NOT Q/(A*eps_o). Thus the average force on a charge q during transfer was NOT q*Q/(A*eps_o) but only half that. Thus, by spooning off the charge from one plate to the other the energy release is NOT Q*d*Q/(A*eps_o) but only half of that. The energy that is released this way (also called potential energy) is identical to the work that YOU would have to do to separate the 2 plates over a distance d.

[@hkl103]

At 23:18 when we substitute the charge and divide by the potential does this hold for both definitions, so one conductor where the potential difference is measured with respect to infinity and two conductors with the same charge where only the charge of one is considered and the potential difference between the two? If yes, why can we do that?

[@lecturesbywalterlewin.they9259]

question unclear. Watch this lecture again. I cannot ad ti the clarity of this lecture

[@hkl103]

​@@lecturesbywalterlewin.they9259do I understand correctly, that the derivation for U you gave only holds for plate capacitors, so U= 1/2QV= 1/2CV^2 only holds for plate capacitors and thus the definition vor C that was used here was obviously the second of the two definitions you gave in the lecture?

[@lecturesbywalterlewin.they9259]

U=0.5xCV^2 always holds

[@hkl103]

​@@lecturesbywalterlewin.they9259thank you very much

[@giplochon]

Walter, At 3:37, F=1/2 QE. Does that mean that all charges in the red volume don't experience the same amount of force? That force varying from F=0 to F=1/2 Q sigma/eps_0. Then we would have to say that the E field penetrates the conductor's surface and decays until it nulls. And statistically, we get half the force. Am I right?

[@lecturesbywalterlewin.they9259]

Infamous factor of 2 Let the charge on each plate of the capacitor be Q=100q (+Q on one plate and --Q on the other). Let the area of each plate be A. The E-field between the plates is then Q/(A*eps_o). Let the distance between the 2 plates be d. Now release q from the + plate to the negative one. The energy that is released is force*distance which is very close to q*d*Q/(A*eps_o). But after this transfer the E-field is reduced, it’s now only 0.99*Q/(A*eps_o). I now release another q thus the energy that is released is now 1% less than with the first q. By the time I have released 10 times one q, the E-field is only 0.9*Q/(A*eps_o) and thus the force on the charge q is only 0.9*q*Q/(A*eps_o). The force keeps going down as I transfer more charge. By the time I have released the last q (of 100) the E-field has become 1% of the original E-field thus the force on the last q is about 1% of the force during the first transfer. Thus by the time that I have released 100q=Q the average E-field was only 0.5*Q/(A*eps_o), NOT Q/(A*eps_o). Thus the average force on a charge q during transfer was NOT q*Q/(A*eps_o) but only half that. Thus, by spooning off the charge from one plate to the other the energy release is NOT Q*d*Q/(A*eps_o) but only half of that. The energy that is released this way (also called potential energy) is identical to the work that YOU would have to do to separate the 2 plates over a distance d.

[@hagaryasser4605]

at 26:37 the lecture is excellent but I do not understand how will the electric field strength remain the same although the distance between the 2 plates increased since E = kq^2/d^2

[@JaiPrakash-bk3uv]

love from my heart Sir.

[@tedhanley1706]

Hello, there is something I do not understand: in moving the two oppositely charged plates apart, Dr. Lewin says (at 1:59) that the Force applied is constant. The equations follow clearly from that notion. My intuition tells me that if the plates are close, then the force to move them apart would be greater. As they are moved further apart, then I would suspect the Force necessary to move them would be less. Force would start large and diminish perhaps as the distance separating them by d squared (to a first approximation as the plates are not points). Am I misunderstanding something?

[@surendrakverma555]

Excellent physics lecture Sir 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏

[@lecturesbywalterlewin.they9259]

Keep watching

[@rinzej]

5:16 Would it be better to write: W/volume = 1/2 * ε E² with ε = K * ε0 ? In this case you get also the electric field energy density if there is a dielectric. By the way, thank you for uploading these videos, I enjoy them a lot!

[@lecturesbywalterlewin.they9259]

Rinze Joustra does this help? inventor.grantadesign.com/en/notes/science/material/S14%20Dielectric%20properties.htm

[@lecturesbywalterlewin.they9259]

Rinze Joustra The dielectric constant, also called k (or ε_r) = ε/ε_o

[@lecturesbywalterlewin.they9259]

Rinze Joustra by writing 1/2 * ε E² (this is the electric field density) with ε = K * ε_o you would get the dielectric constant K if ε is known (and vice versa).

[@praneethreddykoyagura4020]

Sir... please judge the statement ,,,STATEMENT:The work performed by field of conductor which is having a charge of some particular value (field created by only conductor is to be considered) in bringing a unit positive charge is independent of surroundings of conductor ..that is how the charge is actually spread on the surface of conductor.

[@lecturesbywalterlewin.they9259]

the statement is bizarre - crazy and wrong. Please ask quora.

[@sandhyashourie8731]

This broach is best among all of them . : )

[@lecturesbywalterlewin.they9259]

:)