Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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


    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??

  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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook