# Conservative Field Clarification

1. Feb 12, 2010

### Wannabeagenius

Hi All,

As I understand it, a conservative field means that the energy expended by an outside agent in going between any two points is independent of the path so that the closed line integral of Edotdl is zero.

This is presented in the study of electrostatics.

It seems to me that you can have a conservative field under time varying conditions but I'm not sure!

I'm thinking about central force fields which are conservative. As an example, let's take the inverse r squared relationship and assume that the square inverse relation stays the same but the constant in the coulomb law equation increases with time.

Am I correct in saying that, this too is a conservative field?

Thank you,
Bob

2. Feb 12, 2010

### Staff: Mentor

No it would be nonconservative. As you mention, the integral of the work done on any closed path is 0 for a conservative field. Let's say that the constant is increasing over time, then you could do some work to separate two opposite charges, wait a while, and get more work out by bringing them back to the starting position.

This is also related to Noethers theorem.

3. Feb 12, 2010

### Wannabeagenius

I understand your argument. However, with the field that I described the closed line integral of Edotdl is zero.

Bob

4. Feb 12, 2010

### Staff: Mentor

Sorry, I don't know what Edotdl is. But as I described above in such a system the work on a closed path is non-zero so the field is non-conservative.

5. Feb 13, 2010

### Wannabeagenius

The closed line integral of the electrical field around any closed path. This integration would be taken at a frozen moment in time and would be zero.

A conservative field is defined as the work done being independent of path, as your argument reflects. It is also defined as the closed line integration that I have described above being zero.

That is the problem.

Thank you,
Bob

6. Feb 13, 2010

### vin300

Think this way.You know a "A conservative field is defined as the work done being independent of path".
So whatever path you take from point a to b in the field the work done is the same.
Again, whatever path you take back from b to a, the work is the same in magnitude but negative, so the total work around the loop is zero.
Thus you get the second statement from the first

7. Feb 13, 2010

### Staff: Mentor

The two definitions are not equivalent if the field is time varying. However, if the field itself can carry energy then the force can still be conservative even if the field fails to meet one of these criteria.