I see unusually high number of indians in this channel, every second start mesage with "sir". At this rate I expect to see the next 20 years half of the nasa being indian, and I mean the 2nd half.
Professor, I think there might be a mistake. When you have the point charge, which has charge Q (stated at 0:44), inside the sphere that has a uniformly distributed charge of -3Q on the outer surface(started at 0:49), there must also be a -Q charge on the inner surface of the sphere. Thus the total charge inside a Gaussian surface (sphere radius r>b) must be Q-Q-3Q=-3Q So the electric field inside must be -3Q/(4pie_0r^2) You told me I’m wrong, but could you explain how I’m wrong and why? Thanks Professor! (Same question for when we calculate potentials)
Lectures by Walter Lewin. They will make you ♥ Physics. I think you meant to say this response to my other comment professor. I post a lot of questions :)
Dear sir can you please make a video on the conducting spheres when they are earthed or joined using conduction wires and what their potential will be? It will be of great help sir....
Lectures by Walter Lewin. They will make you ♥ Physics. they are not in the list but the paper for iits (jeeadvance) i really tough and have some very good questions you should see them sir if you have time..
If you take a Gaussian sphere between a and b the charge inside that sphere is ZERO (there is minus Q charge induced on the inner surface at a). The E-field between a and b is ZERO.
Excuse me sir, will there be other induced charges on the outer surface that are opposite to the charges induced on the inner surface ? Which makes the total charge on the outer surface of the shell be (-3Q + outer surface induced charge) which is (-3Q + +1Q = -2Q) ?
sir , at 14:23, u said E[electric field] is radially inward and we have already calculated it . my question is , when we calculated that the electric field is inward we also gave -3Q charge on the outer surface.. and at 14:23 we have just gave +Q charge.. would it induce a -Q charge on both [inner surface of cavity and outer surface of sphere] ?
excellent question. YES on the inner surface of the shell there is an induced charge of -Q. That's where the field lines of the +Q charge end. This causes a +Q charge on the outer surface of the shell. Thus the net charge on the outer surface is -2Q. NOTICE that inside my Gaussian surface r>R the net charge is -2Q and that determines the E-field at r=R and r>R.
sir i got that , why -2Q charge was the net charge inside the gaussian surface at 3:07 . .but sir at 14:23 we didn't gave any charge on the outer surface , we just gave +Q charge inside the cavity that induces -Q on in inner surface of cavity and kills the E-field inside cavity. this whole phenomena results in induction of +Q charge on the outer suface so the net charge of the shell should be +Q in this case. and as a result E-field should be outward and Va should be greater than Vp,,..?
The net charge inside my Gaussian surface r>R is +Q - 3Q = -2Q. If I had placed induced -Q on the inner surface of the shell and +Q on the outer surface, the net charge inside my Gaussian surface OF COURSE would ALSO have been -2Q. Thus there was no need for me to mention the induced charges but in retrospect perhaps I should have mentioned it. The directions of my E-fields are all correct. The E-field inside the hollow part is outward the E-field r>R is inward. End of story
at 3:07 I say "Let's assume, it could be wrong, that the E-field is outwards". and I then show that it is wrong as it is MINUS thus it is pointing inward. at 14:23 I obviously give the E filed the correct direction as I showed at 3:04 that it is inwards.
Hello professor, can you recommend me a physics book based on IB HL? Or any physics book suitable for senior high schools student? as always, thank you for uploading these awesome lectures!
That pass will depend on your background in Physics. If you have NO problems with the math in my 8.03 lectures, I suggest you view MIT's OCW lectures of 8.04 by Professor Barton Zwiebach. He is a very good lecturer.
2 equal positive charges will ALWAYS repel each other. If you place negative charges somewhere to overcome the repelling force the -- charges will move to the + charges and the positive charges will move to the -- charges. Your best bet is to nail your 2 positive charges in place so that they cannot move.
Lectures by Walter Lewin. They will make you ♥ Physics. Video describes the electric field around a spherical conducting shell having a point charge placed at its centre. But I want to know the same ( electric field around the shell ) when I place the point charge NOT at the center but at some other point inside the shell.
Sir, just wanted to ask u if we can assume the universe expanding as a quasi-static process since its happening at a small rate? but, again the process needs to be very slightly displaced from equilibrium to be called so... if it is a quasi-static process then what could have been the force making it come to a state of equilibrium? sir,... also wished to know your email address so that i could mail u my doubts on Physics. :)
I can place any charge anywhere in a conductor as long as I hold it in place. In any case the +Q is NOT in a conductor (the conductor is between a and b) and the -3Q is on the outside of a conductor which is perfectly stable.
@@lecturesbywalterlewin.they9259 thank you but what i understand from your explanation is when i need to get for example VA-VB= integration from A to B dot dL and dl is a segment from the path connecting A to B is that right ?
If you mean the Gauss's law from electromagnetism, then, well, you can't "prove" any physical law; you can just collect evidence that it's true, or you can derive it from "more fundamental" laws. Gauss's law and Coulomb's law (which is equivalent in the electrostatic regime) have been tested extremely precisely by experiment and no deviations have ever been found in the regime of validity of classical electrodynamics.
This was extremely helpful, thank you.
“Cold” that’s right. Cold like the back of my
Hands from writing equations for 15+ hours for years and understanding the universe.
Greetings from brazil, you are excellent.
I have seen this problem (very similar, but not exact) on one of my exams in the past.
I see unusually high number of indians in this channel, every second start mesage with "sir".
At this rate I expect to see the next 20 years half of the nasa being indian, and I mean the 2nd half.
Ra Leo we Indians are made to study by our parents whether we like it or not ,and as time passes we start to develop a liking for the subject
Professor, I think there might be a mistake.
When you have the point charge, which has charge Q (stated at 0:44), inside the sphere that has a uniformly distributed charge of -3Q on the outer surface(started at 0:49), there must also be a -Q charge on the inner surface of the sphere.
Thus the total charge inside a Gaussian surface (sphere radius r>b) must be Q-Q-3Q=-3Q
So the electric field inside must be -3Q/(4pie_0r^2)
You told me I’m wrong, but could you explain how I’m wrong and why? Thanks Professor!
(Same question for when we calculate potentials)
+ lamda added to - lamda is zero; that's why the E field r>b is zero
Lectures by Walter Lewin. They will make you ♥ Physics. I think you meant to say this response to my other comment professor.
I post a lot of questions :)
Professor Lewin doesn't talk here about induction Concept so we can do that in other problems
Dear sir can you please make a video on the conducting spheres when they are earthed or joined using conduction wires and what their potential will be? It will be of great help sir....
Prof. Great lecture
easy peasy sir thanks concept crystal clear
I derived! and i got ab/(2a + b) .... Yes yes ... :).. i love physics Sir!
bhaiya can you please hint me. I'm not getting the right value despite of trying hard.
At any point inside the cavity calculated electric field was kq/r*2 but isnt there a electric fied due to induced charge at any point inside cavity
sir what do you think about indian education system and also YOUR THOUGHTS ABOUT THE IIT'S ...
I am not an expert on this. I suggest you google the "100 best Universities in the World" and see whether there is any IIT among them.
Lectures by Walter Lewin. They will make you ♥ Physics. they are not in the list but the paper for iits (jeeadvance) i really tough and have some very good questions you should see them sir if you have time..
@@Random-qu9dz lol so what if they are hard?
Hard doesn't mean better....
Tysm
@ 7:02 Electric Field inside the material of the Shell should be 0. So, shouldn't we take E*[4*pi*(r’2-a2)] = Qinside/epsilon ???
what I have is correct.
Sir, If we take a Gaussian surface between a and b, E would be zero, there must be a charge induced. Where would it be present?
If you take a Gaussian sphere between a and b the charge inside that sphere is ZERO (there is minus Q charge induced on the inner surface at a). The E-field between a and b is ZERO.
Lectures by Walter Lewin. They will make you ♥ Physics. Got it sir, thank you. 😊
Excuse me sir, will there be other induced charges on the outer surface that are opposite to the charges induced on the inner surface ? Which makes the total charge on the outer surface of the shell be (-3Q + outer surface induced charge) which is (-3Q + +1Q = -2Q) ?
correct
sir , at 14:23, u said E[electric field] is radially inward and we have already calculated it . my question is , when we calculated that the electric field is inward we also gave -3Q charge on the outer surface.. and at 14:23 we have just gave +Q charge.. would it induce a -Q charge on both [inner surface of cavity and outer surface of sphere] ?
excellent question. YES on the inner surface of the shell there is an induced charge of -Q. That's where the field lines of the +Q charge end. This causes a +Q charge on the outer surface of the shell. Thus the net charge on the outer surface is -2Q. NOTICE that inside my Gaussian surface r>R the net charge is -2Q and that determines the E-field at r=R and r>R.
sir i got that , why -2Q charge was the net charge inside the gaussian surface at 3:07 .
.but sir at 14:23 we didn't gave any charge on the outer surface , we just gave +Q charge inside the cavity that induces -Q on in inner surface of cavity and kills the E-field inside cavity. this whole phenomena results in induction of +Q charge on the outer suface so the net charge of the shell should be +Q in this case. and as a result E-field should be outward and Va should be greater than Vp,,..?
The net charge inside my Gaussian surface r>R is +Q - 3Q = -2Q.
If I had placed induced -Q on the inner surface of the shell and +Q on the outer surface, the net charge inside my Gaussian surface OF COURSE would ALSO have been -2Q. Thus there was no need for me to mention the induced charges but in retrospect perhaps I should have mentioned it. The directions of my E-fields are all correct. The E-field inside the hollow part is outward the E-field r>R is inward. End of story
sorry sir for taking ur precious time more than that i deserve.. just tell me that at 14:23 and 3:07 minutes u took the same example?
at 3:07 I say "Let's assume, it could be wrong, that the E-field is outwards". and I then show that it is wrong as it is MINUS thus it is pointing inward. at 14:23 I obviously give the E filed the correct direction as I showed at 3:04 that it is inwards.
Hello professor, can you recommend me a physics book based on IB HL? Or any physics book suitable for senior high schools student?
as always, thank you for uploading these awesome lectures!
I suggest you search the web. I would have to do the same.
Lectures by Walter Lewin. They will make you ♥ Physics. Will your book help me with my SAT physics examination?
which book?
Lectures by Walter Lewin. They will make you ♥ Physics. For the Love of Physics.
That will not help you for your SAT exam
please hint me for the evaluation of ab/2a+b, anyone?
Shouldn't the charge on the inner surface of the shell be -q thus net charge muat be -q +q -3q = -3q
question unclear - rephrase and tell me how many minutes into the video?
Sir You looked like Benedict Cumberbatch!
Hello Sir.what is the recommended path to study QM ?and do you recommend any books?Thanks.
That pass will depend on your background in Physics. If you have NO problems with the math in my 8.03 lectures, I suggest you view MIT's OCW lectures of 8.04 by Professor Barton Zwiebach. He is a very good lecturer.
Thank you :)
@@lecturesbywalterlewin.they9259 sir how is 8.04 by allan Adams
Which is better
@@dr.meenakshibhat3141 Allan Adams is cool
one question from electrostatic
where and what charge is to be placed between two equal positive charges to make the whole system come in equillibrium
2 equal positive charges will ALWAYS repel each other. If you place negative charges somewhere to overcome the repelling force the -- charges will move to the + charges and the positive charges will move to the -- charges. Your best bet is to nail your 2 positive charges in place so that they cannot move.
Lectures by Walter Lewin. They will make you ♥ Physics. cant the force be 0 at the center if we aim to put some +charges
what force? In between your 2 positive charges there is a point where the E field is zero. But that was NOT your first question.
Lectures by Walter Lewin. They will make you ♥ Physics. ohhh where is that point?
thanks Sir
Why E inside Cavity is taken 0
bcoz ilt is ZERO! Use Gauss' Law!
@@lecturesbywalterlewin.they9259 sir in conductor E=0
That E=0 can only for inside metal only or whole space(volume) inside metal E=0
I really hope that the heap of paper the professor uses goes to recycling!
may i know what will the electric field distribution, if i place +ve charge not at centre, but at some other point in the sphere.
question unclear.
Lectures by Walter Lewin. They will make you ♥ Physics.
Video describes the electric field around a spherical conducting shell having a point charge placed at its centre.
But I want to know the same ( electric field around the shell ) when I place the point charge NOT at the center but at some other
point inside the shell.
ruclips.net/video/GmXMlg74rg8/видео.html
Thanks sir for such a beautiful question and answer.
And sorry for my previous mistake.
Prof. missing your dotted lines ! :P
Sir, just wanted to ask u if we can assume the universe expanding as a quasi-static process since its happening at a small rate?
but, again the process needs to be very slightly displaced from equilibrium to be called so... if it is a quasi-static process then what could have been the force making it come to a state of equilibrium?
sir,... also wished to know your email address so that i could mail u my doubts on Physics. :)
use google
At 4:52 isnt that inside conducting sphere the electric field is 0
yes inside the solid conductor between a and b the E field is ZERO.
is there charge at the center of any conductor?
If you put a charge inside a conductor (in static situation) it will always go to the outer surface.
Lectures by Walter Lewin. They will make you ♥ Physics. then at 1:01 u put q charge at the center and -3q on the sirface. why?
I can place any charge anywhere in a conductor as long as I hold it in place. In any case the +Q is NOT in a conductor (the conductor is between a and b) and the -3Q is on the outside of a conductor which is perfectly stable.
question how you determine the direction of dr ?
*I always choose dr in the direction of increasing r.* Of course you are free to do it differently but that may get you into trouble.
@@lecturesbywalterlewin.they9259 thank you but what i understand from your explanation is when i need to get for example VA-VB= integration from A to B dot dL and dl is a segment from the path connecting A to B is that right ?
@@lecturesbywalterlewin.they9259 Please if you can make video explaining that because in other books i get confused
I always choose dL in an electric circuit in the direction that I move. In the case of gravity we deal with a quantity r.
how can we prove gauss law
If you mean the Gauss's law from electromagnetism, then, well, you can't "prove" any physical law; you can just collect evidence that it's true, or you can derive it from "more fundamental" laws.
Gauss's law and Coulomb's law (which is equivalent in the electrostatic regime) have been tested extremely precisely by experiment and no deviations have ever been found in the regime of validity of classical electrodynamics.
hey,,,...
is it Walter lewin,???
very young Walter lewin!!!!!
and energetic!!!
www.reddit.com/r/OldSchoolCool/comments/1lmank/young_walter_lewin_teaching_physics_in_rotterdam/
난 한국인
hi
Professor, could I have your email address? I have been watching your lectures for years now.
more practical please. I mean experiments