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Magnitude of en electric field and electrostatic field

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data

    I have a solid, insulating, finite-thickness spherical shell (inner radius a, outer radius b). It's uniformly charged with a total charge of Q. i'm looking for the electric field E, "everywhere in space, including points outside and inside the spherical shell"

    Also, i'm looking for the electrostatic potential of the inner and outer surface of the shell

    2. Relevant equations

    E = (1/4pie0)(Q/r^2)r

    3. The attempt at a solution

    I'm pretty much lost on this problem.
     
  2. jcsd
  3. Sep 28, 2008 #2

    gabbagabbahey

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    [tex]\vec{E}(\vec{r})=\frac{1}{4 \pi {\epsilon}_0} \frac{q}{r^2} \hat{r}[/tex]

    is the electric field due to a point charge at the origin.

    What is the expression for the electric field due to a surface charge [tex]\sigma[/tex]?

    Have you learned Gauss's law yet?
     
  4. Sep 28, 2008 #3
    Yes, we've learned Gauss' law but I'm unsure how to apply it to the problem.

    Due to the surface charge sigma wouldn't the equation be (2*pi*(radius)^2*sigma)/epsilon?
     
  5. Sep 28, 2008 #4

    gabbagabbahey

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    No, the correct formula for calculating the electric field due to a surface charge [itex]\sigma (\vec{r'})[/itex] is:

    [tex] \vec{E}(\vec{r})=\frac{1}{4 \pi {\epsilon}_0} \int_{\mathcal{S}} \frac{\sigma (\vec{r'})}{|\vec{r}-\vec{r'}|^2} \widehat{\vec{r}-\vec{r'}} da'[/tex]

    Where [itex]|\vec{r}-\vec{r'}|[/itex] is the distance between an infinitesimal piece, dq, of the surface charge located at [itex]\vec{r'}[/itex] and the field point [itex]\vec{r}[/itex]. And [itex]\hat{\vec{r}-\vec{r'}}[/itex] is the unit vector pointing from the infinitesimal source dq, to the field point.

    Such integrals are usually very difficult to evaluate, so I recommend using Gauss's Law if you can....what types of symmetry can Gauss's law be used for? What type of symmetry is present in the given problem?
     
    Last edited: Sep 28, 2008
  6. Sep 28, 2008 #5
    Gauss' law can be used for cylindrical, spherical, and planar symmetry. So here there is spherical symmetry. Looking this up gives me the equation of

    E = (QA)/(4*pi*epsilon_o*r^2)
     
  7. Sep 28, 2008 #6

    gabbagabbahey

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    Okay, but what is Q_enclosed inside the spherical shell? How about outside? (I'm assuming you meant Q_enclosed by Q_a)

    Did you just look up this answer or did you calculate yourself using Gauss's law?
     
  8. Sep 28, 2008 #7

    tiny-tim

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    Hi gabbagabbahey ! :smile:

    In the middle of text, the PF server doesn't like a lot of ordinary LaTeX, so use itex and /itex instead of tex and /tex (for "inline LaTeX"): :wink:

     
  9. Sep 28, 2008 #8
    I was able to get the equation through a few equations we have in our notes. Gauss' Law is phi = Q/epsilon_0. The magnitude within each sphere would be constant, so phi is the integral of E*ds. That gives 4*pi*r^2*E.

    Setting the equations equal to each other is 4*pi*r^2*E = Q/epsilon_0

    Solving for E is how I got the equation.

    I meant the QA as the Q enclosed by the inner shell, sorry. And the outer shell would be QB.
     
  10. Sep 28, 2008 #9

    gabbagabbahey

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    Thanks for the tip tiny tim:smile:
     
  11. Sep 28, 2008 #10

    gabbagabbahey

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    Well, Gauss's law in integral form is:

    [tex]\Phi= \oint_{\mathcal{GS}} \vec{E} \cdot \vec{da} =\frac{Q_{enclosed}}{{\epsilon}_0}[/tex]

    Where [itex]\mathcal{GS}[/itex] is your Gaussian surface and [itex]Q_{enclosed}[/itex] is the charge enclosed by your Gaussian surface.

    so, what are you using for your [itex]\mathcal{GS}[/itex]? how much charge does it enclose when its radius [itex]r[/itex] is less than [itex]a[/itex]?
     
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