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Shell's Linear Charge density (electric field mistake?)

  1. Feb 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Figure 23-37a shows a narrow charged solid cylinder that is
    coaxial with a larger charged cylindrical shell. Both are noncon-ducting and thin and have uniform surface charge densities on their outer surfaces. Figure 23-37b gives the radial component E of the electric field versus radial distance r from the common axis, and Es = 3.0*103 N/C.What is the shell’s linear charge density?

    ildTYsX.png


    2. Relevant equations
    Gauss' Law 2λ/4piεr = 2Kλ/r


    3. The attempt at a solution

    Since I know the inside of the shell would be 0 I looked at the graph and determined that Eexternal = -2000N/C and r = .035m

    so then I do -2000N/C = 2kλ/r

    do some magic algebra

    λ = -4 * 10-9 C/m


    The solution manual shows something different and says that Eexternal = -3000N/C and that I should use Eexternal-Einternal to solve. The book gets λ′ = –5.8 × 10−9 C/m.


    This really confuses me on many levels.
     
  2. jcsd
  3. Feb 9, 2014 #2
    To start, just be really clear where you're putting you're Gaussian surfaces -- the first one is a cylinder that you put around the outside of the shell, and this is the equation you derive from it, correct?

    $$E = \frac{\lambda}{2\pi \epsilon_{0} r}$$

    I think the problem you face with using that equation is that when you use it for outside the shell, the charge you're enclosing includes both the charge on the inner and outer shells, so you can't solve for either directly. Looks like both shells have different linear charge densities, [itex]\lambda_{inner}[/itex] and [itex]\lambda_{outer}[/itex] -- which can you solve for directly with the above equation?
     
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