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Electric Field above a cylindrical shell w/ charge density

  1. Feb 12, 2015 #1
    1. The problem statement, all variables and given/known data

    We need to use the equation for the E field of a ring to set up the integral for calculating the E field of a cylindrical shell with a charge density of σ and a radius of R and a height of H, at a point P that is D distance away from the cylinder.


    2. Relevant equations

    E= ∫kdQ/r2

    E field for a ring

    E=∫(k*z*dQ)/(R2 + z2)(3/2)

    3. The attempt at a solution

    See this:
    http://i.imgur.com/wi3oTWV.jpg?1

    I feel like my answer for dQ is incorrect.
     
    Last edited: Feb 12, 2015
  2. jcsd
  3. Feb 12, 2015 #2

    gneill

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    Staff: Mentor

    On your figure it indicates that you can start with the equation for the E-field of a ring. You should have that on hand or you can look it up. It should be one of your Relevant equations ;)

    I notice that your dQ doesn't contain any mention of the charge density σ, or any charge at all. So it won't yield a charge value. The form dQ takes will depend on whether σ is a linear charge density or a surface charge density (based upon the surface area of the cylinder). It's likely the latter because there's a tendency to use σ for "surface" and λ for "linear".
     
  4. Feb 12, 2015 #3

    BiGyElLoWhAt

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    Youre right, it is incorrect. You have units of charge on the left and units of area on the right. What times area gives you charge? Its almost right.
    B is pythagoreans theorem, and that is almost right. You need to find the distance from any point on the shell to d. Each point on the shell has a coordinate of (r, theta, z) and the point d has coordinates equal to what? C and d are both wrong. Before you do the substitution in a, youre integrating over charge, and after the substitution the goal is the same, but youre trying to integrate over the charge via integration over another variable: dz. This means you want to integrate over the values of z that contain the charge. From d to d+h there will be sections of z without any charge in them, or depending on how your coordinate system is set up, there might not be any charge at all along that path. So a is almost right, b is almost right, c and d need to be completely corrected.
     
  5. Feb 12, 2015 #4
    Thanks for the feedback!

    For A) I meant to have dQ = 2πRσdz and i assume sigma is the surface charge density.

    B) What would I want to change my r value too? Wouldnt ( z2 + R2)(1/2) give me the distance to each point on the shell?

    C, D, and E are where I get stumped. It almost seems like the problem should be a double integral...
     
    Last edited: Feb 12, 2015
  6. Feb 12, 2015 #5

    gneill

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    I'm having trouble understanding why you would have 2π without an R, and why h is there. dz is a small element of distance in the same direction as h. If σ is a linear charge density then I wouldn't expect to see any reference to circles; σ dz would give you a ring's charge element. If σ is based on surface area, then you'd need to find the surface area of a small ring of height dz. Then you'll have 2πR involved.
    You'll have a double integral if you don't start with the equation for the field due to a ring of charge. That equation already incorporates the integral that summed the effect of the charge around the ring. With it you can just use the z-distance to the ring elements. No radius or cosines in sight :)
     
  7. Feb 12, 2015 #6
    Ive corrected that mistake, in the work.

    So now
    A) is dQ= 2πRσ dz


    I thin my integral should end up looking like

    E= K2πRσ * ∫ dz/(z2+R2)(3/2) going from d as the lower limit, to d+h as the upper.
     
  8. Feb 12, 2015 #7

    gneill

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    Much better!
    Very close indeed. Recheck your equation for the ring of charge. You may have dropped a 'z' ;)
     
  9. Feb 12, 2015 #8
    it would be the Z on top!

    E= K2πσ ∫ z/(R2 + Z2)(3/2) dz

    so now if I were to answer part B on my paper, I now know that r2= (R2 + Z2)(3/2)

    So r = (R2 + Z2)(3/4) ?
     
  10. Feb 12, 2015 #9

    gneill

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    I'm not certain how to interpret the question. When you sum the contributions of rings the only distance of import is the vertical distance between the point of interest and the given ring element. The radius of the ring and the distribution of charge around the ring was taken care of when the ring equation was derived.

    On the other hand if you do a comparison of your integral to the original one with Coulomb's law you can associate the r2 there with the whole z/(R2 + Z2)(3/2) grouping.

    So I'm not sure what to suggest. My instinct is to just say that for summing the ring elements, z is the "radius"
     
  11. Feb 12, 2015 #10
    Would that not change the integral at the end?
     
  12. Feb 12, 2015 #11

    gneill

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    No. You still integrate over z.
     
  13. Feb 12, 2015 #12
    But it would change what was on the bottom, correct? because now r would be (z+d)?
     
  14. Feb 12, 2015 #13

    gneill

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    No, the integral must remain in its current form.

    When the equation for the field on the axis of a ring of charge was derived there was an obvious way to associate the concept of 'r' with the distance between the point of interest and the individual charge element dq in the ring. But here you're charge element is an entire ring, and in my opinion the only 'natural' distance to it is the vertical distance between the point and the center of the ring.

    If you were to set up the problem as a double integral (starting from scratch, without relying on the ring field equation) then you could return to the distance between the point and the individual charge element interpretation for a radius, and you would again have something of the form ##\sqrt{R^2 + z^2}##.

    So as I said, I'm not entirely sure what the question is asking and this is just my interpretation. You may wish to seek other opinions or just ask your professor how he wants you to interpret the question.
     
  15. Feb 13, 2015 #14

    BiGyElLoWhAt

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    That's not the proper distance. The distance is NOT the value of z at that point. The distance along the z axis is d-z, not z. ##r_{z\to d} = \sqrt{R_{cylinder}^2 + (d_z - z)^2}##
     
  16. Feb 13, 2015 #15

    BiGyElLoWhAt

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    Perhaps it was intended to be ##\Delta z## but I see no delta anywhere, and the relation that I'm seeing thrown around suggests that it doesn't matter where the point d is, the distance between somewhere on the cylinder is only dependant on the radius of the cylinder and the z value of that cross section.
     
  17. Feb 13, 2015 #16

    BiGyElLoWhAt

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    You also need a multiplicitave value in there somewhere, to account for the sin of the angle. This, due to symmetry about the axis of the cylinder that d lies on, should be relatively simple to figure out. More pythagoreans theorem.
     
  18. Feb 13, 2015 #17

    BiGyElLoWhAt

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    Touche (kinda)
     
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