Finding mag E of 2 thin-walled concentric cylindrical shells

  1. Two long, charged, thin-walled, concentric cylindrical shells have radii of 3.0 and 6.0cm. The charge per unit length is 5.0x10^-6 C/m on the inner shell and -7.0x10^-6 C/m on the outer shell. What are the magnitude E and direction radially inward or outward? I figured I could use the following:

    E = [itex] \delta [/itex]/2PIEoR;

    I used E = 5.0x10^-6/(2PIEo)(.03)) = 2997268 inside
    E = -7.0x10^-6/(2PIEo)(.06)) = -2098087 outside
    I added these together and got a wrong answer... it should be
    2.3x10^6 N/C and if the final answer is postive that means its outward, if its negative that means E is inward.
     
  2. jcsd
  3. Your equation is correct, but the E-field that you have to find depends on where you are looking. The E-field far away from the cylinder is smaller in magnitude than close up. Therefore, you (or the quesiton) needs to specify where you are looking for the E-field. Is it in the region outside of the cylinders? Is it in the region between the cylinders? Is it inside the inner cylinder? Are these shells conductors?
     
  4. mukundpa

    mukundpa 507
    Homework Helper

    I think you have missed a line in the question. The distance from the axis of the cylinders at which the field required is given and most probabally it is 4 cm
     
  5. HallsofIvy

    HallsofIvy 40,967
    Staff Emeritus
    Science Advisor

    How did you get that? It certainly wasn't said in the original post. And if it is given, why do you say "most probably"?
     
  6. if shells are conductors i get E=[tex]2.25\cdot10^6[/tex] at r=4 cm too.
     
  7. You are correct, its odd how they worded this problem, it says: What are the (a) magntiude E and (b) direction of the e field at radial distance r= 4.0cm. What are (c) E and (d) direction at r = 8.0cm. How did you figure out E? Sorry about delyaed responce i was out of town.
     
  8. Doc Al

    Staff: Mentor

    You figure out E by using the equation you gave in your first post. [itex] \delta [/itex] is the total charge per unit length contained within the given radius.
     
  9. I tried that and i got the wrong answer, i think i'm not understanding you correctly...

    They say they want the e field at radial distance r= 4.0cm. Radial distance of what? The inner tube, the outer tube? If i use this equation: E = [tex]\delta[/tex]/2PIEoR;
    For R, i'm confused on which R i should use, I already tried using the 2 given radi then add the result together. I also tried using .04m ffor both radi, also wrong....def. lost as usual. :bugeye:
     
  10. mukundpa

    mukundpa 507
    Homework Helper

    Radial distance means the distance along the radius of the cylinders. The point is at a distance of 4 cm from the axis of cylinders. The charge(+) on the inner cylinder induces equal and opposite charge(-) on the inner surface of the outer cylinder and the corresponding (+) charge will go to the outer surface of outer cylinder. Thus the scenario will be such that
    1 There will be +5.0x10^-6 C/m charge on the outer surface of the inner shell
    2 -5.0x10^-6 C/m charge on the inner surface of the outer shell and
    3 -2.0x10^-6 C/m charge on the outer surface of the outer shell.

    As the e field 'inside' a uniform long cylindrical charge distribution is zero, the field at a distance of 4 cm from the axis of the cylinders due to the charges on outer cylinder is zero because the point is within the outer cylinder. The only charge on the surface of inner cylinder is responsible for e field, and that's why it is
    E = delta /2PIEoR =5.0x10^-6/(2PIEo)(.04)) = 2.25 x 10^6 N/C
    (The difference in the result if it is 2.3 ...may be due to the way in which the value of 2PiEo is calculated.)
     
  11. First, view the following drawing of 2 (long, thin-walled, conductive) charged concentric cylindrical shells having radii of R1=3.0 cm and R2=6.0 cm:
    http://physics.uwstout.edu/colphys2/problems/prblm_images/prblm4c.gif

    In the E formula, R is measured from the central axis running down the center of the inner cylindrical shell, just like R1 and R2 are measured in the above drawing. For problem items (a)&(b), R=4.0 cm and is between the inner and outer cylinders, and for problem items (c)&(d), R=8.0 cm and is outside the outer cylinder.

    In the E formula, δ is always the TOTAL NET charge per unit length contained (or "enclosed") within the (imaginary) cylinder of radius R (which is 4 cm for (a)&(b) and 8 cm for (c)&(d), measured from the central axis). Thus, from the original problem specifications:
    For R=4 cm: total net enclosed δ = (+5.0x10^-6 C/m)
    For R=8 cm: total net enclosed δ = (+5.0x10^-6 C/m) + (-7.0x10^-6 C/m) = (-2.0x10^-6 C/m)

    Now, use the E formula for each of the problem items:

    [tex] \mbox{Items (a)and(b) for R=4 cm: } \ \ E \ = \ \frac{\delta}{2 \pi \epsilon_{0} R} \ = \ \frac{(+5.0 \times 10^-6 \ C/m)}{2 \pi \epsilon_{0} (0.04 \ m)} [/tex]

    [tex] \mbox{Items (c)and(d) for R=8 cm: } \ \ E \ = \ \frac{\delta}{2 \pi \epsilon_{0} R} \ = \ \frac{(-2.0 \times 10^-6 \ C/m)}{2 \pi \epsilon_{0} (0.08 \ m)} [/tex]

    Your answers should agree with the textbook.
     
    Last edited: Jul 30, 2005
  12. Doc Al

    Staff: Mentor

    Realize that the value of the field depends on how far you are from the axis of the cylinders. They want you to evaluate the field at R=4.0cm, which is between the two charged cylindrical shells. The only charge that contributes to the field at that point is the charge on the inner shell.

    The equation is correct. They ask for the field at radial distance r = 4.0cm, so R = 0.04m. [itex]\delta[/itex] is the charge per unit length contained within the R = 0.04m, which is just the charge/length of the inner shell. (The outer shell's charge is at R=0.06m, so it doesn't contribute to the field at R=0.04m.)
     
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