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

When is dA not integrated into A for Gauss' law?

  1. Jun 8, 2010 #1
    I have been studying Gauss' law and almost all of the problems I have been doing just have me integrate dA alone into A. I was wondering when do you actually have to do some more in depth integration.
     
  2. jcsd
  3. Jun 8, 2010 #2
    The problems are usually set up with some nice symmetry so you can do the integral trivially. If your area was some crazy shape, or if your b field was not constant then you would actually have to integrate.
     
  4. Jun 8, 2010 #3

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    In general,

    [tex]\int_{\mathcal{S}}\textbf{E}\cdot d\textbf{a}=|\textbf{E}|A[/tex]

    is only true when [itex]\textbf{E}[/itex] is uniform (constant) over the surface [itex]\mathcal{S}[/itex], and normal (orthogonal/perpendicular) to [itex]\mathcal{S}[/itex] at every point on the surface.
     
  5. Jun 8, 2010 #4
    What if it is not perpendicular? Then you will have E[tex]\oint cos \phi dA[/tex]. Then you can just pull out the cosine then because it is constant when compared to dA. If the E is not constant then will you still have to integrate it? Do you have to make it with respect to dA first? The more info the better. Right now, for most occasions I see Gauss' law as EA=q/epsilon naught.
     
  6. Jun 8, 2010 #5

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    That isn't nessecarily true. As a simple counterexample, consider a uniform E-field in the z-direction, [itex]\textbf{E}=E_0\mathbf{\hat{z}}[/itex] and integrate over a spherical surface of radius [itex]R[/itex].

    [tex]\begin{aligned}\int_{\mathcal{S}}\textbf{E}\cdot d\textbf{a} &= E_0\int_0^{\pi}\int_0^{2\pi}\mathbf{\hat{z}}\cdot (R^2\sin\theta d\theta d\phi \mathbf{\hat{r}}) \\ &=E_0\int_0^{\pi}\int_0^{2\pi}R^2\sin\theta \cos\theta d\theta d\phi \\ & = 0 \\ &\neq 4\pi R^2E_0\cos\theta\end{aligned}[/tex]

    Gauss' Law (in integral form) is really only useful (for finding [itex]\textbf{E}[/itex]) in those special cases where, due to symmetry, you can find a surface where [itex]\textbf{E}[/itex] is uniform over the surface and orthogonal to it at every point.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook