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I Electric field of a point charge above a plane conductor

  1. May 16, 2017 #1
    Some things I did not get are,

    1): In the third para, it is said that potential approaches ##\infty## at ##Q##, why ?

    2): In last para, The z component of field by ##Q## is said to be ##-Qh/(r^2 + h^2)^{3/2}##, Why ? should not it be ##+Q## not ##-Q## has the charge itself is positive.

    3): I know ##E_n = 4\pi\sigma##, where ##E_n## is field due to the conductor near its surface and perpendicular to the surface, but here, in last para, the field ##E_z## is induced by the charge ##Q## and ##-Q## not the plane conductor. So how do we get ##\sigma## from this ?

    4): Lastly I don't get the need of placing ##-Q## below the surface, why we did that ?
     
  2. jcsd
  3. May 17, 2017 #2

    haruspex

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    It's rather garbled, but I think I understand most of it.
    The third paragraph I cannot comprehend, but I believe you can largely ignore it. The relevant part is the reference to a uniqueness theorem.

    We wish to find the charge distribution induced on an infinite plane conductor at z=0 by a point charge Q at (0,0,h).
    The potential at the plane will everywhere be zero. If we throw the plane conductor away and instead have a point charge -Q at (0,0,-z) we will also get zero potential everywhere at z=0. The uniqueness theorem says the field at z≥0 is the same in both cases.

    So we can calculate the z component of the field that would be generated at z=0 by the -Q and equate that to the z component of the field that is generated by the induced charge on the plane conductor. A point at distance r from the origin is considered.
     
  4. May 17, 2017 #3
    So as far I get, here a summary . We place ##-Q## charge at ##-h## to simulate the zero potential surface at xy plane. Then we found the total electric field by the both the charges at a point ##r## from the origin. Now if we place the conductor back and kick the added charge then this would be the field at the point ##r## from the origin.

    Since we know that ##E_{\perp}## field at a point is ##E_{\perp} = 4 \pi \sigma## and the field we calculated equals ##E_{\perp}##, hence was the further calculations.

    Am I correct ?
     
  5. May 17, 2017 #4

    haruspex

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    No, just the lower one. The field from the conducting plane must be the same as the field from its -Q replacement.
     
  6. May 18, 2017 #5

    vanhees71

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  7. May 18, 2017 #6
    Yes thanks I finally understand this.
    Can I still do this with a finite plane as long as the point is lying on the plane ?
     
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