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 of an annulus

  1. Feb 14, 2012 #1
    A thin disk with a circular hole at its center, called an annulus, has inner radius R_1 and outer radius R_2 (there's a picture which I will try to describe as best I can at the end of question). The disk has a uniform positive surface charge density δ on its surface. (a) determine the total electric charge on the annulus. (b) The annulus lies in the yz-plane, with its center at the origin. For an arbitrary point on the x-axis (the axis of the annulus), find the magnitude and direction of the electric field E (bold for vectors). Consider points both above and below the annulus.

    there's more but for now I just need help on part b. The picture was this: http://www.cramster.com/answers-aug...ar-hole-center-called-anann_633159.aspx?rec=0


    2. Relevant equations

    δ = Q/A(area)
    E = 1/(4*π*ε_0) * ((δ*A)/(R)^2)

    3. The attempt at a solution
    Part A was easy, I thought, just:
    Q = δ*A = δ*((2π(R_2)^2)-(2π(R_1)^2))

    Then Part B (I'll try to explain in as clear terms as possible).
    I divided the annulus into rings of area (2π(dR)^2)-(2π(R_1)^2). dR being an infinitely small distance for the 'outer radius.'

    I set E = 1/(4*π*ε_0) * ((δ*(2π(dR)^2)-(2π(R_1)^2))/(dR + R_1)^2)

    and then tried to doubled integrate. I used the bounds of the first integral for the ring as [R_1, R_2] and the second integral from [0,dwidth] for the annulus part. (I haven't been taught double integration yet, by the way - just a thing the university does - so I'm trying to figure it out as I go along)

    I believe I have to use a dh (height) term in here somewhere and can I even do (dR)^2???? Would this problem be better solved with thin slices with a d∅ term? I just did this problem: http://www.cramster.com/answers-sep-11/physics/semicircle-radius-quadrant_1485829.aspx?rec=0 .

    So for a annulus, would I just need to use a full circle and double integrate to account for the thickness? That seems logical, but how do you go about adding another dimension to the problem... implicit differentiation??
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 15, 2012 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Here's the image:

    attachment.php?attachmentid=43942&stc=1&d=1329320683.jpg
     

    Attached Files:

  4. Feb 15, 2012 #3

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You should have rings with inner radius R, and outer radius R + dR.

    The area of such a ring is dA = (2πR)dR .

    Have you covered the E field due a ring of charge -- either in class or in an assignment?
     
  5. Feb 15, 2012 #4
    We did the second problem I showed you. But I think I figured it out. My professor showed me the cylindrical coordinate system and it was a little easier to solve it that way.

    Ended up double integrating first for d∅[0,2π] then for dρ[r_1, r_2]. Thanks for you help. I have another question for part (c):

    c) show that at points on the x-axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the center of the annulus and the point. How close is "sufficiently close"?
    d) A point particle with mass m and negative charge -q is free to move along the x-axis (but cannot move off the axis). The particle is originally placed at rest at x = .01R_1 and released. Find the frequency of oscillation of the particle

    My question is how can you show that? Will I need to do some kind of limit as x approaches 0 of the function I found for part b? that would just give me 0. (I used l'hopital's rule cause you can't have infinity - infinity and the derivative of the numerator would = 0 because bot R terms are constant.

    My answer for part b is:

    E = R_2/(2*ε_0*x*sqrt(R_2^2+x^2) - R_1/(2*ε_0*x*sqrt(R_1^2+x^2)

    x is the arbitrary point I chose as my test charge. when deriving the total E for the annulus
     
  6. Feb 15, 2012 #5

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    That can't be correct. It will not have the right units/dimensions for one thing.

    There is no electric charge, and the wrong power of units of length.
     
  7. Feb 16, 2012 #6
    I corrected it:
    E = -(x*δ)/ε_0 * 1/sqrt(R_2^2+x^2) + (x*δ)/ε_0 * 1/sqrt(R_1^2+x^2)
     
  8. Feb 16, 2012 #7
    Now I am trying to take the limit of this function as it approaches 0. However, I just get 0 due to the x term multiplying in the function. Is this the answer? it makes sense to me because there will be no x component to the E vector.
     
  9. Feb 16, 2012 #8

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    [itex]\displaystyle \vec{E}=\frac{x\delta}{\varepsilon_0}\left(\frac{1}{\sqrt{{R_1}^2+x^2}}-\frac{1}{\sqrt{{R_2}^2+x^2}}\right)\hat{i}[/itex]

    Factor out R12 and R22 inside each radical, then take each out of its radical:
    [itex]\displaystyle \vec{E}=\frac{x\delta}{\varepsilon_0}
    \left(\frac{1}{\displaystyle R_1\sqrt{1+\frac{x^2}{{R_1}^2}}}
    -\frac{1}{\displaystyle R_2\sqrt{1+\frac{x^2}{{R_2}^2}}}\right)\hat{i}[/itex]​


    Now consider what this approaches for x << R1 < R2 .
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook