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!

Homework Help: Electromagnetism: Can anyone find the mistake?

  1. Jan 22, 2006 #1
    given: the electric field at a point on the axis a distance x from the plane of a ring is [tex]E = \frac {q*x} {4*pi*E0*(x^2+r^2)^{3/2}}[/tex]

    where E0
    is the permeability coefficient

    The charged ring is replaced by a circular sheet of charge of radius a a surface charge density sigma. The ring can be divided into infinitessimally small rings of radius r and thicknes dr. Show that the electric field is given by [tex] E= \frac {sigma} {2*E0} * [1 - \frac {x} {(x^2 + a^2)^{1/2}}][/tex]

    this is what I did:

    charge on each ring:

    [tex] 2*pi*r*sigma*dr = A*sigma=Q [/tex]

    Electric field on each ring:

    [tex] E = \frac {2*pi*sigma*dr*x*r} {4*pi*E0*(x^2 + r^2)^{3/2}} = \frac {sigma*dr*x*r} {2*E0*(x^2 + r^2)^{3/2}} [/tex]

    Integrate over ring:

    [tex] \frac {sigma} {2*E0} * \int_{0}^{a} \frac {r} {(x^2 + r^2)^{3/2}} dx = \frac {sigma} {2*E0} * [-1/2*\frac{1} {(x^2+a^2)^{0.5}}] (from 0 to a) = \frac {sigma} {4*E0}* [1 - \frac {x} {(x^2+a^2)^{.5}}] [/tex]

    why is that factor 4 here (it's supposed to be 2)? Help's very much appreciated!
    Last edited: Jan 22, 2006
  2. jcsd
  3. Jan 22, 2006 #2


    User Avatar
    Homework Helper
    Gold Member

    Looks like you missed a 'r' in the numerator in when you calculated Electric field on each ring:
  4. Jan 22, 2006 #3
    Yes, sorry...missed to write that one in one line. However, I had it back in the integration the line below, so that it didn't affect the answer. It's now edited.

    Does anyone have any idea about that factor 4?
    Last edited: Jan 22, 2006
  5. Jan 22, 2006 #4


    User Avatar
    Homework Helper
    Gold Member

    The factor is supposed to be '2'. The derivative of [itex] r^2 [/itex] is [itex] 2r [/itex]. So if you take that into account, you will not get '4'.
  6. Jan 22, 2006 #5
    That's exactly what I did and that caused all the trouble:

    substitute: u= x^2 + a^2

    so then you have to multiply by (1/2)...oh yeah....I see! I didn't multiply by two when I did the integration......:yuck:

    Oh dear! :cry:

    Anyways - thank you so much!!!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook