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How did they integrate the charge for Voltage question

  1. Feb 1, 2016 #1
    I have a question on finding potential for a disc of charge when it comes to the simple integration
    upload_2016-2-2_0-18-39.png
    how does the integral of R'dR'/etc have the R'dR' just disappear? i thought it would then be R^2/2
    I am obviously missing something here
     
  2. jcsd
  3. Feb 1, 2016 #2

    gneill

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    Staff: Mentor

    Hi kiwibird4. Please remember to use the formatting template for all questions posted in the homework section.

    The integral in question is not simply of R'dR'. Note that the denominator contains ##\sqrt{z^2 + R'^2}##.
     
  4. Feb 1, 2016 #3

    ehild

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    Homework Helper
    Gold Member

    u-substitution ##u = z^2+R'^2## is applied.
     
  5. Feb 2, 2016 #4
    I'm SORRYYYY :cry::cry::cry::cry::cry::cry::cry::cry::H:H:H:H:H:H:H:H
    I even had to check the lil button that said -I have followed the homework format
    :cry:
     
  6. Feb 2, 2016 #5

    gneill

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    Staff: Mentor

    Somehow we must all struggle through this great tragedy and believe in the promise of a brighter tomorrow :smile:

    Cheers.
     
  7. Feb 2, 2016 #6

    Mark Harder

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    Gold Member

    Assuming you have looked up this integral in tables of same, I think what you're missing is the assumptions about the variables and z that would make the problem trivial if only you knew what they are. Even if you have success using the methods others have suggested earlier, looking for assumptions is a useful exercise that can pay off later on...

    As an aside, it took me a while to get Mathematica to solve your integral. The evaluation hung up until I specified the assumptions required to give an unambiguous answer (The same one, even!). MMA's both interesting and frustrating that way. In order to be as general as possible, it makes few assumptions but provides the user with methods to specify them.
     
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