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

Homework Help: Electric field homework

  1. Oct 22, 2004 #1
    Absolutely stumped here!

    A geiger counter has a metal cylinder of 2.10 dimater with a wire stretched along it's axis whose diamtere is 1.34 x 10^-4 cm in dimater. If 855 V is applied between these two what is the electric field at the surface of the wire and the cylinder??

    lets say lambda = Q / L

    then flux = EA = E 2 pi r L = 4 pi k Qenc = 4 pi k lambda L

    so 2 k lambda / r = E

    then i integrate because V = integrate E dr

    so that V = 2k lambda Log r

    But now i m stumped as to how to proceed please help!!
    thanks a lot
  2. jcsd
  3. Oct 22, 2004 #2


    User Avatar
    Science Advisor
    Homework Helper

    You might try solving Laplace's equation directly since there is no charge in the space between the wire and the cylinder

    [tex]\nabla^2 \phi = 0[/tex]

    using cylindrical coordinates and making use of your symmetry approximation. (It's easier than it looks!)
  4. Oct 22, 2004 #3
    ok thansk a lot BUT i'm not that high in math can u try suggeting a lower math approach?
  5. Oct 22, 2004 #4


    User Avatar
    Science Advisor
    Homework Helper

    Sorry about that -

    You can do it the way you started but you have to careful when you integrate (i.e. the [itex]\ln r[/itex] makes no sense because it can't contain units!)

    What you need to do is to integrate from, say, the radius of the inner wire out to a point of interest. To determine the value of the unknown charge you will need to integrate from the radius of the wire all the way out to the radius of the cylinder in which case the potential difference is known to be 855 Volts.

    You will end up with something like

    [tex]V(r) = \Delta V \frac {\ln r/r_0}{\ln R/r_0}[/tex]

    where [itex]r_0[/itex] is the radius of the wire and [itex]R[/itex] is the radius of the cylinder.

    Technically, there is an arbitrary additive constant in there but it doesn't matter where you define the zero of the electrical potential since you're going to calculate the derivative to find the electric field.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook