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Electric field from a cylindrical shell

  1. Oct 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider a uniformly charged thin-walled right circular cylindrical shell having total charge Q, radius R, and height h. Determine the electric field at a point a distance d from the right side of the cylinder as shown in the figure below.

    db7f47d8d64456bd28cb0002bc64c9d3.jpg

    2. Relevant equations

    E = kQ/r2

    3. The attempt at a solution

    Ok so I started by finding the electric field at a distance d from a ring of uniform charge:

    E = [tex]\stackrel{kdQ}{\sqrt{(d\stackrel{2}{}+R\stackrel{2}{})\stackrel{3}{}}}[/tex]

    Now I want to treat this shell as a collection of infinitely thin rings, so I need to sum the electric field due to the rings and integrate right? This is where Im having trouble:

    E = [tex]\stackrel{k(d+dx)Q}{\sqrt{((d+dx)\stackrel{2}{}+R\stackrel{2}{})\stackrel{3}{}}}[/tex]?

    Then integrate from 0 to h? What about the charge Q? Should it be Q/h?

    Im confused as to how I set this integral up. Any help is greatly appreciated! thanks!
     
  2. jcsd
  3. Oct 17, 2009 #2

    ideasrule

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    Almost. However, a simpler way to think about this problem is to use a single variable, x, to denote the distance to a specific ring making up the cylinder. Instead of dQ, you'd use Qdx/h; instead of d^2, you'd use x. Then you just have to integrate.
     
  4. Oct 17, 2009 #3
    So then I would have something like

    E = [tex]\frac{kQdx}{h\sqrt{(x\stackrel{2}{}+R\stackrel{2}{})}\stackrel{3}{}}[/tex]?

    and integrate from x=d to x=d+h?
     
  5. Oct 17, 2009 #4

    ideasrule

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    Oops, there's one more error I didn't catch. The electric field due to a ring is kdQ/r^2 *cos(theta)=kdQ/r^2 * x/r, where x is the distance to the ring's center. I think you left out the x.
     
  6. Oct 17, 2009 #5
    oh, I had d as the distance to the ring's center, now im confused? what is the d in your equation for a ring?
     
  7. Oct 17, 2009 #6

    ideasrule

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    Sorry for being so confusing.


    Originally you had this:

    You might want to check this equation. It should be kdQ/(d^2+R^2) * cos(theta), and cos(theta)=d/sqrt(d^2+R^2). You just missed the "d". Now, apply the changes I suggested: x=d, dQ=Qdx/h, etc.
     
  8. Oct 18, 2009 #7
    But my text book has a different formula than that, one without cos theta. here:

    IMG_5547.jpg

    So which one do I use?

    In my work I replaced z with d to match the diagram, and that 1/4piepsilon = k
     
  9. Oct 18, 2009 #8
    Oh I see, it should be d2 in the numerator?
     
  10. Oct 18, 2009 #9
    So youre saying d=x, but then you say that d=dx/h. I think youre getting confused because of the dQ. I didnt write that to mean Delta Q or differential of Q or anything like that. the "d" in "dQ" is the same as the one in the denominator.

    EDIT: Ok so this is what I have afer your changes:

    [tex]\int\stackrel{d+h}{d}\frac{kxQ}{h(x\frac{2}{}+R\frac{2}{})\frac{3/2}{}}dx[/tex]

    It came out sloppy, should be integral from d to d+h

    Please tell me if this is right!
     
    Last edited: Oct 18, 2009
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