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Derive the expression for the electric field

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data

    A very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume [tex]\rho[/tex]

    Derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density [tex]\rho[/tex].



    3. The attempt at a solution
    [tex]\rho[/tex]/(r*2*pi*[tex]\epsilon[/tex]_0)
     
  2. jcsd
  3. Oct 28, 2009 #2
    Do you know Gauss's law? For an appropriate gaussian surface (D-field perpendicular to the guassian surface and constant everywhere on the surface) DA = Q or D = Q/A, where Q is the total charge enclosed by the Gaussian surface and A is the surface area of the Gaussian surface.

    Choose a cylindrical Gaussian surface and assume that a negligible amount of flux passes through the end caps, since it is a 'very long' cylinder (meaning the end cap area is small compared to the total area of the cylinder).

    Q is the charge density multiplied by the volume of the Gaussian surface. V = h*pi*r^2, where 'r' is the distance from the axis of the cylinder. So Q = rho*h*pi*r^2.

    A is the area of the gaussian surface (neglecting endcap area). A = 2*pi*r*h.

    So, D = Q/A = (rho*r)/2

    The relationship between D and E is E = D/ep_0, so E = (rho*r)/(2*ep_0)
     
    Last edited: Oct 29, 2009
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