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Electric Field On Charged Cylinder

  1. Oct 2, 2011 #1
    I need to find the electric field at all values of radius for an infintely long cylinder of charge. It's in insulator of radius R, and has a volume charge density
    [tex]\rho = \rho_{0}(\frac{r}{R})[/tex] while r<R.

    I need to find the electric field at all points first off. I'm not entirely sure how to go about doing this. I have gauss's law, but I'm not sure how to go about applying it in this case.
     
  2. jcsd
  3. Oct 2, 2011 #2
    I would start with the integral form of Gauss's law.
    [tex] \oint_S \mathbf{E} \cdot \mathrm{d}\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} [/tex]

    In this case your [itex] Q_{enc} [/itex] will be a function of r, and can be found using the volume integral in cylindrical coordinates

    [tex]Q_{enc}= \int \rho d\tau = \int \rho r dr d\theta dz[/tex]

    and integrating the radius from 0 to r (since you want an expression that will work for any r). I would definitely use cylindrical coordinates and just integrate in the z direction from 0 to some length l (the z length will cancel out later).

    For the left hand side of Gauss's law you want to use a cylindrical Gaussian surface so that the electric field is always perpendicular to the surface. In this case [tex] \oint_S \mathbf{E} \cdot \mathrm{d}\mathbf{A} = |E|A \hat{r}[/tex] where A is the area of the of the cylinder (again just use the variable l for the length in this area). After doing the integral to find [itex] Q_{enc}(r) [/itex] you should be able to equate the two sides and solve for E(r) for r<R. When r>R you need to think about what [itex] Q_{enc} [/itex] will be, but I think you have plenty here to start the problem.
     
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