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Puzzler for newbies to electrostatics

  1. Jul 16, 2004 #1
    Hi...

    This problem is not an uncommon one and you might easily encounter it in any book that has challenging problems on this subject. I am throwing it here for newbies to Electrostatics who would like to try their hand at solving a puzzler...

    If a point charge +q is placed at y = 0 and an imaginary cubical volume is constructed such that the cube has each side equal to a units and y = 0 is one vertex of this cube, what is the electric flux out of a face described by y = a? Does this depend on the location of the charge?

    I will post a hint (a really not so obvious one) in a while...you might want to discuss this one as there seems to be more than one correct solution.

    Cheers
    Vivek

    (I hope this isn't against forum rules :wink: )
     
  2. jcsd
  3. Jul 16, 2004 #2
    Isn't it zero? The electric field vector and the area vector are perpendicular so the flux will turn out to be zero. If the charge is moved, then the electric field vector and the area vector won't be perpendicular anymore so there will be some non-zero flux.
     
    Last edited: Jul 16, 2004
  4. Jul 17, 2004 #3

    Doc Al

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    I presume maverick280857 intends that the charge is at the origin (0,0,0) and one corner of the cube is at the origin as well. The sides of the cube are on the planes x = 0, y = 0, z = 0, x = a, y = a, and z = a.

    If so, then the electric field and area vector are not perpendicular. Make use of Gauss's law and symmetry to find the flux.
     
  5. Jul 17, 2004 #4
    Hi e(ho0n3 and docAl

    Well docAl has given you the hint...so lemme wait some more time before I post the solution.

    Cheers
    Vivek
     
  6. Jul 17, 2004 #5
    I believe the answer (in CGS units) is [itex]\frac {1}{6} \pi q[/itex]
     
  7. Jul 17, 2004 #6
    Oops. What I mean to say was that the eletric field and area vector are perpenduclar on the sides where y = 0 and x = 0. On y = a is another story.
     
    Last edited: Jul 17, 2004
  8. Jul 17, 2004 #7
    I thinks the flux is also zero on the side where z = 0 too.
     
  9. Jul 17, 2004 #8

    Gokul43201

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    e(ho0n3,

    Just use Gauss' Law and the symmetry of the problem. And yes, you are right about the flux being =0 through 3 faces.

    DarkEternal, I think not.
     
  10. Jul 17, 2004 #9
    So we are left with three faces where the flux is not zero. According to Gauss' law, the flux is [itex]q/ \epsilon _0[/itex], so by symmetry the flux out of one of the faces where the flux is not zero is [itex]q/ (3\epsilon _0)[/itex].
     
    Last edited: Jul 17, 2004
  11. Jul 17, 2004 #10
    The correct answer is [tex]\frac{q}{24\epsilon_{0}}[/tex] (in SI Units).

    The solution:

    Supposing the charge q were at the center of a cube of side a. Then, it would be at a distance of (a/2) from either face. The total flux out of the cube would be, as predicted by Gauss's Theorem, [tex]\frac{q}{\epsilon_{0}}[/tex] and the flux through each face would be one sixth of this since by symmetry, each face is identical (and why should the flux out of any particular face be more anyway?). So the answer would have been [tex]\frac{q}{6\epsilon_{0}}[/tex].

    If we now move this charge q the point y = 0--one vertex of this cube--and consider another "Gaussian cube" of side 2a units such that the charge is now at the center of this new cube, the flux out of a side of area [tex]4a^2[/tex] sq. units would still be [tex]\frac{q}{6\epsilon_{0}}[/tex]. Clearly four times a squared is equivalent to four faces of a cube of side a stacked together to form a new face of side 2a.

    So now, the flux through the y = a face of the first cube cube of side a is one fourth of the flux through the y = a face of the second cube of side 2a. Hence, the answer is [tex]\frac{q}{24\epsilon_{0}}[/tex].

    Sorry for this seemingly verbose solution...the best idea would be to draw a diagram and convince yourself that this is so. And yes, the flux out of three faces is zero...but that does not mean that the flux out of the three others will be [tex]\frac{q}{3\epsilon_{0}}[/tex]. Why? Well, the electric field of a point charge is spherically symmetric. The total flux out of a Gaussian surface is indeed [tex]\frac{q}{\epsilon_{0}}[/tex] but that is not so here since the charge isn't quite inside the Gaussian cube of side a you see :-)

    Cheers
    Vivek
     
    Last edited: Jul 17, 2004
  12. Jul 18, 2004 #11
    er...i think so? my answer is still 1/24 of the total flux, just expressed in CGS units, correct?

    here is the way i thought about it:
    the size of the cube involved doesn't matter. consider a cube containing a charge in the center. draw 3 axes through the charge, splitting the cube into 8 regions (similar to how 2 2D axes split the plane into 4 quadrants). so through one of these regions, 1/8 of the total flux passes through. make this region the cube of the problem. the flux is nonzero on 3 of the faces, and by symmetry the flux of each face is equal. so multiply the remaining flux by 1/3 (1/8 x 1/3) to get 1/24 of the total flux. in CGS units, by gauss's law the total flux of the charge is [tex]4 \pi q[/tex] so 1/24 of that is [tex]\frac {1}{6} \pi q[/tex].

    we had this on a pset at MIT, where we used the purcell textbook (which employs CGS units, explaining my preference - the symmetry of maxwell's equations is better expressed in this unit system, in my opinion).
     
    Last edited: Jul 18, 2004
  13. Jul 18, 2004 #12
    Oh, right. I overlooked that fact that Gauss' law applies only when the charge is totally inside a Gaussian surface, not on or out of it. I liked DarkEternal's explanation, but aren't the units wrong (is the unit for flux in CGS different than those of MKS)?
     
  14. Jul 18, 2004 #13
    Hi

    Yes the size of the cube does not matter. I mentioned it since I wanted to make you sure about the position of the charge with respect to the faces of the cube. Darketernal, you split the cube and I enlarged it so there is not a great deal of difference in our solutions to this problem.

    As far as CGS units are concerned, I think it would be a better idea to express it as a fraction of the net flux out of a cube in case the charge were at the center. As long as you are sure about the fraction of that flux (1/24) it makes no difference whether you use CGS or MKS.

    Cheers
    Vivek
     
  15. Jul 18, 2004 #14

    Gokul43201

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    My bad...I misread the question. Thought we wanted the flux out of the cube which would be 1/8 of the total, instead of 1/24.
     
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