# Visualizing field lines with moving current

1. Sep 6, 2012

### Mr Boom

Let's say I have a circle filled with positive charges. Some distance away I have an identical circle filled with negative charges. Since the distribution is uniform in the circles, I can just use the center of the circles as points and calculate the field lines between the two. So far so good. Now let's say I connect the first circle to the second by a resistor. If the sum of the charges are so large that it is essentially unchanged for some length of time (as in I have constant charges and constant current), how can I calculate the electric field?

2. Sep 7, 2012

### Staff: Mentor

No, this just works with spheres (unless you take a modified electrodynamics in a two-dimensional world).

Calculate the field along the connection with usual rules for circuits (constant current in the wire). Use some simulation tool for the electric field, as you probably do not find an analytic solution (except for special resistor profiles where the field is unchanged).

3. Sep 7, 2012

### Mr Boom

I was thinking of spheres, yes, but I was trying to just use a 2D example.

This was my question. I'm wondering how the field lines look visually outside the resistor by allowing current to pass. I realize there will also be a magnetic component. Will the lines straighten or become more arched?

4. Sep 7, 2012

### Staff: Mentor

It depends on the resistor. Without the connection, the field strength is larger close to the spheres and smaller in the middle. If the resistor is a long wire with the same resistance per distance everywhere, you modify this - the field strength close to the spheres gets reduced, the field strength in the middle (and close to the resistor) increases a bit. This should make the field lines close to the resistor a bit "more parallel".

5. Sep 7, 2012

### Mr Boom

OK, that makes sense. I'd like to try to do this problem on my own. Any recommendations on how to start this program? I've plotted the static field lines and I need to superimpose the field due to the resistor?

6. Sep 7, 2012

### Staff: Mentor

That won't work.

You can calculate it with a grid, for example, and I would solve for the potential first:
$\phi(x)=0$ for sphere 1, $\phi(x)=1$ for sphere 2, $\Delta \phi = 0$ in free space, and $\phi(x)=f(x)$ with some function f(x) at the resistor. Instead of a 3-dimensional simulation, it is possible to use the symmetry of the problem (if the connection is symmetric) in a 2-dimensional simulation to reduce the required computing power.