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Deriving Electric Field Inside Plastic Rod Using Gauss's Law

  1. May 3, 2016 #1
    1. The problem statement, all variables and given/known data

    A straight circular plastic cylinder of length L and radius R (where
    R ≪ L)
    is irradiated with a beam of protons so that there is a total excess charge Q distributed uniformly throughout the cylinder. Find the electric field inside the cylinder, a distance r from the center of the cylinder far from the ends, where r < R.


    2. Relevant equations
    Gauss's Law where ∫Edl = q/e0 where q is the charge inside a certain area. Volume of sphere: 4/3πr^3 and volume of a cylinder: 2πr^2L with r being radius of cylinder and L being the length of the cylinder.

    3. The attempt at a solution
    I decided to make my gaussian surface a sphere which was located inside the cylinder. The charge inside that sphere would then be Q(Asphere/Acylinder). I then divided that by e0. I set that equal to the electric field multiplied by the area of the sphere. Like it is supposed to - the surface area of the spheres cancel leaving me with E = Q/(e0*Acylinder). The surface area of the cylinder is given as 2piRL + 2piR^2 so my final answer is Q/(e0*(2piRL + 2piR^2)) but apparently that is not correct. I have a feeling I messed up my charge distribution somehow because I believe that I have to find the charge inside the sphere as a function of volume instead of surface area.
     
  2. jcsd
  3. May 3, 2016 #2

    BvU

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    What is your argument for using a sphere as a gaussian surface ? Does that make use of any symmetry, and of r << R in any way ?
     
  4. May 3, 2016 #3
    I am working under the assumption that since charge is evenly distributed throughout the cylinder, that a gaussian sphere would act like a point charge.
     
  5. May 3, 2016 #4

    BvU

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    Maybe, but there's more charge in the neighborhood, and that charge also contributes to the electric field !
    The gauss theorem is always true, but it is most useful if you can work it around in such a way that the E field is perpendicular to the surface everywhere, so that ##\vec E \cdot d\vec A## becomes ##|\vec E| dA##. For that you need to exploit some symmetry present. The wording of your exercise very clearly points at the symmetry (symmetries) you need to exploit here !
     
  6. May 3, 2016 #5
    How am I supposed to find symmetry if I do not know the pattern of the field? I assume that it is evenly distributed, but I cannot draw any other conclusions.
     
  7. May 3, 2016 #6

    BvU

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    You can assume the cylinder is infinitely long (take it from me). You can pick the origin of a coordinate system anywhere on the axis of the cylinder. Describing features of a cylinder is easiest with cylindrical coordinates . Does it matter where we pick z = 0 ? No. That's a symmetry. Does it matter whch way is the z+ ? No. Thats another symmetry. And so on with ##\phi##. Not with r (sometimes designated ##\rho##), but we chose r = 0 on the axis (for good reasons).
     
  8. May 3, 2016 #7
    Isn't cylindrical coordinates, with phi, theta and rho? What are you referring to as z+? Even with that information how are you supposed to perform a triple integral to evaluate the flux? If so, will the integrand just be 1 or is there a particular function to integrate? If there is, would it be the cylinder represented in cylindrical coordinates?
     
  9. May 4, 2016 #8

    BvU

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    The blue letters indicate a clickable link. There you can find that cylindrical coordinates describe 3d space with a radius, a z and a phi. Here we use r for the radius because he symbol ##\rho## is used for charge density.

    The origin can be chosen anywhere on the axis of the cylinder -- you get the same situation. (Translation symmetry). That means ##|\vec E (r,\phi,z)|## can not depend on ##z##. But there still could be a z component.

    The positive z axis has to point in some direction, e.g. up. The mirror symmetry of this problem is that if you let it point down, then you get the same problem. That means the E field has no z component: if it had, there would be two solutions and there is only one solution for the E field (the solution of the Laplace equation is unique). So far ##\vec E(r,\phi,z) = (E_r(r,\phi),E_\phi(r,\phi), 0)##.

    Digest, research, check etc. these (in physics very important) steps and try to reduce ##\vec E(r,\phi,z) = (E_r(r,\phi),E_\phi(r,\phi), 0)## to ##\vec E(r,\phi,z) = (E_r(r),0, 0)## yourself.
    A triple integral of the constant charge density ##\rho## isn't that complicated.
    And the surface integral isn't complicated either if you can work things around in such a way that the E field is perpendicular to the surface everywhere (see #4) :smile: .

    Don't your notes or textbook include examples of using gauss law ?

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