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

Chrage density

  1. Aug 18, 2010 #1
    There are basically three types of charge densities; linear [tex]\lambda[/tex] (over a thin ring or rod), surface [tex]\sigma [/tex] over a thin flat surface, and over a volume [tex]\rho [/tex]

    The problem is how these densities produce an electric field: applying Coulomb's law for the electrostatic force using appropriate differentials to match the geometry. The first two aren't two bad; but it's the charge in a sphere (as well as out) done in spherical co-ordinates and using Gauss' law in differential form to find the charge density [tex]\rho [/tex], and then in integral form to find the total charge.
    Last edited: Aug 18, 2010
  2. jcsd
  3. Aug 18, 2010 #2
    So what's the question?
  4. Aug 18, 2010 #3
    It's about electric fields set up by these charge densities..
    for example finding the field caused by a charged ring, charged disk or charged sphere

    p.s. can't edit topic title typo :/
  5. Aug 21, 2010 #4
    Did you want general advice?

    Electric field calculations via Coulomb's law are computed via the following:

    [tex]\vec{E}= \frac{1}{4\pi\varepsilon_0}\int_{q}\frac{dq}{r_{12}^2}\hat{r}_{12}[/tex]

    where [tex]r_{12} = \vec{r}' - \vec{r}[/tex], and primed coordinates represent locations to the source (charge carrier), and unprimed is the observation point (where you wish to find the field). Also, the quantity [tex]q[/tex] is the total charge in the configuration, while [tex]dq[/tex] is a differential element of charge.

    The choice of which charge density to use is (or will be in time) natural:

    [tex] dq = \lambda d\ell' = \sigma da' = \rho dV'[/tex]

    We only integrate over the source coordinates because that is the only place where there is any charge.

    You may find it convenient to use each representation above, when you are computing charges residing on: lines, surfaces, and solids respectively.

    So, for a sphere, it is useful to use the volume charge density, because it is a solid.

    But, I am unable to understand what the question is here. Please advise if I have not answered it properly.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook