2.3.4 Example 7

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  • Опубліковано 1 гру 2024

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  • @jg394
    @jg394  12 років тому +4

    I've got some videos on the problems from chapter 2, but I am hesitant to share. I want people to struggle. Let me know what you think.

  • @hershyfishman2929
    @hershyfishman2929 3 роки тому +1

    For anyone confused, we're integrating the contribution of potential from every point on the surface of the sphere to some point of interest P. The distance 𝓇 to P from each point on the surface = the square root in the denominator of the integral. z = 0 and (obviously) the radius R are both centered at the center of the sphere. Cos θ and in turn the distance 𝓇 will vary as the radius R is moved around to point at different points on the surface.

  • @manis404
    @manis404 12 років тому +1

    Great work you doing here man. This is seriously awesome!

  • @jg394
    @jg394  11 років тому +1

    I tried making some videos, but I decided not to release them. One, they are very long, much longer than the lectures. Two, you really need to use brute force to work through the problems so that you get a good sense of how to do math in physics.
    I am thinking about doing some kind of tutor series, but asking for remuneration for access. That way, students won't turn to it until they have exhausted all other avenues.

  • @rexxar31337
    @rexxar31337 4 роки тому +1

    explanation for solving the integral was very helpful! Thanks alot!

  • @musicislife665
    @musicislife665 6 років тому +4

    Try with u = z^2 + R^2 - 2Rzcos theta. It is a little bit easier. Good job

  • @LeCepheid
    @LeCepheid 10 років тому +2

    Hi, I was simply wondering if you may upload the videos that show more examples. I understand that your hesitant to show these videos, but I'm looking for more sources on how to solve more problems, and I love to hearing your explanations throughout the problems.
    Aside from this, I was wondering how you've decided to choose such substitutions when solving this integral. After taking several classes in Calculus, I've never seen such substitutions. Can you point me in the right direction? Thank you.

  • @jg394
    @jg394  12 років тому +1

    No, YOU'RE awesome!

  • @airsheepgames
    @airsheepgames 11 років тому

    thanks for the explanation. it is way better than reading the book. just one question, why do we have to flip sign when we calculate the inside part?

  • @jimmyti9cer
    @jimmyti9cer 12 років тому

    this is seriously a dream come true.

  • @jg394
    @jg394  11 років тому

    Good catch.

  • @امبسامابوسمره
    @امبسامابوسمره 4 місяці тому

    لك وحششش والله ♥️♥️♥️

  • @JulieanGalak
    @JulieanGalak 6 років тому +1

    Welp, I'm struggling....
    Most of the problems I'm ok on, but I'm stuck on 2.26 from 4th edition (the conical surface one).
    I tried to do an area double integral in cylindrical coordinates, with phi:0->2pi and s=z:0->h
    In the solution manual, he seems to do a single integral with respect to curly-r.
    I'm stuck on two things. First, how can he do this as a single integral. I don't understand that at all. I don't think this was covered earlier in part 1?
    Second, is my method not valid? Or am I just making errors? I don't get the same answers. Close, but not same.
    He got V(a) = (sigma *h)/(2 epsilon0) whereas I got (sigma*h)/(sqrt(2)*epsilon0)
    But for V(b) I got that same (sigma *h)/(2 epsilon0) whereas he has (sigma *h)/(2 epsilon0) *ln(1+sqrt(2))

    • @jg394
      @jg394  6 років тому

      With MV calculus, you look for symmetry as much as possible so you don't have to do messy triple integrals. In this case, it's a uniform surface charge so if you can find the potential for a line of charge from the top to the vertex at the bottom, you can multiply that by 2*pi*r. You are right that he skips a TON of stuff in chapter 1 -- there is no room to condense ALL of MV calculus into a single chapter.

    • @jg394
      @jg394  6 років тому

      BTW, where did you find the solutions? I vaguely recall seeing the solutions for the 2nd Ed and I found them very terse and not at all helpful, except to confirm that you did it right.

    • @JulieanGalak
      @JulieanGalak 6 років тому

      Real Physics - I'll try to find the link

    • @JulieanGalak
      @JulieanGalak 6 років тому

      Real Physics - sent you an email.

  • @AllisinMusic
    @AllisinMusic 7 років тому

    Thanks so much!

  • @kareemmuhammed3370
    @kareemmuhammed3370 6 років тому

    3:16 R SQUARED NOT R