# Is there a converse of uniqueness theorem

1. Jul 24, 2007

### pardesi

is there a converse of uniqueness theorem for circuits have for charged conductors.

or atleast is there such a thing in case of circuit analysis ..

2. Jul 24, 2007

### olgranpappy

Could you state the theorem a bit more precisely for us?

3. Jul 24, 2007

### pardesi

if in a given volume u know the charge on each conductor(u may notknow the charge distribution) and if u know the charge distribution in space( $$\rho$$) then u know everything about that region potential,field....

4. Jul 24, 2007

### LHarriger

So you want to know if the following is true:
If you know the potential, electric field, then you know the charge distribution.
I would say the converse is not true. Afterall, we use the method of images to calculate the potential based on a ficticious distribution of charge.

5. Jul 24, 2007

### Parlyne

If you know the potential and where there are boundaries, then you can determine the charge distribution uniquely.

6. Jul 25, 2007

### pardesi

@LHarriger
well i don't want to know the charge distribution but the charges

@Parlyne
can u please explain that with an example

7. Jul 25, 2007

### LHarriger

This is both true and a good point. However, technically this statement is not the converse of the uniqueness theorem (though the techincality is extremely trivial.)

There are actually two statements:

First Uniqueness Theorem: The potential in a volume V is uniquely determined if the charge density throughout the region and the value of the potential V on all boundaries are specified.
Converse: In a volume V, the charge density throughout the region and the value of the potential V on all boundaries are uniquely determined for a given potential V.

I said that this was not true for the reason stated earlier, namely for a given boundary and charge distribution we employ the method of images by defining a fictitious boundary and charge distribution that gives a potential satisfying both the actual and fictitious cases. This is in direct violation of the converse statement. However, I want to change my mind on this. The reason is that when we use the method of images, we are changing our volume of interest. (Infact we must always put our fictitious charge in the expanded volume, otherwise we would be changing our density and would solve Possoin's Equation for the wrong source charge.)
I now am of the mind that the converse should be true. First, if you know the entire potential, then of course you know it on the boundary (like I said, the technicality was trivial.) Second, using Possoin's Equation you can calculate the unique charge density for that potential.

Now that I think about it though, you were probably refering to the second theroem,

Second Uniqueness Theorem Given a volume V that conains conductors of known charge and also contains a known fixed charge density between the conductors, the electric field is uniquely determined. (The region as a whole can be unbounded or surrounded by a conductor).
Converse Given a volume V in which the electric field is known, the charge on conductors inside the volume as well as any charge density between the conductors is uniquely determined.

I would respond on this as well but I have to leave and I want to make sure my response is watertight (unlike last time)

Last edited: Jul 25, 2007
8. Jul 25, 2007

### pardesi

so is the converse to the second unique theorem true.also i think 1 and it's converse holds for 1 though the theorem is stronger than the converse i think

9. Jul 25, 2007

### pardesi

i don't think the terme charge distribution in the converse theorem can be unique because clearly by the second uniqueness theorem once the charge is fixed no matter how it is distributed the field everywhere is fixed provided u know the charge density everywhere.
what i think is true is if u know the field evrywhere then u know the charge density in free space and total charge on each conductor but not how it is distributed

10. Jul 25, 2007

### LHarriger

Gauss's Law in differential form should gaurentee that the converse is true.

11. Jul 26, 2007

### pardesi

yes got that
gauss law in differential+gauss law in integral