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Gaussian Cylinder help

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data

    A cylindrical shell of radius 7.00 cm and length 260 cm has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 20.0 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. Find values for the following.

    (a) the net charge on the shell

    (b)the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell

    2. Relevant equations

    E = Q/(4π r² εo)

    3. The attempt at a solution

    This problem is giving me such a headache. I thought you just have to rearrange the above equation to solve for Q as that is the total charge but I'm wrong. I don't understand what's going on conceptually with all these gaussian problems. I looked at online and book figures and it's just not clicking yet. PLEASE help!
  2. jcsd
  3. Oct 25, 2008 #2
    the equation you're using is not Gauss' Law - it's the electric field due to a point charge or spherical distribution of charge.

    Gauss' Law:

    [tex] \int{E dA} = \frac{Q}{\epsilon o}[/tex]

    The integral might look complicated, but for basic problems you can rewrite it for the following symmetries:

    Spherical: E * 4 π R²
    Cylindrical: E * 2 π R L
    Planar: E * A
  4. Oct 25, 2008 #3
    Thank you so much!!!!

    I am so confused with all this electric stuff...it's like a foreign language to me. It's the first time I'm completely lost in my Physics class.
  5. Oct 26, 2008 #4
    Ugh still a little confused.

    So I use E * 4pie R^2 --> 36000 * 4pie (.07^2)

    How do I find the net charge though?
  6. Oct 26, 2008 #5
    So the equation I just used helps me find the net electric flux which equals the net charge/e0.

    So to find the net electric charge, I should multiply epsilon0 by the net electric flux...but I get a wrong answer.
  7. Oct 26, 2008 #6
    What expression did you get for the net electric flux through a cylinder?
  8. Oct 26, 2008 #7
    That would be E*4pieR^2 right?

    and my answer to that was 2216.708 using a radius of .07m. Then multiplying that number by epsilon0 should give me the total charge right?
  9. Oct 26, 2008 #8
    No, that's the flux through a sphere. The area of the length side of a cylinder if 2*pi*r*l.
  10. Oct 26, 2008 #9
    Ok, so q = (E*2*pi*r*l) * (epsilon0)

    That's q= 36000*2*pi*.07*.260 * (8.88542 x 10^-12) = 3.66 x 10^-8.

    It's still the wrong answer.
  11. Oct 26, 2008 #10
    The electric field you're given is located at a distance of 20 cm from the radius of the cylinder, not 7 cm! Also, 260 cm = 2.6 m.

    I think you're not picturing flux correctly - what you're doing is imagining a hypothetical cylinder wrapped around your charged cylinder at a certain distance (in this case 20 cm from the axis), and given a constant flux (E field times area of the length of the cylinder) Gauss' Law is telling you the amount of charge contained in your imaginary cylinder (which, by the problem statement, is the charge on the real cylinder).
  12. Oct 26, 2008 #11
    Ah yes you are absolutely correct. Your description is perfect! So am I understanding that the net charge I am solving for in this problem is that of the imaginary cylinder which is the SAME as the real cylinder?

    But....even with the mistakes I made, I'm still getting the wrong answer for some reason. I replaced .07 with .20m and I changed .26 to 2.6m.

    q= 36000*2*pi*.20*2.60 * (8.88542 x 10^-12) = 1.05 * 10^-6.

    edit: nevermind, I realized I had the wrong units. This is correct. I'm sorry I haven't had any sleep and I'm a little off lol
  13. Oct 26, 2008 #12
    I don't understand why the answer to part b would be 0 though.

    It's asking to find the electric field.

    That means E= q/(epsilon0*2*pi*R*l)

    The problem states that it is 4.0 cm from the axis which means that r=.04m.

    Using the net charge found from part a (1.05 * 10^-6)/(epsilon0 * 2 * pi * .04m * 2.6m)...this doesn't give me 0 though.
  14. Oct 26, 2008 #13
    The "q" in that formula is the total charge inside an imaginary cylinder of radius 4 cm. And that charge is...?
  15. Oct 26, 2008 #14
    Apparently 0 lol. But I don't understand why...it's the same concept as part a isn't it? Why is the net charge 0 in this case?
  16. Oct 26, 2008 #15
    How much charge is there inside the radius of 4cm? Where is the charge you calculated in part a located?
  17. Oct 26, 2008 #16
    That charge is located 20cm away from the axis.

    Is it because 4cm is within the 7cm of the real cylinder, thus making it 0 since the inside charge of a closed surface is always 0?
  18. Oct 26, 2008 #17
    The charge is located 7cm away from the axis, not 20cm. But in either case, yes, inside a radius of 4cm there is no charge!

    The inside charge of a closed surface can be anything, by the way. But in THIS case, it's zero for a cylinder of radius of 4cm.
  19. Oct 26, 2008 #18
    So anything up to 7cm would be 0, anything above would be the equation I used earlier?
  20. Oct 26, 2008 #19
  21. Oct 26, 2008 #20
    Thank you all for your help! Seriously, I cannot tell you how grateful I am. This forum is really appreciated, especially with members like you!

    Thanks again.
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