# Charge Density

1. Jul 15, 2009

### Shelnutt2

How would one determine the charge density of a fluid?

I'm using the equation of I = qnvA . Amperes = charge of electron * charge density * velocity * Area. The only thing I'm not sure of is how to determine charge density. I've read several websites and I have a few textbooks here. Everything talks about it differently, and nothing specifically references a fluid. Many talk about solids, but not fluids. Are the properties between solids and fluids for charge density calculations the same? Can someone point me to a formula?

Thanks

2. Jul 20, 2009

### Shelnutt2

Is this the wrong forum? Should I ask for it to be moved? I really wasn't 100% sure what forum this belonged in but I figured EE would cover it.

3. Jul 20, 2009

### Bob S

Do you really need to know charge density, or is the product of some of your factors enough? For example,
Charge density times velocity = Coulombs / cm3 times cm/sec = Coulombs/cm2-sec = amps/cm2.
Would the electrical resistivity of the fluid be enough? What kind of fluid are you interested in? In many cases, like seawater, just knowing the resistivity is enough.

4. Jul 20, 2009

### Shelnutt2

I need to find amps, so it's really all or nothing for that equation. Without the charge density, I'm left with Coulomb * m2 * m/s = Coulomb m3/s or Amp * m3. I need the charge density.

As for the fluid, well I'm looking for how to find the change density for generic purposes. I don't have a specific fluid I'm working with, but I'm looking at a broad range of fluids. Water/Saltwater sure. But I really need a generic way to calculate this all.

5. Jul 27, 2009

### Shelnutt2

No one?

6. Jul 27, 2009

### Bob S

Are these the units you are working with?

q Coulombs
n number of charges per cm3
v velocity cm/sec
A Area cm2

The product has to equal amps = Coulombs/sec

What kind of fluid are you talking about? plasma? seawater? Cross-field propulsion (ion currents)?

7. Jul 27, 2009

### Shelnutt2

Yes those are the units (expect I'm using meter not cm, but thats not overly important).

For the fluid, seawater, brack/brine water, tap water. I guess maybe it's not as generic as I though, in terms of the equation and charge density in general. I'm looking at liquids specifically.

8. Jul 27, 2009

### Bob S

Last edited by a moderator: Apr 24, 2017
9. Jul 27, 2009

### Shelnutt2

Interesting, thanks for the link, it's quite detailed so I'm going to take my time reading through it.

I guess I should explain to you exactly what I'm trying to figure out. What I want is to find the amperage of a flowing fluid. As long as there is a net charge in said fluid, and the fluid is moving, it can be said to have a certain amperage, no? I am trying to work on the theory behind a MHD generator. In all the books on it I've read, amps are the only thing that isn't really clear. Voltage is simple, based on the hall effect. Amps, however, that is conveniently left out in any detail.

10. Jul 27, 2009

### Bob S

I thought so. First, you need a strong magnet at right angles to the current. You need reasonable sized electodes to conduct the current. You need a dc current source to get the amps needed. The current is not flowing in the fluid, but is supplied by you. The MHD force is perpendicular to BOTH the magnetic field AND the supplied current. This based on the Lorentz force law. The voltage on the electrodes depends on the conductivity of the water.brine/whatever.

11. Jul 27, 2009

### Shelnutt2

Yes and no. You are talking about MHD propulsion, while I am talking about a generator to produce electricity. Same principles, but by no means do I supply the amps. The amps are created by the flow/movement of the fluid. Not the movement by the amps (which is what you are talking about). When it's inside the magnetic field, we get voltage, and we have electricity. When the fluid is flowing outside the magnetic field, there is still amps, but with no volts it means nothing.

Voltage is simply V = B * L * u, where B is magnetic field strength in tesla's, L is width (meters) between anode and cathode, and u is velocity (m/s). (It starts out as an integral but everything simplifies down to straight multiplication because it's all assumed to be perpendicular).

The amperage is just something I'm not entirely sure about. After discussing it with different professors, the closest someone could help me was give me the formula of I = q * n * v * A .