1. Not finding help here? Sign up for a free 30min 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 due to a charged disk

  1. Aug 1, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm reading Griffith-Introduction to electrodynamics, In chapter 2 about electrostatics, I've encounter few problems that I've managed, to solve (luck !!), I'm asked to calculate the electric field due to a charged disk of radius R in a point P above the center (Pic)

    2. Relevant equations
    Electric field due to uniform continuous charge distrubition
    [tex] \vec E = \frac {1}{4 \pi \epsilon _0} \int \int \frac {\sigma \hat{r}}{r^2}dS[/tex]

    3. The attempt at a solution
    I've tried to use the usual approach, that is cutting the disk into little circles, [itex] \lambda = \frac {Q}{L} = \frac{\sigma A }{L} = \frac{\sigma 2 \pi r dr}{2 \pi r} = \sigma \cdot dr [/itex]
    The electric field due to that circle is [itex] |\vec E_z| = \frac {\lambda}{4\pi\epsilon_0} \int \frac{dArc}{(R^2 + z^2)^\frac{3}{2}} = \frac {\lambda}{4\pi\epsilon_0} \int_0^{2\pi}\frac {zR \cdot d\theta}{(R^2 + z^2)^\frac{3}{2}} [/itex],You just multiply by 2 [itex]\pi[/itex] and sub the [itex]\lambda[/itex] , [itex] |\vec E_z| = \frac{\sigma zR \cdot dr}{2\pi\epsilon_0}\frac{1}{(R^2 + z^2)^\frac{3}{2}} [/itex], I should have written[itex] d\vec E_z [/itex] because we'll integrate that from 0 up to R, so [tex] \vec E_z = \frac{\sigma}{2\pi\epsilon_0(R^2 + z^2)^\frac{3}{2}} \int_0^zR\cdot dr = \frac{\sigma zR^2}{2\pi\epsilon_0(R^2 + z^2)^\frac{3}{2}}[/tex], I know that this is wrong, so can someone tell me where I've messed things up ?
    One more Request, where can I find more problems about electrostatics, I need some practice Thanks !!
     

    Attached Files:

    Last edited: Aug 1, 2015
  2. jcsd
  3. Aug 1, 2015 #2

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    First, in deriving the E field due to the circle, why is there ##r## when you have already defined the circle radius to be ##R##? Second you should check again the correct expression for the integrand of the E field due to a circle, for instance I guess you know that all E field components perpendicular to the circle's axis cancel out for any point on the axis. This gives us a cosine of the half-angle of the cone formed by the observation point and the circle, what's the expression for this cosine?
     
  4. Aug 1, 2015 #3
    I I've missed that, thanks ! I will Edit that right away, I've included the expression of the cosine in ##\frac{1}{(R^2 + z^2)^{\frac{3}{2}}}##
     
  5. Aug 1, 2015 #4

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    Yes I know that but don't you think you missed one thing in the numerator?
     
  6. Aug 1, 2015 #5
    Aw there a z missing,
     
  7. Aug 1, 2015 #6

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    And finally in the last integral, the integration element should be ##dR## right, which means any ##R## must be inside the integral and to be integrated along.
     
  8. Aug 1, 2015 #7
    I think it should be ##dr##, because I've followed the approach that cuts the circle into little parts, from ##r \text{ to } r+dr##,
     
  9. Aug 1, 2015 #8

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    Nope, remember in your approach initially you wanted to divide the disk into co-centric rings and calculate the electric field due to a single ring with radius ##R##, let's denote this circle's E field ##E_{ring}##. Having found this (hopefully you did it correctly), you still have to add contributions due to the other rings constituting the original disk. Which means you have to add up rings with varying radius from zero out to the radius of the original disk, let's denote it by ##R_{disk}##.
    May be what keeps confusing you is that you haven't changed the ##r## in the expression of ##\lambda##.
     
  10. Aug 1, 2015 #9
    Aw, I've missed that too, In fact I intended to make ##r## the radius of the rings but I wrote ##R## and sticked with it, Okey I'm going to edit that two, I think I wrote a wrong ##\lambda##, it's dR not dr too
     
  11. Aug 1, 2015 #10
    ##\lambda = \sigma dR##
    ##\vec E= \frac{\sigma z}{2\pi\epsilon_0} \int_0^r \frac{RdR}{(R^2 + z^2)^{\frac{3}{2}}} = \frac{-1}{\sqrt{R^2 + z^2}}## evaluate from 0 to r and I get ##|\vec E| = \frac{\sigma z}{2\pi\epsilon_0}\cdot(\frac{1}{\sqrt{r^2 + z^2}} - \frac{1}{z})##
     
    Last edited: Aug 1, 2015
  12. Aug 1, 2015 #11

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    Only some small mistakes left there, first you should check again whether ##\pi## should be present and why the magnitude of the E field is negative.
     
  13. Aug 1, 2015 #12
    ##|\vec E| = \frac{\sigma}{2\epsilon_0}\cdot(1 - \frac{z}{\sqrt{r^2 + z^2}}) ## Waow thank you very for your assistance :)
     
  14. Aug 1, 2015 #13

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    You are welcome.
     
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 due to a charged disk
Loading...