The Electric Field Due to a Ring of Charge (See note in description)
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- Опубліковано 10 лип 2024
- Explains the calculation of the electric field due to a ring of charge. This is at the AP Physics level. (important note: at 6:38 there is suppose to be an "a" in the numerator. I don't catch this mistake until much later in the video.)
- Наука та технологія
I just want to thank you and other youtubers who give their time to do instructionals like these. They save lives....and my grades :D
so true :)
omg your comment is 6 years ago wow
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I love the way you teach physics. You assume we know nothing (and I assure you - I know nothing) and you hit every key point. I could not imagine taking physics without people like you to help me on youtube!
I really dig the sound effects :D
great explanation, completely shat on my physics prof in 10 minutes
i laughed at the sound effects at 0:40
where does the "a" go at the end?? it just disappeared!
please tell me he forgot to add it
wait.... he just explained that....lmaooo
but y he forgot the "a" he says it later
so when R becomes 0, (a^2)^3/2 becomes a^3 and one of the a's cancel out with the a on top leaving a^2 on the bottom
Thank you, very systematically explained, and not boring like the way some of my teachers would explain. Digging this, keep it up!
Hi Mr. M! I still love to watch these videos while reviewing for my college e+m classes. Cheers, from your class of 2013
Thanks for all your videos mate, I greatly appreciate your fantastic explanations.
Dude, you rock. All there is to it. I sincerely wish that my professors lectured in such a clear and concise manner.
Thank you for catching the 'a' at the end, I was confused for a little bit xD
Fantastic video, thank you so much for posting this. I have a phyics midterm in a few hours and this is a life-saver.
Thank you for a clear, concise and to the point explanation of a classical physics 2 problem
Thanks sir! You just helped out a college student from Egypt! - Arab Academy Of Since, Technology and Maritime Transport.
Thank you so much for this video! I was wondering about how the integral worked (my Professor didn't explain it well and the book, "University Calculus" by Young and others, skipped steps)! Please keep making these videos as it has really helped this college student out! -Branden, Hayward CA
Thanks man. It was One of the best explanation of derivation in phyics. You saved 5 marks of mine in class 12th
That was explained really well...thanks!
Thank you so much for making these videos. I would not have passed my midterm if it was not for these videos.
Did you leave that a out on purpose as an instructional piece? I noticed it and then when you noticed it it really brought the whole thing back around full circle for me. My mind was struggling a little bit with this concept and your video really helped dial it in. I’m sure I’ll forget it though if I don’t practice it haha! Thank you!
love the sound effects at the start!
Thank you so much! You made this question seem easier.
thank you for your clear teaching!
You are the king professor!
I say it again, you teach physics far better than my prof !
thank you very much
I absolutely love your "Oh, I missed an a here" hehehehe. Thanks man, made understanding the continuous distribution easier than my textbook. Cheers!
thank you so much, your videos might just give me a chance at passing this course
Your teaching method is the best in world
i like the way how the integral of the ring ,, in much simpler than the integral of the line in ur last example,, i just think its cool how it works out like that
Dude, May God bless you so much. thanks
Thanks for your excellent videos :) from Turkey
thanks for light....hope to see more of your videos....
Thanks a lot Sir.. Hope to get more helpful videos of yours.. God bless u..
Well your effort of bringing a ring and stick really matters
Thank you sir, its very helpful, although in the begening , the *a* is missed, but it ok at the end. thank you so much.
Thank you so so much for this video!
Dunno why you sound amazing to me 😅💞 btw thanks man you made it very simple
Subscribe well earned. Great video!
really helpful. Thank you vary much !!!
But if we are indeed calculating the e field of the ring of charge at the point, why is Q changing if it is a uniformly distributed charge? Would the bounds of the integral not end up as being the circumference of the ring or are you just trying to show how the ring of charge behaves like a point charge from a far distance? Thanks a lot by the way this explanation of the integral is very easy to follow.
love the sound effects :D
problem is, when you are in the middle of the ring, ie a=0, you don't get zero as you should.
Yea you do. a is in the numerator so it all becomes zero
Yes you do. For the case that a = 0, the numerator will be (aQ) = (0*Q) = 0. Assuming that r is some arbitrary non-zero value, Ex = (0/sqrt(R^2+0^2)) = 0, thus the magnitude of the electric field at the origin is 0 N/C.
amazing sound effects
ı hadnt understood this until ı found u. thank u.
Seeing these videos really is making me realize that my professor is way over complicating the explanations for these. I found this so confusing for the past few weeks and this really helped!
Thank you very much..... That was helpful
thank u so much .... u really helped me alot ....i really appreciate what u did
thank you you are better than my second semester physics lecture
You are awesome thank you for the video
Thanks a lot ur methods r helping me a lot
you know he's a genius when he does his work in marker, cause he dont make mistakes
thanks ssooo much it was really helpful
Amazing!
Thank you so much!
A thin circular ring, of radius 20cm, is charged, with a uniform charge density(rho). If a small section, of 1cm length, is removed from the ring. Calculate the electric field intensity at the center of the ring.
Thank God for this video. You saved me
Great video thank you
This channel would have saved me some heartbreak in Physics 1
Its funny how people commenting about he forgot the "a" in the equation before they finished the whole video.
Try to be smart ass but they have to watch this video just because they dont understand their professor's lecture in class.
Anyway, good job on the explanation lasseviren1, its way easier understand your explanation than my professor's.
Even funnier that 2 years after your comment, people are still doing it
Exactly!!
Well, not all the people are a college students with a professor. Some people are still in high school and doing self-study
Roman Samulyak Lets go! 4 days til the test!
he found out and corrected it so why tf ur still complaining about it
Thank you!
What if you had an electron at point P and you let it go. How would you find the speed of the particle when it was at the origin?
Great explained
thnx. Youre videos are very helpfull. Thanks alot.
"dE sub x" OMG ITS A Y! was bothering me, you fixed it though :) on a more serious note, great video, you explain this very simplisticly and you helped a lot thanks. make more physics videos :)
What happened to the x after you multiplied it with dQ( the triangular definition of cos theta)
Thank you!!
are there any video for the charged disk ?
How do you calculate the force between the point charge and a circular charge if the Circle itself isn't perpendicularly orientated to the point charge but it was just simply positioned in a flat 2D plane respectively to the point? This is very important to me, make a video about it please.
Wow! That was great
why did you take the the indefinite integral of dq, instead of the definite integral of dq from 0 to 2pi*r, as you did in the Straight Uniformly Charged Wire video?
How do you know there are no limits of integration?
Great video by the way
You just made me subscribe your channel😉
I don't know what happened, but i didn't get this until I saw your video. Thanks mate :)
The equation at 6:36 does not contain 'a' in it..and it also doesn't cancel out with anything else..so please take this into consideration
watch til end, he caught it
Thanks allot, that's really helps ❤
I understood the canceling of y-components conceptually. But I didn't quite get it mathematically. The integral of the y-components should be zero right?
I integrated the (dEsinθ) and got to ( kQR/(a^2+R^2)^3/2 )It should be zero. Where did I made a mistake in my calculations?
When integrating dq at the end, why is there no integration limits like (from 0 to 2Pi)?
a dipole of moment p=Qa coul-m is aligned parallel to an electric field along the x axis. The field is non-uniform and varies in magnitude lineary along the x axis with a rate of change dE/dx=K. find the force on the dipole.
Are you Finnish? If yes, then your pronounciation is impeccable.
thanks sir, appreciated
sir, when you are using Pythagorean to find the R, isn't that under the premise that the angle between the opposite and the adjacent is 90?? but when you change R, and move around the circle, the angle between those two no longer remains as 90, which means R cannot be sqrt (a^2+ x^2). Could you please tell me why we are allowed to put R as that despite the change of angle? Thank you.
wondering the same thing
Subscribed for sound effects.
Brilliant video. I'm in eng phys and the explanations are nothing compared to what you have here.
I have a question, when you say that dQ doesnt change.. what do you mean? I thought the charge Q is evenly distributed throughout the ring, so each segment has the same dQ. so how can dQ change?
Is there anyway to do this using a gaussian surface. If so, how?
thx for helping me 🌹
What if place the ring flat on paper and find electric field at point in same plane at some distance x? I need answer please
Thank you sir!
I think the dq is throughout the central axis, not along the ring itself. the further you are from the ring, the lower the charge. if im right...
How could we do this in polar coordinates?
Could someone please tell what would be the feild at exactly the center of the loop?
what happens if there is a force at the point that is away from the ring? does this change the equation at all or is the force of that point charge ignored
That is a test charge and a test charge do not cause any effect on the electric field it's magnitude is considered to be 1
how come you can integrate dQ for a ring but you have to convert it to dy or dL for a wire?
Favorite UA-camr lmao
Awesome Thanks!
Hi sir, what if the point of charge is in value of infinity? What should we do with it?
Thanks for this, my professors at a "top engineering university" (quotation emphasized) are terrible at teaching physics.
Man said ok bye. 🥺
Best video ever found ❤..
thanks man!
Thank you very much good sir
i have a question
does the charge of a rod is the same as line of charge ?
thank you so much
Yes, I think the charge of a rod is the same as a line of charge. Here is a video I made about a charge on a rod: ua-cam.com/video/WmZ3G2DWHlg/v-deo.html