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Electric Field of a Cylinder (Hollow/Solid) at the face

  1. Sep 12, 2015 #1
    1. The problem statement, all variables and given/known data
    Part a.
    A cylinder (with no face) is centered symmetrically around the z axis, going from the origin to infinity. It has charge density σ and radius R. Find the electric field at the origin.

    Part b.
    Same problem, except this time instead of a hollow cylinder with no face we have a solid cylinder. It also now has charge density ρ. Find the electric field at the origin.

    2. Relevant equations

    Ering = kQz/(z2+ R2)3/2

    σ =dq/2πRdz

    Edisk = σ/o * (1 - z/(z2+R2)1/2)

    ρ =dq/πR2dz

    3. The attempt at a solution

    I understand that to complete part a I must integrate the electric field of a charged ring (a distance z from its center) from z = 0 to z = ∞. I understand that to complete part b I must integrate the electric field of a charged disk (a distance z from its center) from z = 0 to z = ∞. Here is my solution to part a (I'm not sure if it's correct):

    EHollow Cylinder = ∫0 kQz/(z2+ R2)3/2 dz = kσ2π/R

    I don't know how to set up the integral for part b. When I integrate the equation for Edisk, with a dz tacked onto the end (not so sure about this), I get undefined. Is this a part of the problem, or is my math incorrect?
  2. jcsd
  3. Sep 13, 2015 #2
    Anyone have any ideas?
  4. Sep 13, 2015 #3


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    If you check the dimensions of your answer, you can see that it has the wrong dimensions for electric field.

    The integral should represent a "sum" of fields of infinitesimally thin rings. The charge in one of the rings will be infinitesimal and should be represented by dq rather than Q. You can express dq in terms of the the charge density σ and dz. (This is how dz gets into the integral.)
  5. Sep 13, 2015 #4
    Thanks for the reply!
    I think I made a mistake typing that in; the equation I integrated (which I did not correctly transcribe, sorry for the confusion) is with dQ = σ2πRdz (that's why my final answer has σ2π in it).

    Assuming this is the correct answer, here's where I get stuck with part b. Rather than being able to manipulate Q into an expression of σ and area, the equation for the electric field of a disk already has σ (which I assume can just be replaced with ρ). I don't see anything I can manipulate but the ρ to get a dz into the equation, but when I do that I get a dq and a dz.
  6. Sep 13, 2015 #5


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    Yes, dQ = σ2πRdz. But, your answer to part (a) is not correct. You can see that your result has the wrong dimensions for an electric field. I think you are probably just making a minor error in not keeping track of the factors of R.

    You cannot replace σ with ρ. They do not have the same meaning or the same dimensions. If dQ represents the total charge in one of the disks of thickness dz, try expressing dQ in terms of ρ, R, and dz. Then, get an expression for σ in terms of ρ and dz by noting that σ = dQ/(area of disk).
  7. Sep 13, 2015 #6
    Part a.
    If I integrate kσ2πRz/(z2+ R2)3/2 dz from 0 to ∞ I get kσ2π, which checks out with dimensions. I believe I was forgetting to integrate with the extra z in the numerator (which does not get replaced by dz).

    Part b.

    Thank you for explaining the difference between ρ and σ! I thought they were interchangeable.

    dQ = ρπR2dz

    σ = dQ / 2πR = ρπR2dz / 2πR = ρRdz / 2

    However, integrating with this substitution from 0 to ∞ still gets me an undefined solution.

    0 (ρR / 4εo) * (1 - z / (z2+R2)1/2) dz = [(ρzr / 4εo) - (ρr (z2 + r2)1/2 / 4εo)]0
  8. Sep 13, 2015 #7


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    OK. That looks good. Of course, you could express your answer in terms of ε0 instead of k.
    Is 2πR the correct expression for the area of the disk?

    To evaluate "at infinity" you need to take the limit of the expression as z → ∞. You should find that the limit is finite.
  9. Sep 13, 2015 #8
    Expressing my electric field for a hollow cylinder in terms of εo, I get σ / 2εo. So, this is the same as the electric field for an infinite sheet of charge at distance z away?

    For part b., I changed the expression for σ (fixed the area of a disk) and got σ = ρdz. I did what you said and expressed the integral evaluation as E = limb → ∞- [E(b) - E(0)]. I get E = ρR / 2εo. This checks for the dimensions of E, I believe (though clearly I'm prone to simple mistakes).
  10. Sep 13, 2015 #9


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    Yes. That's an interesting coincidence!

    That all looks correct to me. Good work.
  11. Sep 13, 2015 #10
    Thank you so much for your help! You're a great teacher.
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