David Griffiths Electrodynamics | Problem 2.7 Solution

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  • เผยแพร่เมื่อ 16 ม.ค. 2025

ความคิดเห็น • 48

  • @Ainz18
    @Ainz18 4 ปีที่แล้ว +13

    Brandon i really appreciate your work! you are awesome!

  • @imcharenbaatsongchanger7902
    @imcharenbaatsongchanger7902 2 ปีที่แล้ว +5

    Thanks for showing how to solve the integrals. I was successful in solving the first term but the second term was a wall for me. Thank you again

  • @sairaafzal4863
    @sairaafzal4863 3 ปีที่แล้ว +5

    By the way sir you are doing absolutely extraordinary job....👌👌

  • @JaS-rm6zn
    @JaS-rm6zn 3 ปีที่แล้ว +2

    very clear explanation , thank you so much!!!

  • @kooshva8193
    @kooshva8193 ปีที่แล้ว

    Thank you for your solution! ❤

  • @bulentdemir953
    @bulentdemir953 2 ปีที่แล้ว +1

    great effort, thanks a lot :)

  • @alexisnunez8871
    @alexisnunez8871 หลายเดือนก่อน

    Thanks!!

  • @vanshtomar486
    @vanshtomar486 3 ปีที่แล้ว +1

    Thanks a lot . Especially for solving the integration 😃😃

    • @brandonberisford
      @brandonberisford  3 ปีที่แล้ว +4

      Always!! Don't want skip the math like the professors 😂

    • @sairaafzal4863
      @sairaafzal4863 3 ปีที่แล้ว

      @@brandonberisford yah you are great at it.

  • @jaweriafayaz3376
    @jaweriafayaz3376 3 ปีที่แล้ว

    Thank you sir. All steps are very clear it help me alot..

  • @fathii2172
    @fathii2172 4 ปีที่แล้ว +2

    thanks a lot

  • @bruna.corcino
    @bruna.corcino 5 หลายเดือนก่อน

    Obrigada pela explicação! Meu esposo não quis me ajudar na resolução da integral 😢

  • @imcharenbaatsongchanger7902
    @imcharenbaatsongchanger7902 2 ปีที่แล้ว +2

    I think you made a mistake. Actually it should be |z-R| instead of |R-z| because we are trying to fing E outside the charge. If it is inside. .ie. |z-R| ,then we can directly write E =0 as there is no charge enclosed

  • @iota885
    @iota885 3 ปีที่แล้ว

    8:53 why isn't it a right triangle

    • @zzzzzzzz4284
      @zzzzzzzz4284 3 ปีที่แล้ว

      Bruh xD

    • @brandonberisford
      @brandonberisford  3 ปีที่แล้ว +1

      In general you could choose any point outside the sphere and any point on the sphere and in general it is not a right triangle.

    • @iota885
      @iota885 3 ปีที่แล้ว

      O okay thank you :)

  • @emanjanjua8022
    @emanjanjua8022 3 ปีที่แล้ว

    Sir!!great job🤩🤩can u explain why u have taken mod of (R+Z) while applying limits?

    • @imcharenbaatsongchanger7902
      @imcharenbaatsongchanger7902 2 ปีที่แล้ว

      I think he made a mistake. Actually it should be |z-R| instead of |R-z| because we are trying to fing E outside the charge. If it is inside. .ie. |z-R| ,then we can directly write E =0 as there is no charge enclosed

  • @aroojfatima6555
    @aroojfatima6555 2 ปีที่แล้ว

    sir your awesome 😍

  • @sairaafzal4863
    @sairaafzal4863 3 ปีที่แล้ว

    We are waiting.....

  • @gabelemmie8103
    @gabelemmie8103 9 หลายเดือนก่อน

    I understand the z component of the electric field but what about the radial dependence there should be a non-zero z component in the radial component at least I was not able to get it to go to zero.

    • @gabelemmie8103
      @gabelemmie8103 9 หลายเดือนก่อน

      nvm I figured it out

    • @shaunakmishra1430
      @shaunakmishra1430 9 หลายเดือนก่อน

      ​@@gabelemmie8103 Could you share what you figured

    • @gabelemmie8103
      @gabelemmie8103 9 หลายเดือนก่อน

      @@shaunakmishra1430 th-cam.com/video/d689JZuNcKo/w-d-xo.html this is a good video to watch. So the guy in the current video just takes the z component of the electric field without getting into the specifications of script vector r. In the video I posted it shows you how to get the z component through the vector. Hope this was helpful.

  • @hi_my_name_is_daniel
    @hi_my_name_is_daniel 10 หลายเดือนก่อน

    2:59

  • @samlocr388
    @samlocr388 2 ปีที่แล้ว

    Thank yooouuuuu

  • @youtubegoldmines
    @youtubegoldmines 9 หลายเดือนก่อน

    Makes you appreciate gauss

  • @AnshulSharma1997
    @AnshulSharma1997 2 ปีที่แล้ว

    Whats the difference between spherical surface and a solid sphere. I am not able to understand the element you took as according to me spherical surface is hollow inside having some thickness on the surface having surface charge density sigma. But in solid sphere we have material inside. Also how da=volume element of a spherical coordinate

    • @brandonberisford
      @brandonberisford  2 ปีที่แล้ว +1

      A spherical surface (in reference to this problem) is a spherical shell that contains some surface charge density sigma (no thickness) and is hollow inside. A solid sphere is say like a metal ball. It is not hollow inside, it contains material throughout.

  • @sairaafzal4863
    @sairaafzal4863 3 ปีที่แล้ว

    Sir why did'nt you post lectures of ch 3 problems

    • @brandonberisford
      @brandonberisford  3 ปีที่แล้ว +2

      I have started too, however it is hard to find time to record as these videos take alot of time too make as I have to find the time to solve them myself first, plus full time job etc.

    • @sairaafzal4863
      @sairaafzal4863 3 ปีที่แล้ว

      @@brandonberisford ok best of luck. 👌👌

  • @israelcamposromero2602
    @israelcamposromero2602 2 ปีที่แล้ว +1

    Excuse me, why \sigma da = \sigma R2 sin\theta d\phi?

  • @rahilafzalnihal4723
    @rahilafzalnihal4723 3 ปีที่แล้ว

    Thank you for explaining the integration. I was stuck there.

  • @wtj11
    @wtj11 3 ปีที่แล้ว

    ohh gosh thats some ridiculous stuffff !!!!!!!!!!

  • @okiedokie9344
    @okiedokie9344 3 ปีที่แล้ว

    You should leave the company you are working in and join my univeristy as a professor :(

  • @officersmiles9114
    @officersmiles9114 2 ปีที่แล้ว

    this problem was evil

    • @brandonberisford
      @brandonberisford  2 ปีที่แล้ว

      LOL. Yeah its an insane problem to throw at the beginning. Much worse is to come though....