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Gauss's Law and a uniformly charged cube

  1. Jul 21, 2003 #1


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    I can't seem to figure out how to apply Gauss's law to a uniformly charged cube(side length a), charge density = rho = Q/a^3. Here's the problem: I have a uniformly charge cube, of side length 'a,' with total charge Q which has a conductor on two oppossing faces. I want to know the electric field along a line which travels from the center of one of the conductior faces straight through the cube to the center of the opposing face which has a conductor on it. The text (Liboff's quantum mechanics text) says to use Gauss's law. But I don't see how. Let's assume that instead of a cube I have an infinite sheet of charge of finite thickness 'a.' I then use a pill box with one side in the conductor and the other side in the medium. But this gives me the incorrect answer of E = rho*x/epsilon. I think I should get a potential proportional to z^2 where z is the distance from the center along a line perpendicular to the conducting surface.

    Any thoughts?


  2. jcsd
  3. Jul 21, 2003 #2
    Working on it

    I am thinking, it seems like you asking for something that could be described as something like the electric field inside a capacitor, yet the space inside the capacitor is filled with a uniform charge distribution.

    The only problem is, because it is actually a cube, the capacitor's plates are finite in extent (so you can't ignore the fringing field).

    This may not help...
  4. Jul 21, 2003 #3


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    Re: Working on it

    Just the person I'm looking for!!! :-)

    This has to do with Problem 7.16 in Liboff

  5. Jul 21, 2003 #4
    Liboff Ed. 2

    Here is the question as put in the 2nd Ed. of Liboff:

    A large dielectric bue with edge length L is uniformly charged throughout its volume so that it carries a total charge Q. It fills the space between condenser plates, which have a potential difference, pho_0, across them. An electron is free to move in a small canal drilled in the dielectric normal to the plates.

    The Hamiltonian for the electron is (with x measured from the center of the canal) ...

    H=(p^2)/(2m) + K*x^2/2 + e*phi_0*x/L

    (a) ...

    (b) What are the eigenenergies and eigenfunction of H? HINT: Rewrite the potential energy of the electron as:

    V=K/2(x^2 + 2*gamma*x)=K/2*[(x+gamma)^2 - gamma^2]

    gamma = e*phi_0/(L*K)

    then change variables to z = x + gamma. To evaluate K, use Gauss's law (neglecting edge effects).

    Are you answering this question (b)?
    Last edited: Jul 21, 2003
  6. Jul 21, 2003 #5


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    Re: Liboff Ed. 2

    The entire question both (a) and (b)

  7. Jul 21, 2003 #6
    OK, I'll think about it.

    I have to work this into my day schedule, so it may take me some time.

    However, I can see that "neglecting edge effects" means assume you are inside a capacitor (with infinitely extending plates). Let me know if you agree with this hypothesis.
  8. Jul 21, 2003 #7


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    I *think* I know but I'm not sure. The E-Field inside the cube is not an E-Field with the field lines on one direction. That much seems certain. The geometry of the charged cube implies that the field inside at different points are are not pointing in the same direction. For example: Le the faces of the cube be parallel the x, y and z planes. If I'm at the center of the cube and I move in the +x direction then the field at my new location must point in the -x direction. If, instead, I move in the +y direction, the the field at my new location must point in the -y direction. Same with the z-direction. However if yhou stick to one direction (say the x-direction) and treat that as a wall of charge instead of a cube of charged the the above notions (i.e. moving off the center, E-field pointing in the opposite direction that I move) then I think that's what he means by "nelgect edge effects.

    Sound right?

    Man!! What a tough little problem! I guess it's been a while since I've dabbled with Gauss's Law!! :-)

  9. Jul 21, 2003 #8

    You dudes are making the problem too damn difficult. Just use Gauss' Law before I get pissed off.

    CircIntgral[ D dot ds ] = Qenc,

    where D = electric flux density, s = surface area, and Qenc = enclosed charge.

    You have a cube with total enclosed charge Qenc = Q.

    You know that the circular integral over the surface of the cube is:

    CircIntgral[ D dot ds ] = 6 * (a^2) * D.

    Therefore, the flux density at the center of a face is just:

    D = Q / (6 * a^2).

    Convert electric flux density D to electric field E using the permittivity constant.

    E = D / epsilon.

    E = Q / (6 * epsilon * a^2).

  10. Jul 21, 2003 #9


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    Re: Re: Gauss's Law and a uniformly charged cube

    If you're that impatient with us then simply ignore this thread.

    And the problem is not to find the field at the center of the cube. If you read the original question then you'd note that I asked
    What you just did was what I tried in the first 5-seconds of tryin to solve this. The form of the answer must be that the field increases as the distance from the center increases. The field will always be directed away from the center. The problem is to find the field magnitude as a function of the distance from the center. The result should be (in the given approximation of neglecting fringing) E ~ x where x is the distance from the center. The region of interest is only along a line which runs from the center of one face to through the center of the cube, to the center of the other face. Each of these faces being covered with a conductor

    See -- http://www.geocities.com/physics_world/Fig-7.12.gif

    This is the diagram in the text for this problem.

  11. Jul 21, 2003 #10
    Re: Re: Re: Gauss's Law and a uniformly charged cube

    I was only joking before... but now you've seriously pissed me off. I gave you the answer you should be looking for.

    E = Q / (6 * epsilon * a^2)

    This is the electric field at the face center of a cube with side a.

    If you want the field as a function of distance along a line then it's just:

    E = Q / (6 * epsilon * (d/2)^2) = (2/3) * Q / (epsilon * d^2), when d > a;


    E = 0, when d <= a.

    All you have to do is change the side length of the cube to get to put the face of the cube at the point where you want to calculate electric field. Of course, the electric field inside the box will always be zero because there will be no enclosed charge.

  12. Jul 21, 2003 #11


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    Re: Re: Re: Re: Gauss's Law and a uniformly charged cube

    Relax. Life's too short my friend.

    First off - I misread your comment. I thought I read center of cube but I now see that you wrote center of face. (dyslexia kicking in again).

    I'm well aware of what I should be looking for. And I'm also well aware of how to use Gauss's Law. And I'm not sure that what you've done is valid. Upon close inspection it does not appear to have been executed correctly. You can't take the integral and evaluate it like that. The value of D varies all over the surface of the cube. So you can't pull it out of the integral like that.

    Let me get back to this later.


  13. Jul 21, 2003 #12
    (sdeliver645 thinks...)

    eNtRopY is increasing the disorder of this thread.

    Boltzmann said:

    Now he is dead.
  14. Jul 21, 2003 #13


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    Does he do this often?

  15. Jul 21, 2003 #14
    Re: Liboff Ed. 2

    sorry if I interfere here, but...

    Please note that the electric field of an 'uniformly charged cube' is not the same as that of an 'uniformly charged cube between condenser (=conductor) plates'.
    The conducting plates require the E-field to be at right angles with the plates' surfaces. So the field we talk about here is that of an uniformly charged cube plus the cube's mirror images wrt. to each plate, plus images of images, and so on. It's indeed the field of an infinitely long charged bar with square cross-section.
    Well, if the Hamiltonian is given, then why worry about the geometry of the field? OK, you must still find the value of K, but that looks like a small side-problem to me - the main problem seems to be the quantum-mechanical one (after all, it's a QM book, not an electrostatics book).
    Last edited: Jul 21, 2003
  16. Jul 21, 2003 #15
    Re: Liboff Ed. 2

    ...and note this:
  17. Jul 21, 2003 #16


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    Re: Re: Liboff Ed. 2

    Nope. That's never been how I study. I do the problems no matter how painful. Along the way I find my weak spots and fix them.

    And yes. I know the field is different with the plate - but in this thread that fact sometimes it slipped my mind when I was writing/explaining.

  18. Jul 22, 2003 #17
    Re: Re: Re: Re: Gauss's Law and a uniformly charged cube

    Again, the answer outside the cube is:

    E = Q / (6 * epsilon * (d/2)^2) = (2/3) * Q / (epsilon * d^2), when d > a and Q = rho * a^3.

    Initially, I had misred the problem and thought that all of the charge was on the outside of the cube, but this is still the correct answer for d > a.

    There is no problem with evaluate the integral this way. That is the beauty of Stoke's relation... which is what the point of such a homework problem is... so that you gain true understanding of Gauss' Law.

    pmb, you say that the electric field isn't uniform over the surface of the cube (which it isn't), but this doesn't matter. "D dot ds" ensures that the vector components of the D field are normal to the surface. Since the 100% of the D magnitude is normal to the surface at the center of a cube face, then the derived expression gives you the right answer along a line intersecting the center of the cube and a cube face.

  19. Jul 30, 2003 #18


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    Science Advisor

  20. Jul 30, 2003 #19


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    Re: Re: Gauss's Law and a uniformly charged cube

    Is there a reason why you keep quoting me jeff?

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