1. ## reasonning questions

I have some question that I am not able to answer.
here they are:

Find the electric potential function Φ(x) at a point that is a distance x>0 from an infinite uniform planar sheet of charge of density ρso located at x=0, when the reference point (where potential is zero) is at a distance x0>0 from the sheet. I found Φ(x)=-(ρso/2ε0)*[x-x0]

Now the questions:
1-Is it possible to locate the refernce point at x0=infinity? if not why?
2- Is it possible to locate the reference point at x0=0? if not why not?
3- describe the surfaces on which the potential is constant. Can the potential-integral formula be used for this problem? If not, why not?

Can someone help me reason?

Thank you for the help.

I have some question that I am not able to answer.
here they are:

Find the electric potential function Φ(x) at a point that is a distance x>0 from an infinite uniform planar sheet of charge of density ρso located at x=0, when the reference point (where potential is zero) is at a distance x0>0 from the sheet. I found Φ(x)=-(ρso/2ε0)*[x-x0]

Now the questions:
1-Is it possible to locate the refernce point at x0=infinity? if not why?
2- Is it possible to locate the reference point at x0=0? if not why not?
3- describe the surfaces on which the potential is constant. Can the potential-integral formula be used for this problem? If not, why not?

Can someone help me reason?

Thank you for the help.
1. Why not? The "only" difficulty you will have is that your potential is infinite. If you are prepared to do the Math to subtract infinite constants then okay. (I'm not being sarcastic: in Quantum Field Theory this is done all the time.) However to be practical about it, I wouldn't recommend this procedure.

2. This reference point makes MUCH more sense and is a standard, especially when the plane of charge is a conducting surface.

3. You should be able to get this one. What equation does x satisfy when phi is constant? x = constant, of course. So the equipotential surfaces are planes parallel to the plane of charge, as you could argue simply from symmetry.

-Dan

3. Thank you topsquark,
As you said, I found the third question I just wanted to be sure. I understand now the first two questions.

B