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Why is the electric field just outside a conductor twice...

  1. Feb 14, 2016 #1
    Why is the electric field just outside a conductor twice the field produced by a uniform sheet of charge?

    My textbook's explanation is that you can imagine that near a point P, the charge at the surface of a conductor looks like a small uniformly charged disk centered at P, giving an electric field of magnitude σ/(2ε0) pointing away from the surface both inside and outside the surface. Inside the conductor, this field points away from point P in the opposite direction. (I understand all of this part.) It continues to say that because the net field inside the conductor is zero, the rest of the charges in the universe must produce a field of magnitude σ/(2ε0) in the outward direction. Therefore, the field cancels out the inward field produced by the disk mentioned earlier, and additionally adds on the the outward field produced by the disk to make the total field σ/(ε0). This last part is what I don't understand. Since the field produced by the rest of the charges is used to cancel out the inward field produced by the disk, how is it able to continue penetrating outward with magnitude σ/(2ε0) and add on to the outward field of the disk? Shouldn't it have already been canceled out by the disk's inward field?

    I have uploaded the picture in the textbook below.
     

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  3. Feb 14, 2016 #2

    Drakkith

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    Staff: Mentor

    Hi jt200. I'm currently in my first E&M class and that does seem like a strange way of explaining it to me. I like hyperphysics' way of explaining it: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gausur.html#c3

    Basically they say that the e-field inside the conductor is zero, as we all know, and then they find the flux through a gaussian cylinder with its axis oriented perpendicular to the surface. Gauss' law tells us that the flux exiting a gaussian surface is proportional to the enclosed charge. Since the e-field inside the conductor is zero, and the sides and bottom end of the cylinder are inside the conductor, ALL of the flux from the charges enclosed by the cylinder must exit through the top surface (which is just outside the conductor's surface).

    Another way to think about it: the field lines from each positive charge on the surface of the conductor must run from the positive charge to a negative charge. Since there aren't enough negative charges for all the excess positive charges, their field lines have nowhere to run to inside the conductor. If you've seen the field line pattern from two like charges placed next to each other, you'll remember that they bend away from each charge. Now imagine a positive charge on the surface of the conductor. It's surrounded by positive charges on all sides, and the only direction for its own field lines to run is outwards, away from the conductor. The same is true for every other positive charge. So the e-field from the charges outside the disk in your book's example isn't "used up", it simply goes elsewhere.

    Hopefully that's all correct. If not, someone let me know.
     
  4. Feb 15, 2016 #3
    I too had trouble there, but the explanation in Griffiths is satisfactory. It goes like this : the electric field near a large charged sheet is E=σ/2ε° on both sides and they both point in the opposite directions, so as you cross the surface there is a discontinuity in the E field by σ/ε° (can you see why?)
    Now, since the E field inside the conductor is zero, the field outside it is σ/ε° as there must be a discontinuity by this amount.
     
  5. Feb 16, 2016 #4
    It may help if look at the case of a parallel plate capacitor (the ideal case, with infinite plates). For the usual case, with opposite charges, you have a situation somehow reversed: the fields from the two plates add up inside the capacitor (so the field is σ/εo) and cancel each other outside the capacitor.

    You can imagine that the two plates have the same type of charge too, and draw again the fields to see how they add and subtract.
     
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