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Electric field vector & surface integral

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data

    See figure attached.

    2. Relevant equations



    3. The attempt at a solution

    See figure attached.

    The solution shows that,

    [itex]\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}[/itex]

    How did they obtain this?
     

    Attached Files:

  2. jcsd
  3. Sep 14, 2011 #2
    First of all, you did not write correctly the expression of the electric field in Cartesian coordinates. (Remember that |r| is not -usually- equals to 1)

    Then, how do you usually evaluate the surface integral (the flux) over a certain surface?
     
  4. Sep 14, 2011 #3
    It's not a vector, it's a unit vector, that's what the hat signifies.

    For the flux since it's a closed surface I can apply divergence theorem,

    [itex]\oint_{S} \vec{F} \cdot \hat{n}dS = \int\int\int_{V}\vec{\nabla} \cdot \vec{F}dV[/itex]

    I'm still confused as how to get [itex]\vec{r}[/itex].
     
  5. Sep 15, 2011 #4
    Bump, still looking for some help on this one!
     
  6. Sep 15, 2011 #5
    Sorry for misunderstanding r, but everyone has his own conventions and symbols :D

    In Cartesian coordinates, how do you write the position occupied by an object? By giving the three coordinates. You can write both [itex] \vec{r}_P=\left[\begin{array}{c}x\\y\\z\end{array}\right][/itex] and [itex]\vec{r}_P=x\hat{i}+y\hat{j}+z\hat{k}[/itex] since[itex]\left[\begin{array}{c}x\\y\\z\end{array}\right]=x\left[\begin{array}{c}1\\0\\0\end{array}\right]+y\left[\begin{array}{c}0\\1\\0\end{array}\right]+z\left[\begin{array}{c}0\\0\\1\end{array}\right][/itex]

    Was this your problem?

    Now, I think that for the first part of the question you cannot use the flux theorem because the problem asks to calculate the flux through one face of the cube, not through the whole surface: I fear you need to calculate explicitly the flux by evaluating the surface integral or something like that (it shouldn't be too difficult, however).
    You'd better express the electric field in term of its cartesian components and then try to integrate.
     
  7. Sep 15, 2011 #6
    So for the first part of the question, "Express the electric field vector in its rectangular coordinate components" does this have any significance with the cube at all?

    The electric field vector is going to generate a radial vector field from the point charge outward everywhere in space.

    The radius depends on which point in space you are observing (i.e. any point (x,y,z)).

    Thus,

    [itex]\vec{E} = \frac{q \vec{r}}{4\pi \epsilon_{o}r^{3}}[/itex]

    Where,

    [tex]\vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \quad \text{Giving,} \quad r = \sqrt{x^{2} + y^{2} + z^{2}} [/tex]

    I'm moving onto the cube portion now, I'll post my results.
     
  8. Sep 15, 2011 #7
    Here's what I've got so far, can't remember how to evaluate such an integral,
     

    Attached Files:

  9. Sep 16, 2011 #8
    I think it is right.

    To evaluate the integral, you can use http://en.wikipedia.org/wiki/List_of_integrals_of_irrational_functions" [Broken] > List of integrals involving [itex]R=\sqrt{ax^2+bx+c}[/itex] > [itex]\int \frac{dx}{R^3}[/itex]
     
    Last edited by a moderator: May 5, 2017
  10. Sep 16, 2011 #9
    How would I do that? I don't have

    [itex]ax^{2} + bx + c[/itex], we would be missing the term with the b cofficient.
     
    Last edited by a moderator: May 5, 2017
  11. Sep 16, 2011 #10
    put b=0 :D
     
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