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Homework Help: What is the magnitude of the electric field inside the insulator

  1. Sep 6, 2004 #1
    Consider a long uniformly charged, cylindrical insulator of radius R with charge density 1.1 micro-coulombs/m^3. (The volume of a cylinder with radius r and length l is V = pi*r^2*l)

    What is the magnitude of the electric field inside the insulator at a distance 2.7 cm from the axis (2.7 cm < R)? Answer in units of N/C.

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    The axis they are referring to in the problem runs through the cylinder from top to bottom...

    i dont really know where to go with this problem. any pointers/tips/starting points would be great.
     
  2. jcsd
  3. Sep 6, 2004 #2

    HallsofIvy

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    The electrical field from a single point charge q is the vector -qr/|r|3 where r is the vector from the charge to the point and |r| is the length of that vector. (The r in the numerator just gives the direction of the vector. The cube, rather than a square, in the denominator is to cancel the length of that vector.)

    Set up a cylindrical coordinate system with origin at the center of one base and positive z-axis along the axis of the cylinder. The "differential of volume" in cylindrical coordinates is r dr d&theta;dz and the "differential of charge" is &rho;r drd&theta;dz where &rho; is the charge density.
    Integrate -(&rho;r/|rho|3) rdrd&theta;dz (r is the vector from the given point (x,y,z) to the point in the cylinder and r is the distance from the origin to to the point in the cylinder) over the cylinder.
     
  4. Sep 7, 2004 #3

    nrqed

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    The simplest way to answer this is to use Gauss' law. Have you learned this?

    Set up a cylindrical gaussian surface or radius r< R and length L (with its axis coincident with the axis of the real cylinder). The net flux through your gaussian surface will be the magnitude of the E field at a distance r times the surface area of the curved side of the gaussian surface, namely [itex] \Phi = E 2 \pi r L [/itex]. On the other hand, the net flux is also the total charged contained inside your gaussian surface divided by [itex] \epsilon_0 [/itex], according to Gauss' law, i.e. [itex] \Phi = q_{in}/ \epsilon_0[/itex]. The charge contained inside your gaussian surface is [itex] q_{in} = \rho \times \pi r^2 L [/itex]. Now set the two expressions for the flux equal to one another (the length L of your gaussian surface will cancel out) and solve for E. Sub in the values for r (the 2.7 cm), [itex] \rho[/itex] and you're done.

    Pat
     
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