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Potential due to infinite sheet

  1. Oct 6, 2007 #1

    nicksauce

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    1. The problem statement, all variables and given/known data
    Calculate the potential V(z), a height z above an infinite sheet with surface charge density [tex]\sigma[/tex] by integrating over the surface.


    2. Relevant equations
    [tex]V(z)=\frac{1}{4\pi\epsilon_0}\int_s{\frac{\sigma dA}{r}}[/tex]


    3. The attempt at a solution
    So [tex]V(z)=\frac{\sigma}{4\pi\epsilon_0}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}{\frac{dx dy}{\sqrt{x^2+y^2+z^2}}}
    [/tex]

    However, unless I am wrong, this integral does not converge.
    We know the E-field due a infinite sheet is [tex]E=\frac{\sigma}{2\epsilon_0}[/tex], so the potential should be [tex]V=-\frac{z\sigma}{2\epsilon_0}[/tex], right? So where is the error?
     
  2. jcsd
  3. Oct 6, 2007 #2

    learningphysics

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    You're right. This integral doesn't converge... the reason is because V=kq/r takes the voltage at infinity = 0... in other words this integral will give you the voltage at z relativie to z=infinity... since the field is constant, this value will be infinite...

    [tex]V=-\frac{z\sigma}{2\epsilon_0}[/tex]

    takes the voltage to be 0 at the sheet itself...

    Do you have to solve the problem using that particular integral? I don't think it is solvable that way...

    Maybe the question wants you to derive the field by integrating over the surface... and then just use V = -integral.dz just like you did?
     
  4. Oct 6, 2007 #3

    nicksauce

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    Well the follow up question, is "Use Coulomb's law to calculate the E field above the plane. Compare the electric field to the potential. Why did you get the value you did for the potential?

    So it sounds like I'm supposed to get infinity for the potential, and then say it's because the potential at infinity is not zero.

    The question specifically says to obtain the potential by integrating over the surface, so I can't think of another way to solve it.
     
  5. Oct 6, 2007 #4

    learningphysics

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    Rather if the potential at infinity is 0, then the potential at z is infinite... this makes sense because we have a constant field multiplied by an infinite distance... that'll give infinite voltage...

    ie [tex]V_z - V_{infinity} = -\int_{\infty}^z \vec{E}\cdot{\vec{dz}}[/tex]

    so [tex]V_z - 0 = \frac{\sigma}{2\epsilon}(infinity - z)[/tex]

    so Vz is infinite.

    Yeah, I guess they just want you to explain why you get that answer... did they specifically ask you to take potential at infinity to be 0?
     
  6. Oct 6, 2007 #5

    nicksauce

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    Err right... I got mixed up a bit there.

    They did not.
     
  7. Jan 5, 2011 #6
    Conclusion of the story"potential is found by electric field","electric field of sheet(which is found by gauss` law) is constant"and "gauss` law is totally based upon inverse square dependence of coulomb`s law",which implies that coulomb`s law is not obeyed(as said in gauss` theory).More-over it not possible to experience same force due to electric field just few mm away and meters away from charged sheet.
     
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