1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric field of a Ring

  1. Sep 6, 2013 #1
    1. The problem statement, all variables and given/known data

    I am suppose to find an expression for the electric field of a ring.


    2. Relevant equations

    [tex]E =\frac{Kq}{r^2}[/tex]


    3. The attempt at a solution

    I calculated my results and I reached up to this:

    [tex]\frac{Kx\Delta q}{(R^2 + x^2)}^{3/2}[/tex] where R = radius, x = distance, K = constant, q = charge.

    And then I looked at my book and noticed they integrated with respect to [tex]\Delta q [/tex] which got me confused because [tex]\Delta q [/tex] is not a geometric property. And the final expression the book got was [tex]\frac{Kx q}{(R^2 + x^2)}^{3/2}[/tex].

    How come they can integrate with respect to [tex]\Delta q [/tex]?
     
  2. jcsd
  3. Sep 6, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You can integrate over the total charge in the same way you can integrate over the length of the ring, or over the total mass to calculate the moment of inertia. The integration variable does not have to be a length, it can be nearly everything.
     
  4. Sep 6, 2013 #3
    Integration is just summation of infinitely small elements of something. Here you are adding up all the Electric field due to all the Δq of the ring at a point along the axis at a distance x. So according to your equation

    [itex]\sum \Delta E =\sum \frac{K x\Delta q}{(x^2+y^2)^{3/2}}[/itex]
    becomes,
    [itex]\int \ dE =\int \frac{K x dq}{(x^2+y^2)^{3/2}}[/itex]
    when Δq→0
     
  5. Sep 6, 2013 #4
    But when I was measuring the electric field of a line of charge I got the expression [tex]\sum \frac{d \Delta Q}{(y_1^2 + d^2)^{3/2}}[/tex] but with that expression I cant integrate over [tex]\Delta Q[/tex] because its not a geometric quantity so I had to replace [tex]\Delta Q[/tex] with [tex]\Delta Q = Q/L \Delta y.[/tex] I don't understand why we had to change it with the field of a line but not with a disk?
     
    Last edited: Sep 6, 2013
  6. Sep 6, 2013 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Integration variables don't have to be geometric quantities.

    You don't have to in both cases.
     
  7. Sep 6, 2013 #6
    I am not understanding, then why did my book try and trick me?
     
  8. Sep 6, 2013 #7
    They did those two problems different with the same reasoning I mentioned above. And they did the same thing with an electric field of a disk as they did with the line of charge by replacing [tex]\Delta Q[/tex] with the density over the surface area [tex] 2r\pi dr[/tex].
     
  9. Sep 7, 2013 #8

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I don't think it did.
    If there are multiple ways to solve a problem, you have to choose one. There is no "right" or "wrong" choice.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Electric field of a Ring
Loading...