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Choosing Integrating constants for Electric Field

  1. Sep 9, 2015 #1
    1. The problem statement, all variables and given/known data
    A nonconducting disk of radius R has a uniform positive surface charge density sigma. Find the Electric field at a point along the axis of the disk at a distance x from its center. Assume that x is positive

    2. Relevant equations

    3. The attempt at a solution
    I know I'm suppose to find dEX for one ring and then integrate to find the field due to all the rings.
    dEx= (k) (2πσrx)dr /(x^2 +r^2) ^3/2
    Why should you integrate this component from 0 to R and not R to -R
  2. jcsd
  3. Sep 9, 2015 #2


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    Where exactly is the part of the disk from r=0 to r=-R? That is, where are the negative radius locations?

    By the way, it is not an integrating constant as you suggest in the title. This is a definite integral so there is no integrating constant.
  4. Sep 9, 2015 #3
    For any future reference,what you mean is called "interval of integration" and not constants.
  5. Sep 9, 2015 #4
    sorry you're right I should've said that I don't understand why the limits of integration are 0 to R and not -R to R. The center of the disk is located at (0,0) and so the negative radius is at (0,-R) and the positive is at (0,R). The radius of the first ring I'm integrating is r and so then I have to integrate for the entire disk.
  6. Sep 9, 2015 #5
    Thank you, I will be much more clear next time
  7. Sep 9, 2015 #6
    Think about it logically.You are integrating radially outwards from 0.
  8. Sep 9, 2015 #7
    So dEx=(1/4πε)*((2πσrx)/(x^2+r^2)^3/2)) is the electric field component in the x direction and so when you integrate to obtain the electric field for all the small rings in the disk , you are working your way out towards R, the radius of the entire disk. So that's why you integrate from 0 to R and not -R to R?
  9. Sep 9, 2015 #8
    It's easy,see?
    You've forgotten the dr in your formula.
  10. Sep 9, 2015 #9
    Ahhh ok I see thanks. Yeah, super clear now. Thanks I'll add the dr. Thanks again!
  11. Sep 9, 2015 #10
    You're welcome.I'm studying for an E-M exam right now anyway,so it's good use of my time.
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