Potential due to infinite sheet

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SUMMARY

The discussion focuses on calculating the electric potential V(z) above an infinite sheet with surface charge density σ using the integral V(z) = (σ / (4πε₀)) ∫∫ (dx dy / √(x² + y² + z²)). Participants confirm that this integral does not converge, leading to the conclusion that the potential at height z is infinite if the potential at infinity is set to zero. The correct expression for the potential is V = - (zσ / (2ε₀)), derived from the constant electric field E = (σ / (2ε₀)) above the sheet. The discussion emphasizes the relationship between electric field and potential, highlighting the implications of Gauss's law.

PREREQUISITES
  • Understanding of electric potential and electric fields
  • Familiarity with Gauss's law
  • Knowledge of calculus, particularly double integrals
  • Concept of surface charge density
NEXT STEPS
  • Study the derivation of electric fields using Gauss's law
  • Learn about the implications of infinite charge distributions
  • Investigate the relationship between electric potential and electric field
  • Explore advanced calculus techniques for evaluating improper integrals
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Students of electromagnetism, physicists, and educators seeking to understand the implications of electric fields and potentials in relation to infinite charge distributions.

nicksauce
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Homework Statement


Calculate the potential V(z), a height z above an infinite sheet with surface charge density \sigma by integrating over the surface.

Homework Equations


V(z)=\frac{1}{4\pi\epsilon_0}\int_s{\frac{\sigma dA}{r}}

The Attempt at a Solution


So V(z)=\frac{\sigma}{4\pi\epsilon_0}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}{\frac{dx dy}{\sqrt{x^2+y^2+z^2}}}<br />

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

V=-\frac{z\sigma}{2\epsilon_0}

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?
 
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.
 
nicksauce said:
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.

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 V_z - V_{infinity} = -\int_{\infty}^z \vec{E}\cdot{\vec{dz}}

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

so Vz is infinite.

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

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

did they specifically ask you to take potential at infinity to be 0?
They did not.
 
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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