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Calculating Electric Flux Φ w/o typical Gaussian surface

  1. Jun 19, 2010 #1
    Hello all, new here.

    In the past few weeks, I have been trying to gain a basic understanding of classical electricity and magnetism through the fantastic lectures at http://ocw.mit.edu, specifically the physics course 8.02 taught by Dr. Walter Lewin.

    In his explanation of electric flux (lecture 3), he derives Φ for an arbitrary sphere of radius R containing charge Q, and comes up with Φ = Q/ε0, as expected per Gauss's Law. He then explains that any closed surface containing charge Q will have that same flux.

    Exploiting symmetry with a sphere, the calculation is fairly easy, so I wanted to challenge myself and see if I could calculate the flux for a uniform cube with a charge Q at its center. I am finding the problem unexpectedly difficult and I'm not even sure what kind of integral to construct.

    Is this too ambitious to calculate given my little experience in this area? My calculus skills are rusty, but I am eager to learn and enjoy a challenge.

    My only reasoning so far is that the total flux through the cube would be 6 times the flux through a single plane, so I am approaching the problem by trying to calculate the flux through a square on the xy-plane with 1 corner at the origin and charge Q under it. While the normal at every point is the same, simply constructing the correct expression for the changing E-field vector and the correct integral is kicking my butt :P

    Any tips or suggestions to point me in the right direction are appreciated!
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Jun 19, 2010 #2


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    The fact that the electrostatic flux through any closed surface is equal to the charge enclosed divided by epsilon-naught can be easily proven using the divergence theorem and Coulomb's law (or Gauss' law in differential form).

    Start with Coulomb's law:

    [tex]\textbf{E}(\textbf{x})=\frac{1}{4\pi\epsilon_0}\int \rho(\textbf{x}')\frac{\textbf{x}-\textbf{x}'}{|\textbf{x}-\textbf{x}'|^3}d^3x'[/tex]

    Take the divergence of both sides of the equation, use the Leibniz integral rule, and the fact that


    and you find that

    [tex]\mathbf{\nabla}\cdot\textbf{E}(\textbbf{x})=\frac{1}{\epsilon_0}\int \rho(\textbf{x}')\delta(\textbf{x}-\textbf{x}')d^3x'=\frac{\rho(\textbf{x})}{\epsilon_0}[/tex]

    Which is Gauss' law in differential form (one of Maxwell's equations). From there, just apply the divergence theorem, along with the definition of charge density and you find

    [tex]\begin{aligned}\Phi &= \oint_{\mathcal{S}}\textbf{E}\cdot d\textbf{a} \\ &= \int_{\mathcal{V}}(\mathbf{\nabla}\cdot\textbf{E})d^3x \\ &= \frac{1}{\epsilon_0}\int_{\mathcal{V}}\rho(\textbf{x})d^3x \\ &= \frac{Q_{\text{enc}}}{\epsilon_0}\end{aligned}[/tex]

    If this kind of vector calulus is too advanced for your current level, and you simply want to convince yourself with your cube example, just begin by expressing the electric field of the point charge Q at the origin in Cartesian coordinates by using the fact that the radial unit vector is


    From there, just compute [itex]\textbf{E}\cdot d\textbf{a}[/itex] for one face of the cube and integrate over that face.
  4. Jun 20, 2010 #3
    You want to calculate the flux through the sides of a cube of the field of a charge placed at its center.
    You were right while reasoning that the total flux through the cube would be 6 times the flux through a single side.
    So, you have got to calculate the flux through one side.

    Take, say, the top surface.
    Divide it into rectangular strips.
    Let x be constant for each strip. Let each strip be of thickness, dx.
    (the length of the strip = a is along y and the width = dx is along x).

    Choose one strip. Let it be at position 'x'.
    You'll have to further divide this strip (differential elements with sides dy & dx).
    Choose one differential element. Let it be at the point (x,y).
    Find out the value of E at this point. To make calculations simple choose the center of the square to be the origin. To calculate the flux, you'll need to find the normal component of E at this point. You should get [tex]{(\frac{{a}^{2}}{4} + {x}^{2} + {y}^{2}})}^{\frac{3}{2}}[/tex] in the denominator.
    Integrate this w.r.t y to find the flux through the strip.
    Integrate this again w.r.t x to find the flux through the side.
  5. Jun 20, 2010 #4
    Thank you both for the replies!

    To gabbagabbahey, yeah that vector calculus is a bit beyond my skill, though I did still look into it and it gives me an idea of what math I can look forward to learning :)

    I realize now this thread really is more mathematical than physics, so I'll definitely be re-working my approach.

    Thank you graphene for that push in the right direction. Seems like I'm almost there!
  6. Jun 29, 2010 #5
    If you take a look online for 'divergence of a vector' you'll find that the case of flux from a cube is a fundamental concept called 'divergence', which is pretty self-explanatory.

    It's not a difficult bit of math (because the cube considered is infinitesimal you can forget about all the complication of taking strips and sub-elements - just consider a simple cube with an average flux through the centre of each face. All the complications vanish in the limit)

    If you can master that concept you'll be a long way down the road.
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