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Electric Field of an Infinite Plane

  1. May 7, 2010 #1
    1. We are given a static system of conductors and charge. We know that x=y is a plane of equal potential (for example, a system that can give such result is an infinite plate in that plane). Which of the following fields can represent an electric field of such system?

    a. [tex]\vec{E}[/tex] = az[tex]\hat{x}[/tex] - az[tex]\hat{y}[/tex]
    b. [tex]\vec{E}[/tex] = a[tex]\hat{x}[/tex]
    c. [tex]\vec{E}[/tex] = a[tex]\hat{x}[/tex] - a[tex]\hat{y}[/tex]
    d. [tex]\vec{E}[/tex] = ay[tex]\hat{x}[/tex] - ax[tex]\hat{y}[/tex]
    e. [tex]\vec{E}[/tex] = a[tex]\hat{x}[/tex] + a[tex]\hat{y}[/tex] + a[tex]\hat{z}[/tex]


    The attempt at a solution:
    I figured that since the plane extends infinitely in the z direction, no field can be present there. This is also because of the first axiom of electrostatics relating to the flux being outward normal to the plane. Therefore, choice e is eliminated. In fact, the z axis should have no effect on the field. a is also eliminated.
    Also, choice b has no dependency on the y axis, but the E field should be partially dependent on it...b is eliminated.
    We also know that the electric field due to an infinitely charged plane is constant. I want to say that the answer is therefore c...I am lead me to believe that choice d can be eliminated, but I'm not sure. I'm stuck between c and d.
    Any suggestions?
     
  2. jcsd
  3. May 7, 2010 #2
    Some quick things you should know:

    You must have,

    [tex]\vec{\nabla}\times\vec{E} = 0[/tex]

    if there even exists a potential.

    Also, you know the electric field will be perpendicular to an equipotential. So take the cross product of the electric field with a vector perp to the x=y plane (cross products of parallel vectors should be 0).
     
    Last edited: May 7, 2010
  4. May 8, 2010 #3
    If we say that x=y, then the method of taking a cross product between a vector perpendicular to the plate and the E-field works for both c and d, doesn't it? Or did I just miss something when I did it?
     
  5. May 8, 2010 #4
    It can't work for both. Just work them out and you will see.
     
  6. May 8, 2010 #5
    The electric fied points away from the plane. The unit normal to the plane is given by "c" with arbitrary constant a. C is the correct answer.
     
  7. May 8, 2010 #6
    It is best if he used the cross product to show the answer is (c) and not (d) since the E-field is perp to the plane, and not give him the answer right away.
     
  8. May 8, 2010 #7
    I agree. I only confirmed it a different way for him since he had already deduced the correct answer but not by the cross product or the unit normal methods.
     
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