Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.


    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook