Electric potential for an infinite plane charge distribution

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Homework Help Overview

The discussion revolves around calculating the electric potential for a charge placed at a distance from an infinite plane charge distribution. The subject area includes concepts of electric fields and potentials in electrostatics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the integration of the electric field to find the potential and express concerns about the unknown constant that arises in the process. There are questions about the implications of choosing different reference points for the potential and the potential for infinite values when considering circular charge distributions.

Discussion Status

Participants are exploring various interpretations of the problem, particularly regarding the integration process and the significance of the reference point for the potential. Some guidance has been offered about the nature of the unknown constant and the importance of boundary conditions in determining a unique solution.

Contextual Notes

There is mention of the potential being infinite when considering a circular plane distribution and extending it to infinity, raising questions about the appropriateness of the reference point chosen for the potential calculation.

p3rry
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Hello everybody,
I have to calculate the electric field and the potential for a charge [tex]q[/tex] placed at distance [tex]d[/tex] from an infinite plane charge distribution [tex]\sigma[/tex].

For the electric field there's no problem, but how I can get the electric potential for an infinite charge distribution?
 
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Integrate the field.
 
ok, but I get an unknown constant.

The field is [tex]E=\frac{\sigma}{2\epsilon_{0}}[/tex] in the x direction. So if I integrate it I get

[tex]V(x)=V(0)-\frac{\sigma}{2\epsilon_{0}}x[/tex]

Where the constant is unknown...maybe I don't need to know it

Further, if I calculate the potential starting from a circular plane distribution and then pulling the radius to infinity I get an infinite potential
 
Last edited:
p3rry said:
ok, but I get an unknown constant.

The field is [tex]E=\frac{\sigma}{2\epsilon_{0}}[/tex] in the x direction.


No, that's just the field of a uniformly charged plate with surface charge density [itex]\sigma[/itex]... what about the field of the point charge?

So if I integrate it I get

[tex]V(x)=V(0)-\frac{\sigma}{2\epsilon_{0}}x[/tex]

Where the constant is unknown...maybe I don't need to know it

There is always an "unknown constant" when calculating the potential. This is because it is determined from the differential equation [itex]\textbf{E}=-\mathbf{\nabla}V[/itex], and as you should know, first order DE's need at least one boundary/initial condition to find a unique solution...You are free to choose any value for your constant, simply by choosing a suitable reference point (a point where you define the potential to be zero)...in this case, choosing the origin as a reference point makes things simple (so that V(0)=0).


Further, if I calculate the potential starting from a circular plane distribution and then pulling the radius to infinity I get an infinite potential

Unless you show your calculation, I cannot be certain of your error, but I suspect you are unknowingly choosing your reference point to be at r=infinity...r=infinity is usually a bad choice of reference point when dealing with charge distributions the extend to infinity.
 

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