# Electric flux Confusion

1. Feb 3, 2006

### aalmighty

The electric flux, by a definiton, is the number of electric field lines per unit area. Yet, when we apply gauss' law for a 1 Coulumb charge, we get a non-integral value. What then, is the physical Significance of the value we get?

2. Feb 3, 2006

### Staff: Mentor

Electric field lines are just visual aides for picturing the field. The actual number of field lines emanating from a point charge is arbitrary, but will be proportional to the charge. The field strength will be proportional to the density of field lines; flux would be a number of lines passing through an area.

Rather than get hung up on counting field lines, think in terms of the actual field strength.

3. Feb 4, 2006

### rbj

in addition, since the Coulomb is a completely artificial and human-defined unit of charge, if there ever was a real physical reality to these "field lines" (there isn't, as Doc said, it's a visual aid to visualize flux density and why it is an inverse-square phenomenon), the number of lines would certainly not care about with what units humans decided measure charge with.

perhaps, instead, try to think of each electron or proton in that Coulomb of charge emitting a field line (say, for each electron or proton, having the direction of the field line be totally random, so that all directions are equally covered). then, as you draw successive concentric spheres around that coulomb of charge, you can see the density of field lines getting lower according to the inverse square law.

4. Feb 4, 2006

### Staff: Mentor

No. In the "field lines" picture, the magnitude of the electric field (i.e. its "strength" or "intensity") corresponds to the number of field lines that pierce a unit area oriented perpendicular to the field. The flux of the electric field through a surface corresponds to the total number of field lines that pierce the surface, regardless of the area of the surface or the angle that the lines make with respect to the surface.

The flux of the electric field through a surface is defined as

$$\int {\vec E \cdot \hat n dA}$$

where the integral is taken over the surface and $\hat n$ is the unit vector perpendicular to the surface at each point. If $\vec E$ is perpendicular to the surface, and has the same magnitude, everywhere on the surface, then the flux is just the magnitude of $\vec E$ times the area of the surface.

Last edited: Feb 4, 2006
5. Sep 14, 2008

### VinnyCee

So... what is the unit for electromagnetic flux?

6. Sep 14, 2008

### rbj

the unit for electrostatic flux density would be unit of charge per unit of area (length^2). in SI, it would be "coulombs per meter squared". if you multiply that by $1/\epsilon_0$, you get the electric field (volts/meter), which if you multiply by the remote charge (coulombs), you get the force (newtons).

oh, but you were asking about "flux", not flux density. the total flux that's emanating from the charge Q, is Q. that's distriibuted across a surface area $4 \pi r^2$ to give you a flux density of $Q/(4 \pi r^2)$ and is the origin of the inverse-square law for 3-dim space.

7. Sep 14, 2008

### Defennder

Is there such a thing as electromagnetic flux? There is electric flux and magnetic flux, but electromagnetic?

8. Sep 14, 2008

### DrZoidberg

Electric flux is defined as the product of the electric field strength and the area.
So the unit is V/m * m^2 = V*m.

If you compare that with magnetic flux you can easily get confused as magnetic flux has the unit Vs. Therefore you might expect electric flux to be As, but that is not true because electric flux is defined completely different then magnetic flux.
In other words - electric flux is not the electric equivalent of magnetic flux.
Of course magnetic and electric flux should have been defined in an equivalent way but that didn't happen and so - for historic reasons - we are stuck with these definitions.

9. Sep 14, 2008

### rbj

no. that's off by a constant factor of $\epsilon_0$. "flux density" is defined to be flux divided by area. but flux density, represented as this D vector, is not the same as field strength, represented as this E vector. the relationship between D and E is D=$\epsilon_0$E. in SI units, the unit for electrostatic flux is coulombs. 5 coulombs of charge emit a total of 5 coulombs of electrostatic flux.

it's a matter of units. if you do it sorta like the cgs system, the units for electrostatic flux and field are the same, but there's a dimensionless factor of $4 \pi = 1/\epsilon_0$ between them. and, in the Lorentz force equation, the E field and B field are the same units.

10. Sep 14, 2008

### Defennder

Was he talking about flux due to electric displacement or electric field?

11. Sep 15, 2008

### DrZoidberg

Strange. Whereever I look - books, wikipedia, even online lectures from the MIT website - it says that electric flux is E*A.
Are there 2 different definitions of electric flux? E*A and D*A?

12. Sep 15, 2008

### rbj

well, i guess there are semantic issues. i looked it up at Wikipedia and they use both "flux density" and "displacement field". if you consider the simple Coulombs inverse-square law:

$$F = \frac{1}{4 \pi \epsilon_0} \ \frac{Q \ q}{r^2}$$

and consider this F to be the force that charge Q is exerting on charge q (it's also the force that q exerts on Q), then the electric field (created by Q) at the location where charge q is, is the force per unit charge

$$E = \frac{F}{q} = \frac{1}{\epsilon_0} \ \frac{Q}{4 \pi r^2}$$

that E field is in units newtons/coulomb (which is the same as volts/meter).

the latter factor $Q/(4 \pi r^2)$ is what i meant as "flux density". imagine a sphere centered at location Q with radius r, so that charge q sit on the surface of that sphere. the effect of that charge Q, which doubles if Q doubles, is distributed over the surface area of that sphere. if we call the constant of proportionality of that "effect due to charge Q" to be 1, and we call that the flux emanating from charge Q, then the total flux is simply Q. that total flux is divided among an area of $4 \pi r^2$ (which is what q can sense).

that is what that "D" quantity is. but our freedom to arbitrarily call that constant of proportionality "1" does not extend as far as the field strength, E. then we need to multiply by $1/\epsilon_0$, both to get our units right (and to get the right value in terms of those units). personally, i wish that both cgs and Planck Units chose their unit of charge so that it was $\epsilon_0$ was normalized to 1 rather than normalizing $4 \pi \epsilon_0$ as they did. if they did that, flux density (D) and field strength (E) would be exactly the same thing in a vacuum, measured in the same units and being the same value. but they didn't do that.

shucks.

13. Oct 6, 2008

### Santosh K

You said that "Electric flux is defined as the product of the electric field strength and the area.
So the unit is V/m * m^2 = V*m."

every ehere i see electric flux is defined as E*A.Where did you get it as D*A??????

14. Oct 6, 2008

### ZapperZ

Staff Emeritus
If you are in a dielectric media, then the "displacement field" D is what you have to use. Since $D=\epsilon E$, it doesn't change the dimensional analysis.

Zz.