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Electric field of an infinitely large positively charged plane sheet

  1. Aug 4, 2010 #1
    On drawing electric field vectors around an infinitely large positively charged plane sheet, we find that the length of the vectors (which represent the magnitude of the field at that point) on both sides of the sheet decreases with distance, and is directed away from both sides of the sheet.

    We know that on joining these vectors to get electric field lines, the magnitude of the field at a point is now indicated by the number of electric field lines per unit area passing through the area element at the point perpendicular to the plane of paper on which the lines are drawn.
    However, we see that the number of electric field lines per unit area passing through the perpendicular plane is the same at a point near the sheet as well as a point far away from the sheet. Clearly, this indicates that the magnitude of the electric field at the points near and far from the sheet is the same.

    However, this is contrary to the observations stated in the first paragraph. Can anyone explain this, please???
     
  2. jcsd
  3. Aug 4, 2010 #2

    Doc Al

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    Staff: Mentor

    Sure. Your first paragraph is wrong. The field from an infinite sheet of charge does not decrease with distance.
     
  4. Aug 4, 2010 #3
    but y? i think the length of the vectors will decrease?
     
  5. Aug 4, 2010 #4

    Doc Al

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    On what basis do you think that?

    You can derive the field either using Gauss's law (the easy way) or by direct integration. You'll find that the field is independent of distance.
     
  6. Aug 5, 2010 #5
    Consider two points A near the sheet on a field line and B far from the sheet on the same field line.

    Clearly, by geometry, the distance of unit positive charge placed at A from all the elements and points on the sheet is smaller than when placed at B. Thus, by the Coulumb's Law, the forces due to various elements of the sheet on the unit positive charge at A is greater than the corresponding forces on the charge, when placed at B. The cosine components of the various forces will add up and the resultant force on the charge at A will be greater than the resultant force on the charge at B. This has led me to the conclusion that the field at A will be greater than that at B. The generalization based on this statement leads me to conclude that the field will decrease with distance.

    I know that this is contrary to the observations of the Gauss Law, but that is why the question has risen in my mind. Please help me!!!!!!!
     
  7. Aug 5, 2010 #6

    Doc Al

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    OK.

    OK. The force from any particular charge element in the plane on A is greater than on B. (But the angle that the force acts changes.)

    This is where you are wrong. You have to actually do the calculation and see how they add up. While the force on B (from any particular charge element) is less, the angle changes, which changes the vertical component.

    You'll need to set up the integral and actually do the calculation to see that no matter where you are above the plane, the total field is the same. (Treat the plane as a set of concentric rings of charge.)
     
  8. Aug 5, 2010 #7
    Thank you very much for your valuable reply. I have got my answer!!!!!
     
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