# Electric potential for an infinite plane charge distribution

1. Oct 5, 2009

### p3rry

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?

2. Oct 5, 2009

Staff Emeritus
Integrate the field.

3. Oct 5, 2009

### p3rry

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

Last edited: Oct 5, 2009
4. Oct 5, 2009

### gabbagabbahey

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

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).

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.