Electric potential for an infinite plane charge distribution

AI Thread Summary
To calculate the electric potential for a charge at distance d from an infinite plane charge distribution with surface charge density σ, the electric field is given by E=σ/(2ε₀). Integrating this field yields V(x)=V(0)-σ/(2ε₀)x, but the constant V(0) remains unknown. It is emphasized that the potential can be defined by selecting a reference point, typically where the potential is set to zero. Choosing an inappropriate reference point, such as r=infinity, can lead to misleading results, particularly for infinite charge distributions. Properly defining the reference point is crucial for accurate potential calculations.
p3rry
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Hello everybody,
I have to calculate the electric field and the potential for a charge q placed at distance d from an infinite plane charge distribution \sigma.

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 E=\frac{\sigma}{2\epsilon_{0}} in the x direction. So if I integrate it I get

V(x)=V(0)-\frac{\sigma}{2\epsilon_{0}}x

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
 
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p3rry said:
ok, but I get an unknown constant.

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


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

So if I integrate it I get

V(x)=V(0)-\frac{\sigma}{2\epsilon_{0}}x

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 \textbf{E}=-\mathbf{\nabla}V, 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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