Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Where am I wrong? (Electrostatic Problem)

  1. Jul 3, 2008 #1
    A 1-D metal put inside a E0 electric field just like shown in following figure.
    --> |- <-- +| --> 
    --> |- <-- +| --> 
    --> |- <-- +| --> E0
    --> |- <-- +| --> 
    --> |- <-- +| -->
    Suppose the positive charge density of δ1, the negative charge density δ2, δ1=-δ2.
    The inducing electric field Ei=Epos+Eneg=-2δ1/ε0
    so, δ1=δ1*ε0/2

    for positive side, using Guass's Law,
    the E inside the metal equals to 0.
    So, Eout=q/dA*1/ε000=1/2E0

    but it should be E0

    Where am I wrong.
  2. jcsd
  3. Jul 4, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    BiBByLin: You might want to draw a nicer picture and use TeX for forumulas, at least I can't see what you are doing and want to calculate.
    Last edited by a moderator: Jul 4, 2008
  4. Jul 4, 2008 #3

    Doc Al

    User Avatar

    Staff: Mentor

    The field from each sheet of charge is [itex]\sigma/2\epsilon_0[/itex], so the induced field within the conductor is [itex]-\sigma/\epsilon_0[/itex].
  5. Jul 4, 2008 #4
    It's easy to get confused between applying Gauss' theorem and simple superposition of effect. Let me try to help...


    No. You can see the total Ei due to both δ1 and δ2 but each Epos and Eneg are not equal to δ1/ε... they are in fact δ1/2ε.

    If you use superposition of effects (i.e. calculating Epos, Eneg and them sum them) you must apply Gauss to each surface as if the other surface was not there at all... if you instead use the knowledge that there is another field, then you end up summing the same field twice.

    That is what you do when you solve the problem by considering the conductor as a whole, apply Gauss only once (around one of its two surfaces) and imposing that E=0 inside the conductor. Having only one "side" of the integral's surface, you immediately get that E(external)=δ1/ε. That is already the final E outside the conductor, and you know it since the start that it will be E0, so you immediately derive that δ1=E*ε.

    Alternatively, you could solve the problem by superposition of all fields: Epos+Eneg+E0. But you need to derive Epos and Eneg by using Gauss twice, once per surface (each application of Gauss is applied only on the E field in question, not the total), without considering the existance of the other and without considering the presence of E0. What you get in this case, is that (assuming all fields are positive when pointing towards the RIGHT):

    Epos=δ1/2ε on the right side of the right surface
    Epos=-δ1/2ε on the left side of the right surface
    Eneg=-δ1/2ε on the right side of the left surface
    Epos=δ1/2ε on the left side of the left surface

    Add them properly in each region and you'll get that between the surfaces Epos+Eneg=-δ1/ε (and must be exactly opposite of E0, which is therefore δ1/ε), while outside the conductor Epos+Eneg=0, therefore there is only E0.
    Last edited: Jul 4, 2008
  6. Jul 7, 2008 #5
    thank you very much!

    But if I apply the Laplace Equation:

    curl2φ=0, because φ is independence of x and y. So:


    when Z increase to infinitude. E=-curlφ=-C1 =E0.

    Suppose the midst of conductor layer is Z=0. when z=0, it always E=E0. But it should be 0, where am I wrong?
  7. Jul 7, 2008 #6
    First, don't write "curl[tex]\phi[/tex]" because that's an alternative name for "rot[tex]\phi[/tex]", i.e. [tex]\nabla [/tex]X[tex] \phi[/tex], instead Laplace equation is

    [tex]\nabla^2 \phi = 0[/tex]

    [tex]\nabla^2 \phi[/tex] is not the same as [tex]\nabla [/tex]X[tex] \nabla [/tex]X[tex]\phi[/tex]

    The solution for [tex]\phi[/tex] outside the conductor is correct, it is in fact linear and descending towards the direction of E.

    Inside the conductor, [tex]\phi[/tex] = constant, which in turn means E=0.

    You are wrong at least when you say "when z=0, it always E=E0": why? Where did you get this from? The only place where you see "z" is in the expression of [tex]\phi[/tex], set it to 0 there and you get [tex]\phi=C_2[/tex], derive it and you get exactly E=0, not E=E0.

    Anyway you didn't really derive [tex]\phi[/tex] from Laplace equation at all...

    Looks like you're jumping too fast without doing the job, try to slow down and put some clean to the mess ;)
  8. Jul 7, 2008 #7
    because E=-[tex]\nabla[/tex][tex]\varphi[/tex].
    from upper equation [tex]\varphi[/tex]=C1Z+C2;
    so E=-C1=E0, it is independence with the position. So, I can said when Z=0 or somewhere z<1/2a, E=E0.

    by the way, maybe it could be [tex]\epsilon[/tex]E0 because of the continuity at the boundary.

    So sorry, lots of confusion.
  9. Jul 7, 2008 #8
    Actually I wrote something wrong as well, but anyway the fact is that E is indeed discontinuous, and \phi is not derivable in z=0. The linear expression for \phi is valid only outside the conductor, not inside it. You cannot say E=-C1 inside, because \phi is not = to C1 Z + C2 inside! (or at least C1 inside is not equal to C1 outside).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook