# Volume enclosed by isosurfaces in electrostatics

1. Jul 10, 2009

### woertgner

Imagine that you have N points scattered in the plane. Each point is charged and contributes to the total potential the same Coulomb potential proportional to 1/r. Now consider some equipotential line of the total electrostatic potential. Is there an analytical expression for the surface enclosed by this line and for the length of this line?

2. Jul 10, 2009

### vin300

If you are saying you want to find out the area enclosed by this line then by simple geometry you know the circle has maximum area.
The potential due to a sheet is diectly proportional to r.
that due to line is proportional to ln(r)
You wrote about the "volume" enclosed by isosurfaces

3. Jul 13, 2009

### woertgner

I am not sure if I understood your answer. I rephrase my question in more detail. Imagine you have one charged point in the plane. The electrostatic potential due to this point charge is up to a multiplicative factor: phi=1/r. Consider an equipotential line by setting phi=c. This equipotential line clearly is a circle and it is easy to compute its perimeter and surface. Now consider two equally charged points in the plane. The electrostatic potential due to these two charged points is up to a multiplicative factor phi=1/r1+1/r2, where r1 is the distance from the first point and r2 is the distance from the second one. Consider an equipotential line again by setting phi=c. If c is small enough you obtain two slightly deformed circles, if c is larger you obtain two merged circles forming something like number eight. How to compute the perimeter and surface now? What about a general case when we have N points in the plane and the potential is given by phi=1/r1+1/r2+1/r3+...+1/rN? By setting phi=c we obtain a set of deformed circles around points, which are sufficiently far away from other points and merged circles from others that are not sufficiently isolated. How to compute the perimeter and surface now? The generalisation of this problem to 3D is similar.