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: Gauss's law

  1. Feb 17, 2004 #1
    I have a HW problem for non-calculus based physics that I can't seem to figure out. Please help me if you can:

    Consider a sphere of radius R woth a uniform positive charge distribution (rho)=q/V. Using Gauss's law, prove (a) that the magnitude of the electric field outside the spherical volume is E=q/4(pi)(permittivity of free space)r^2= (rho) R^3 / 3 (permittivity of free space)r^2, (b) that the magnitude of the electric field insode the sphere at a distance r from the center is E=(rho)r/(permittivity), (c) compare the results of a and b when r=R.


    I have Gauss's law: E(4(pi)r^2=4(pi)koQ and that for r>R, E=ko Q/r^2
    and i substituted in (rho)V for Q and 4/3 (pi)r^3 for V and simplified, but this did not prove part A.
    And for r<R, Q'=Q(r/R)^3 because the inner sphere encloses a volume 4/3 (pi)r^3 which is (r/R)^3 times the total volume. but this doesn't work out either. And how do i reconcile ko from coulomb with the permittivity?
     
    Last edited: Feb 17, 2004
  2. jcsd
  3. Feb 17, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    All you need is Gauss's law. Gauss's law in terms of permitivity of free space (pfs) is:
    E*(Area) = Q/(pfs). The area is 4&pi;r2; the volume of charge is 4/3*&pi;R3. This should give you the answer you seek.
    For r<R, use the same idea. The area is 4&pi;r2; the volume of charge is 4/3*&pi;r3. This should work.
    You don't need to use Coulomb's law, only Gauss's law. But k0 = 1/(4&pi;(pfs)).
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook