# Proof for why flux is proportional to number of field lines

What is the proof for this (assuming that we draw infinite field lines).

I understand why flux through some area is proportional to the number of field lines through that area only in the case of an isolated singular point charge. However, I dont understand why this would be true in other cases, like two unequal charges.

To make it clear, when an area has twice the amount of field lines going through it than another area, why is the flux through it exactly twice the other area. Why not thrice or half?

Also, why do field lines that originate isotropically from one charge always end isotropically at another charge.

• Do you understand that an infinite number of field lines pass through the surface? Commented Aug 7, 2023 at 0:56

1. Field lines are just a way to visualize the force and its direction, they do not "exist"! It is arbitrary how many field lines you draw per cm^2, so only the relative density of them tell you something of the strength of the field. Your latter statement is not true, only a isolated charge has isotropic field lines. Since the describe the force between the two charges they have to start and end between them
• But when considering areas inside the same diagram, flux is still proportional to the field lines right? And by isotropic I mean isotropic when originating or ending, as in the field lines which originate uniformly from a postive charge also end uniformly at the negative charge. Commented Aug 6, 2023 at 12:48
• If your picture is in 2d it is not true, since it is the field lines in 3d. Commented Aug 6, 2023 at 13:42

For many field line (lines of force) are a useful visual aid and it is not necessary to count all the field lines when using the concept, for example, when discussing the electric field (strength) due to a point charge.

Suppose that the strength of an electric field is measure in terms of field lines passing through unit area.

Please note that my graphic skills were not up to drawing areas $$A$$ and $$B$$ as curved surfaces.

Assuming spherical symmetry and the radial nature of the $$N$$ field lines starting out from charge $$Q$$ and passing through area $$A$$, the same $$N$$ field lines must pass through area $$B$$ which has four times the area of area $$A$$.

Thus the electric field strength at a distance $$2a$$ from from the charge is a quarter that of the electric field strength at distance $$A$$.
This is the inverse square law.

So when you work out the magnitude field at two positions as $$2 \,\rm N/C$$ and $$10 \,\rm N/C$$ you could imagine that five times as many field lines pass through an elemental area at one positions as compared with the other position.

As has been pointed out one must be wary when drawing a 2-dimentional diagram which shows field lines and then inferring a numerical value of the field strength by counting the field lines as we live in a 3-dimensional world.

Note that in a 2-dimensional world the electric field from a point charge would not follow the inverse square law.

• I am able to understand why this is is true for a point charge. However, what about in the general case? For example, why is it true in the case of a dipole? Commented Aug 6, 2023 at 15:58

I would argue that you have the logic backwards. Field lines are a convenient pictorial representation of vector fields; there is no a priori reason for the number of field lines to be proportional to the flux. You could, for example, arbitrarily decide to draw twice as many field lines on the "left side" of a point charge as on the "right side."

In order to (a) remove some ambiguity in how to draw field lines and (b) produce a meaningful picture, we adopt a convention when drawing field lines that the density of field lines at a point $$x$$ is proportional to the value of the field at $$x$$. This is a convention -- not something intrinsic about field lines that you can prove.

Once we adopt this convention, then it is the case that the number of field lines poking through a small surface with oriented area $$d\vec{A}$$ is proportional to the flux through $$d\vec{A}$$. Loosely speaking, the reason is that the flux is $$\vec{E} \cdot d\vec{A}$$, and by our convention above the number of field lines at the location of the area element will be proportional to $$E$$ at the area element. (Since the number of field lines will be the density of the field lines times the infinitesimal area).

• However, if we adopt that convention and draw the field lines in such a way that their density is proportional to the field, the field lines arise isotropically from the charge. Alternatively, when we draw them originating isotropically, their density becomes directly proportional to the field. Why is this so? Commented Aug 7, 2023 at 14:32
• @MarcCarlsan In the case you have two point charges, then if you are close to the second charge, you can ignore the effect of the first charge. The reason is just Coulomb's law; if you get close enough to the second charge, then the factor of $1/r^2$ will mean the field due to the second charge is much larger than the field from the first charge. In this window where we are "close enough" to the second charge, the field is approximately that of a single point charge; isotropic. The field lines inherit this property of the underlying field because of the convention we use to draw field lines. Commented Aug 8, 2023 at 1:10