8.02x - Lect 7 - Capacitance, Electric Field Energy
ฝัง
- เผยแพร่เมื่อ 30 มิ.ย. 2024
- Capacitance, Field Energy, Frightning (explosive) Demo
Lecture Notes, Capacitance of Spheres, freepdfhosting.com/86b291986d.pdf
Assignments Lecture 6, 7 and 8: freepdfhosting.com/73aa29b41a.pdf
Solutions Lecture 6, 7 and 8: freepdfhosting.com/7e11dd8e3f.pdf - วิทยาศาสตร์และเทคโนโลยี
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.
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Thank You Sir! You are seriously a Steve Jobs for physics.
This is really a place to experience Physics.
thanks for your kind words. You can write me a personal msg on my channel just as you wrote your last 2 msgs
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.
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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 .
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??
So many years later your lectures still help students like me. Thank you very much! Greetings from Argentina.
:)
@@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?
@@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.
@@lecturesbywalterlewin.they9259 Sir, I am from India, I am preparing for iit jee. Sir, how can I contact you plzzz sirr
Me too 😊 from NEPAL
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!
Thanks
Is it okay? I'm not taking the assignments associated in each lecture. Because I'm only here for understanding the concepts
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
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.
You are the teacher of the world!! In some years, everyone will watch thesee classes because no teacher can do what YOU DO!!!
+Yuri Aps That would be very cool!
Yeh I am seeing
@@lecturesbywalterlewin.they9259 That's right.
Love from India sir...
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
:)
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 :)
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!
+Mike Odie Thanks Mike
AS SOON AS A PROBLEM ARISES WHILE STUDYING PHYSICS, WALTER LEWIN SIR IS ALWAYS THERE.....💯💯💯❤️❤️
Thank you, Professor Levin, for this class.
Veronica Melo You are very welcome Veronica
29:47 that experiment was just beautiful!
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.
i just wished i had a math teacher that had interested me in physics like this, finally learning physics at 66yo
Go and check Professor Leonard, he is the Walter Lewin of Maths.. you won't regret it
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...
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.
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 ;)
I have already studied all these topics in my school but I watch these lectures again and again because they are just too interesting.
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....
:)
Mandar Kulkarni when you’ve taken through differential equations and the calculus in this class still scares you
Complete mastery of the topic and lecturing in general. Helps greatly with my physics teaching.
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!
I really love these lectures, Thank you very much
+Mohamed Eldegwy Thank you and have a great 2016
WOOOOOOW!! THAT'S SO AMAZING!! thank you Walter Lewin for this amazing lectures!!
DAMNNN I LITERALLY CAN SAY THAT THIS WAS THE BEST LECTURE OF MY LIFE TILL NOW!
Really cool experiments.. physics never fails to fascinate!!!
Huge fan professor...
YOU ARE THE BEST TEACHER EVER MR. LEWIN!
Every time I watch these lectures I gain a new perspective. Kudos to you sir.
Muchas Gracias profesor Walter, sus clases son lo máximo! Saludos desde Perú
javier suarez Gracias por sur amables palabras.
\\/\////////@lter
Wish you were my teacher at school... The concepts are so simple when you explain them.. Thanks sir.. You are my hero now.
Dear Professor Lewin. Besides being a great physicist and teacher you are great at drawing. Specially on perspectives. (in case no one else noticed)
You will be forever my hero!! You peel off all my difficulties.
:)
I'm loving the practical experiments from your lectures too.. love from india 🥰
love you professor. really you are great, you are the storehouse of knowledge and logics. I'm really greatful to get lecture from you.
You are always welcome professor. Professor do you have visited Nepal?
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.
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.
You're very welcome!
How can a person sleeps in his lectures.. he is taking this precious gift of taking his classes live granted...not gd.
The only person who could tell me what a capacitor actually is🙏🙏
This is really incredible. Love from South Korea 👏🏻
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.
:)
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!
+Ayan Gangopadhyay That's great Ayan. Are you now going to watch my Electricity & Magnetism (8.02) lectures?
+Lectures by Walter Lewin. They will make you ♥ Physics. yes I'm already at lecture 7 thank you again!
+Ayan Gangopadhyay superrrrrr!
+Lectures by Walter Lewin. They will make you ♥ Physics. Do you explain uniqueness theorems in any of these lectures, Professor?
+Ayan Gangopadhyay no I don't
Very well explained sir .Completely soothing to here 😌
This lecture helped me a lottt.❤
Sir you are my favorite lecturer. Truly inspired by you sir....i love your lectures
:)
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.
so kind of you
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 😁
Thank you Shuby for your kind words
I loved the demonstrations! Wish I could have you as a professor for physics.
I'm a high school kid prof.lewin you're videos are really beautiful and engaging. Inshorts "You made me love Physics!!!"
✌️✌️✌️
Thank you very much Professor. Your lectures are a pure joy!
You are very welcome
Than you, sir the way you present the concepts with demonstrations is fun to learn
Really, You made me to love physics more & more
Sir, you are a great teacher in my mind.
You're the boss, Lewin!
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
Has this been confirmed as sustainable proof?
@@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!
@@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
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.
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.
love from my heart Sir.
Thank you! Greetings from the Virgin Islands.
< 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 > :)
Great lecture Prof. Lewin.
thank you professor, that's very kind of you .
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🇱🇰🇱🇰
Every teacher should be like you🙏
Love from india
sir I will definitely become a quantum physicist. sir I like you. I find me very happy when chat with you
That really made me fall in love with Physics... Greetings from India Sir ❤️❤️
Glad to hear that!
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🤗
Keep it up
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.
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.
the best professor!!!
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?
+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.
@@lecturesbywalterlewin.they9259 for some reason it is difficult to wrap my head around
@@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.
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?
@@arqampatel5122 yes that is right. Charge one plate only, it does not create a force on itself.
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.
:)
fall in love with physics.
love from india .
I love your teaching dear teacher ❤❤
Amazing classes. Thank you very much for your lectures.
Glad you like them!
no teacher can do what YOU DO!!!
See you friday, professor! :)
I think Professor Walter Lewin should give a lecture on teaching too. He's a living Legend
Greetings from Bangladesh 🇧🇩
Favourite one 💙💙
Loved your lectures ☺️
Glad you like them!
I'm an eighth grade student and I can understand everything you teach
You are a really good teacher
Love from india
Thank you very much fr this interesting lecture.
I bow you sir.
:)
Professor, I still do not understand why the force is 1/2QE instead of QE?
Why you used the average electric field ?
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.
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.
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...
+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.
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?
37:54 "AAAH we did it, it broke!!! lolol.
I don't understand how anyone can sleep through your lectures
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.
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,
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?
Are these spheres for which capacitance calculations are done HOLLOW or SOLID or "EITHER" ??
THANKS FOR THESE 'PRICELESS VIDEO LECTURES' by Walter Lewin. They have brought me FAR!
30:20
Would have loved to see a fully loaded capacitor discharge. Once without work being done on the capacitor and once with work having been done on it. You should be able to see that in the intensity of e.g. a spark
What happens if I insert my hand in between the parallel plates of a capacitor if a). I am standing on an insulating stand and b). I am in contact with the ground? I know it sounds crazy but does it do any harm? And are there any other cases similar to a and b?
+Anery Patel Anery, Let's assume you stand on a good insulator. There will be charge polarization in your hand due to induction. Thus one side of your hand will become + charged and the other -. Your net charge on your hand will remain zero. If you now step off your insulator the same thing will happen. One side of your hand will become + charged and the other side -. There will be no current going through you in either case.
+Lectures by Walter Lewin. They will make you ♥ Physics.
Okay. But what if I touch the hand which is inserted, with other hand? And I think if I touch the plates of the capacitor directly then there are chances of current flowing through me. Isn't it?
+Anery Patel If you have one hand between the plates and you touch that hand with your other hand, the same thing will happen - charge polarization. If you touch one plate and if you are not insulated all charge will flow to the ground. You discharge the capacitor. You have created 2 capacitors in series. One of them is the Earth. If you are insulated, then you have created again 2 capacitors in series and charge will flow on you - you are one of the two capacitors. Which fraction of the available charge (on the plates) will move to you depends on the size of the plate capacitor and on your capacitance.
Can you please suggest a resource for better understanding the 1/2 factor in the expression for the electric field at 3:30?
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.
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?
From Afghanistan I inspire my student like you
Sir since potential energy of pair of opposite charges is positive due to the attractive nature of force, should not be the potential energy of capacitor be positive as the force between the capacitor plates is attractive?
One more question. Is workdone = negative of potential energy stands true for all conservative forces?
If I stretch a spring over a distance x I have to do positive work 0.5kx^2. That is called PE. If I lift an object vertically over a distance h. I have to do positive work mgh. That is called PE.
PE of a capacitor is 0/5*C*V^2. It's always positive.
Professor ...in the last lecture you discussed about the high voltage break down and you have said that two parallel plates can cause electric field pass through air when the electrons have enough KE to ionize the molecules of o2 and nitrogen in case of very high voltage so then the electric field between them is not zero ....but why did you take E=0 between these parallel plates at the starting of the lecture in case of high potential difference between them the electric field is not 0 right?
Hi sir, thank you very much for the lecture. I have yet another question :)
in 3:00, you discuss how the charge being built up specifically on the boundary of the capacitor causes the energy stored to be (1/2)QE. I was wondering where was the connection of this to the alternative reason given for the 1/2 coefficient, namely that the energy stored is (1/2)QV, (which you shortly after derive) because half of the energy is dissipated in the circuit due to joule heating. I tried thinking about it and lost sleep over it, but I couldn't put my finger on it. Thank you :)
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.
still in 2024 it helps me a lot thank you sir
greetings from india
Which book you are following for this lecture and course?
Please tell me sir.
I'm following them using Young & Freedman - University Physics but you can look at the syllabus and they'll have more options there.
Why was the charging of the 100 microF capacitor (for the fuse demonstration) took much longer than for the circuit of 12 capacitors (1000 microF)? As far as I know the charging process takes approximately 5 or 6RC, so does that mean that the “monster” capacitor of 100 microF had a big resistance?
Professor, when you say that we have to take the average of the electric field inside and outside the plate, can't it be explained by saying that the plate won't experience it's own electric field?
yes you can do that.
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
Hi Sir,
When you are moving the upper plate over a distance x then the upper plate will experience a electric field due to negative charge which is present on the bottom plate right ( i.e electric field due to an infinite sheet of charge E = ( Sigma / 2* epsilon zero) ) that's why you are taking 1/2 right ?
yes that is correct.
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.
Proffesor I love You and your lectures, they really made me love physics. But the force on metal plate is equal to E field generated by ONE plate times charge Q of the second plate, rather than average inside/outside field of both times charge of one plate.
The charge on one plate is Q. The E-field right at the surface of that plate is E=Q/(A*eps0). Yet the electric force on that plate is not QE, it's QE/2, that is a bit strange, students would expect that the force is QE, Thus I explained why it is QE/2. The thought that each plate contributes E/2 to the Field is a way to arrive at the same result. Gauss' Law applied near one plate ==> E=Q/(A*eps0. Thus the notion that I should take the field "contribution" (E/2) of one plate to calculate the force on that plate does not appeal to me; it's not kosher but you get the right answer.
Such a great lecture, but I have an inquiry please... In case Of DBD ( dielectric barrier discharge).. after using the dielectric that will reduce the Enet in the discharge gap Enet =Efree/K... it means lowering the discharge comparing with if both the electrodes were bare .. Is that right?
So can I say the DBD and Capacitor is using almost the same concept but in DBD low dielectric constant is better to give more Electric field for ionization.. in capacitor higher
??dielectric constant so high stored charge.
Thanks in advance professor
Electric discharge E-fields in various materials (including gasses) can be found on the web. It's not simply related to K.
But I would love to understand the implications of energy at cosmology scale! I mean Alan guth said that energy of an electric field(which we calculated from a capacitor) means we can create electric field If we input the equal amount of energy. Similarly, at the time of early universe, what do we mean by these values of potential energy as there was no external agent present there to do external work.... So what consequences/implications of these energy values have at cosmic scale (at the beginning of tim itself)…,??