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Cylindrical Shells and Gauss' Law

  1. Feb 4, 2007 #1
    1. The problem statement, all variables and given/known data

    An infinitely long cylindrical shell of radius 6.0 cm carries a uniform surface charge density sigma = 12 nC/m^2. The electric field at r = 5.9 cm is approximately


    a.0.81 kN/C

    b.zero.

    c.1.3 kN/C.

    d.12 kN/C.

    e.0.56 kN/C.



    2. Relevant equations

    See below.

    3. The attempt at a solution

    Will the correct choice be b. 0? I am assuming that this is a conducting cylindrical shell. When r < R, E will be 0.




    1. The problem statement, all variables and given/known data

    An infinitely long cylinder of radius 4.0 cm carries a uniform volume charge density rho = 200 nC/m^3. What is the electric field at r = 8.0 cm?


    a.0.23 kN/C

    b.0.11 kN/C

    c.57 kN/C

    d.0.44 kN/C

    e.zero



    2. Relevant equations

    electric flux = Integral [E*dA] = Q_enclosed/epsilon_0

    3. The attempt at a solution

    I sloshed through this one because I mixed up the radii (r and R) at first. Did I eventually get the correct answer?

    Q_enclosed = rho*2*pi*R^2*l

    flux = E*2*pi*r*l

    E*2*pi*r*l = (rho*2*pi*R^2*l)/(epsilon_0)

    E = (rho*R^2)/(r*epsilon_0) = (200*10^-9 c/m^3)(0.04 m)^2/[0.08 m*8.85*10^-12] = 452 N/C??



    Thanks.
     
  2. jcsd
  3. Feb 4, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    No reason to think the shell has any thickness or that it's a conductor. Nonetheless, since the charge is on the surface of the shell (and we presume no other charge exists) the field will be zero for r < R. (Since the charge enclosed within any cylindrical shell with such a radius would be zero.)


    I didn't check your arithmetic, but your method is correct.
     
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