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Electrostatics question help

  1. Apr 13, 2009 #1
    A point charge Q is placed at the centre of a cube. What is the electric flux through each face of the cube.

    ok so the answer is to say that since the cube is a closed surface, Gauss' Law tells us that the total flux through the cube is [itex]\frac{Q}{\epsilon_0}[/itex] and then from symmetry 1/6 of that is through each face. so the final answer is [itex]\Phi=\frac{Q}{\epsilon_0}[/itex].

    I was wondering why i dont get the same answer when i do it this way:

    set it up with cartesian coordinates. let's find the flux through the face with outward normal [itex]\vec{e_x}[/itex].

    [itex]\vec{E}=\frac{Q}{4 \pi \epsilon_0 (\frac{a}{4})^2} \vec{r}[/itex]

    now [itex]\vec{r}=(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},)[/itex] and [itex]\vec{e_x}=(1,0,0)[/itex]
    so [itex]\Phi=\int_S \vec{E} \cdot \vec{dA} = \frac{Q}{\pi \epsilon_0 a^2} \vec{r} \cdot \vec{e_x} \int_S dA = \frac{Q}{\pi \epsilon_0 a^2} \frac{1}{\sqrt{3}} a^2[/itex]

    which ends up as [itex]\Phi=\frac{Q}{\sqrt{3} \pi \epsilon_0}[/itex]

    obviously the first way is much easier but surely it should be possible both ways?
     
  2. jcsd
  3. Apr 13, 2009 #2

    Redbelly98

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    Re: Electrostatics

    There are two problems that I can find:

    • The cube's surface is not a constant distance a/4 from the charge, as you appear to assume
    • The unit vector r^ is not (1/√3,1/√3,1/√3) everywhere on the surface.
     
  4. Apr 14, 2009 #3
    Re: Electrostatics

    ok yeah so i see now that its not a constant a/2 from each surface

    what is [itex]\hat{\mathbf{r}}[/itex] then (i really should know this!)

    is it possible to do this problem this way or is gauss' law the only way?
     
  5. Apr 14, 2009 #4

    Redbelly98

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    Re: Electrostatics

    The vector (x,y,z) points in the same direction as r^, everywhere. Normalize that vector, and you'll have r^.

    It should be possible, just not necessarily easy.
     
  6. Apr 14, 2009 #5
    Re: Electrostatics

    yeah so [itex]({x \over |x|},{y \over |y|},{z \over |z|})[/itex]

    why doesn't the one i wrote earlier work though???
     
  7. Apr 14, 2009 #6

    Redbelly98

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    Re: Electrostatics

    Not quite, that vector just has components (±1,±1,±1). Try dividing by |r| instead of |x|, |y|, and |z|.

    Not sure if this will help, but: the integral you're trying to do contains an r-hat term. What you wrote earlier is not equal to r-hat, therefore using it will give a wrong answer when you do the integral. (Or did I misunderstand your question here?)
     
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