# Need Help with Applying the Point Form of Ohm's Law

1. Aug 15, 2012

### Prince Rilian

Need Help with Applying the "Point Form" of Ohm's Law

I am quite familiar with the standard form of Ohm's Law

V = IR​

as I have been using it for years in circuit analysis. However, there is another form of Ohm's Law that is known as the "Point Form" of Ohm's Law:

E = Jρ​

Where E is a vector quantity expressing Electric Field, E is a vector quantity expressing Current Density, and ρ is a scalar quantity expressing Resistivity.

It seems that the "Point Form" of Ohm's Law would be quite useful for extending my knowledge of circuit analysis. I feel that if I use it right, I could tell what E is at any point in an electrical circuit. However, I do not have an "intuitive feel" on how to use it yet in circuit analysis. Could I have a few pointers on how to do this?

2. Aug 16, 2012

### CWatters

Re: Need Help with Applying the "Point Form" of Ohm's Law

Long time since I did this but..

I don't think it will be very helpful for typical electronic circuits unless perhaps you just want to understand and work with conductance rather than resistance.

Might be more useful if you are analysing the electrical properties of materials or gasses?

3. Aug 16, 2012

### Prince Rilian

Re: Need Help with Applying the "Point Form" of Ohm's Law

I had noticed a major difference between the two forms of Ohm's Law. Despite the similarity of appearance of the two forms, the standard form needs two points in space to operate off of, as V in the equation is the difference in voltage between the two ends of the resistor. This equation may better be written as

V2-V1 = IR​

where V2 and V1 are the nodal voltages at the two ends of the resistor. However, in the point form of Ohm's Law, there is only a single point in space in consideration.

4. Aug 16, 2012

### Studiot

Re: Need Help with Applying the "Point Form" of Ohm's Law

Everything you have said is correct.

It is, however more conventional to use conductivity, σ, rather than resistivity so

J = σE

Don't forget that 'voltage' is the unit for two distinct quantities potential and potential difference.

The E above is potential, (and subject to Faraday's laws)

The voltage across a resistor is potential difference, not potential, although we often foget this when referencing to zero in a circuit.

As to uses

Current density J is the given by multiplying the drift velocity(vector) by the charge density(scalar).

Thus

J = vdρ

Thus σ = ρ vd / E

= ρμ

Where μ is the charge mobility.

5. Aug 16, 2012

### Prince Rilian

Re: Need Help with Applying the "Point Form" of Ohm's Law

Hmmm... it looks like the quantities of charge density and resistivity use the same symbol (ρ). For some reason that never hit me before. I wonder if that isn't the reason why conductivity is used in the Point Form of Ohm's Law more often than resistivity is?

6. Aug 16, 2012

### Studiot

Re: Need Help with Applying the "Point Form" of Ohm's Law

Conductivity is used because it is more general.

It is used when you split the sources eg in a dielectric and ionic solution, plasma etc.