# How electric flux is analog to water?

The definition of electric flux is very often understood through the analogy of water. In the water example, water flux is easy to understand. Water flux is how much water flows per second in a given area. If you know the speed and area, it is all good.

The analogy doesn't work for me too well. $$E$$ then must be considered as a speed in water analogy. Since $$E$$ is not speed, nothing actually flows in a given area per second. $$E$$ field lines are just there (wouldn't call this "flow").

If so, then I am facing difficulty in understanding what exactly E times A actually yields? It cannot give us the number of field lines per second, since nothing is flowing and there is no "second" unit. Perhaps it gives us how many of the total E field lines actually are at the intersection of the given area - but if so, then I don't get why E times A gives us this.

I am looking for a logical explanation to this. I understand Gauss's Law, but it doesn't help me understand this.

• In electric current, charge carriers (like electrons, ions, holes in semiconductors, etc) is "flowing" in a wire, that's why current density is related to change drift speed. So your understanding that nothing is directionally moving in a current is wrong. Commented Jun 8, 2023 at 13:07
• I'm not talking about current. i'm talking about single point charge(stationary) and surface into which its electric field lines "pass through". Commented Jun 8, 2023 at 13:14
• Sorry, I have misunderstood you, my fault. Commented Jun 8, 2023 at 13:22

The analogy doesn't work for me too well.

Well, it is only an analogy but not an equality, because $$\mathbf{E}$$ and $$\mathbf{v}$$ are different things. The analogy is just motivated by some mathematical similarities:

• Both are vector fields.
• Both satisfy Gauss's law. The velocity field satisfies $$\nabla\cdot\mathbf{v}=0$$ because water is incompressible and cannot be created or destroyed. The electric field satifies $$\nabla\cdot\mathbf{E}=0$$, if there are no charges.

$$E$$ field lines are just there (wouldn't call this "flow").

Correct.

I'm having a difficulty what exactly E times A actually gives ?

$$\mathbf{E}$$ gives the density of the field lines (i.e number of lines per area). Hence $$\mathbf{E}\cdot\mathbf{A}$$ gives the number of field lines passing though the area $$\mathbf{A}$$.

• To me, E is the electric field strength at each point of the surface. This is easier to me. but I also get your point. for one charge, there'll be one electric field line at each point, but for 2 charge, there'll be 2 electric field lines at each point on the surface, hence, density of field lines. correct ? Commented Jun 8, 2023 at 13:38
• @Chemistry There is just one electric field generated by all the charges together. But to calculate this total field you can calculate the contributions from each charge separately and then add these. (This is just the principle of superposition.) Commented Jun 8, 2023 at 13:45

The flux does not depend on the velocity of the water, only on the "amount" of water, or the amount of field lines passing through the surface. It's important to remember that it is just an analogy and has its limitations. Remember that flux is equal to $$\phi = \vec{E} \cdot \vec{A} = |\vec{E}| |\vec{A}| \cos(\theta)$$, and so suppose you have water, or field lines, flowing through a surface at an angle then only the water, or the field lines, that is perpendicular to the surface is relevant.

• I didn't include cosa for simplicity, but that is understood. you mention flux is the amount of water(well, if water didn't have speed, nothing would come out of the surface so it truly depends on speed), but in case of Electric flux, no speed. Commented Jun 8, 2023 at 12:26
• if you say number of field lines passing through the surface, then let me know the following: imagine there're 100 total field lines in space, and area of surface is 2m^2(not all field lines pass through it). so flux would be now 100 * 2(imagine cosa = 1). What does 200 mean now? field lines passing through surface has to be less than 100, so what does 200 tell us then Commented Jun 8, 2023 at 12:28
• True, In the context of electric flux, the electric field lines are static and do not have a speed associated with them. They represent the direction and magnitude of the electric field at different points in space. Therefore, the concept of speed is not directly applicable to electric field lines in the same way it is for water flow. Electric flux focuses on the density or concentration of electric field lines passing through a given surface area. Commented Jun 8, 2023 at 12:30
• could you try to answer the following: "if you say number of field lines passing through the surface, then let me know the following: imagine there're 100 total field lines in space, and area of surface is 2m^2(not all field lines pass through it). so flux would be now 100 * 2(imagine cosa = 1). What does 200 mean now? field lines passing through surface has to be less than 100, so what does 200 tell us then" Commented Jun 8, 2023 at 12:36
• Suppose 100 field lines equals an electric field strength of $100 V/m$, you will first need to know how many of the field lines actually pass through the surface. Often the assumption is made that a field is homogeneous near the source, and thus 100 field lines actually pass through the surface. Otherwise John's explination is correct. Commented Jun 8, 2023 at 12:45

It's just math. The same mathematical pattern shows up in both models. But it's no more physically significant than the mathematical fact that 1+1=2, which applies to entirely different physical objects, like teacups and stars.

• if you say number of field lines passing through the surface, then let me know the following: imagine there're 100 total field lines in space, and area of surface is 2m^2(not all field lines pass through it). so flux would be now 100 * 2(imagine cosa = 1). What does 200 mean now? field lines passing through surface has to be less than 100, so what does 200 tell us then Commented Jun 8, 2023 at 12:37
• @Chemistry If there are 100 total field lines spread out through all of space, the number that pass through a finite area is zero. Commented Jun 8, 2023 at 12:42
• well, with electric flux formula, you would not say this as flux is E times A which won't be 0 Commented Jun 8, 2023 at 12:46
• @Chemistry E is not the number of field lines in all of space. Commented Jun 8, 2023 at 12:55
• ah, I think I know the reason of confusion. So in E*A, E is the strength of electric field at one of the point of the surface then and we multiply it by A, to get the whole strength through the surface, but this means we treat every point on the surface to be the same strength of E. Why the same ? on this image ibb.co/CvsPNR9 , you can see d1 and d2 are different so strength would be different at those points of surface, then why we treat it the same ? Commented Jun 8, 2023 at 13:11

The idea of flux and field was mainly derived from fluid dynamics. The flow of a fluid as represented by its velocity field is depicted via streamlines. Volume flux (Av) is the volume of fluid passing through a point per unit of time. When the frame is tilted up at angle@, some of the flow misses the frame. The frame's effective area now corresponds to A(effective) = A cos@ with as many flow lines passing through the tilted A as through A