Is electric flux dependent on geometry of gaussian surface?

We say, Electric Flux, $$\phi_E = ES \cos (\theta)$$ where $$\theta$$ is the angle between E and S. However, there is another definition for the same given by Gauss Law - $${Q_{enclosed} \over \epsilon _o}$$, where $$Q_{enclosed}$$ is the charge inside the gaussian surface. Effectively one can say that electric flux depends on the electric field, Surface Area (i.e. geometry) and charge enclosed (by gaussian surface).

However, I have seen a lot of books say that flux only depends on charge enclosed. The math is clearly against that statement. Can someone please clarify this confusion for me?

• why is the math "against this statement"? The integral of your $ES\cos \theta$ gives $\int_S {\bf E}\cdot d{\bf S}$ which depends only on the enclosed charge. Nov 21, 2021 at 18:23
• Gauss' Law is not a definition of flux. It is a mathematical relationship concerning the electric field. It describes a property of the electric field in the presence of charges. Flux for a flat area is defined by your first equation. Nov 21, 2021 at 20:41

"We say, Electric Flux, ππΈ=πΈπcos(π) where π is the angle between E and S. However, there is another definition for the same..."

No. There is not another definition.

The electric flux, $$d\Phi$$, through a surface element, $$d\mathbf S$$, is defined by $$d\Phi=\mathbf E.d\mathbf S$$ Gauss's law states that the net flux emerging through a closed surface S is related to the algebraic sum, $$Q$$, of the charge enclosed by S, according to $$\int_S d\Phi = \frac 1 {\epsilon_0} Q$$ This is not a definition but a law. If the charges are stationary, it may be deduced from Coulomb's law, but it also applies if the charges are moving, so it is a law in its own right.

It is, indeed, remarkable that if we compute the integral over different surfaces enclosing $$Q$$, we get the same resultant flux for each surface. You should try starting with a very simple case: spherical Gaussian surfaces of different radii with a single stationary charge at the centre.

"Effectively one can say that electric flux depends on the electric field, Surface Area (i.e. geometry) and charge enclosed (by gaussian surface)."

Perhaps one could say this, but it would be clearer to say that if we know the electric field strength at all points on a closed surface, S, we can calculate the resultant flux through S by applying the definition of $$d\Phi$$ and integrating over S, OR if we know the charges inside S, we can calculate the resultant flux through S using Gauss's law. 'OR' is more appropriate than 'AND'.

Your statement is rather like saying that the total number of limbs possessed by a group of able-bodied people depends on the number of arms + the number of legs AND on the number of people.

[The definition of a limb is an arm or a leg. Wood's law of limbs is that an able-bodied person has 4 limbs. We can calculate the total number of limbs either from the definition (by direct counting of arms and legs) or the law (simply by counting people and multiplying by 4).]

• So can I say that "Gauss's law" doesnt depend on the geometry but flux in general can be calculated with other parameters like E and S ? Nov 22, 2021 at 4:46
• I think that's fine, though I'd substitute "shape of the closed surface containing the charges" for "geometry". Nov 22, 2021 at 10:44
• Yeah that's a lot more specific. Thanks for clarifying. I really appreciate it. Nov 22, 2021 at 10:56
• Thank you. In my answer I did emphasise the distinction between a law and a definition, because I find that many students seem not to be clear about it. A definition establishes the meaning of a term. A law says something about how things are in the world, and often takes the form of a relationship between different quantities. Laws are often worded using terms that have special definitions; for example Gauss's law relates charge and flux, both terms with special definitions β though I wouldn't, off the top of my head, like to give the definition of charge! Nov 22, 2021 at 11:08
• Yeah right. I gotcha :D Nov 22, 2021 at 14:23

The electric flux through a small surface is defined to be $$\Delta\Phi=\mathbf{E}\cdot\Delta\mathbf{S}=E\ \Delta S\ \cos(\theta).$$ For larger surfaces the more general definition is $$\Phi=\int \mathbf{E}(\mathbf{r})\cdot d\mathbf{S}.$$ This definition applies to arbitrary surfaces (open or closed, plane or curved) and arbitrary electric fields (homogenous or inhomogenous). So far this is just a definition of a new quantity $$\Phi$$, there is no physical content in it.

Now you were comparing two very different situations.

1. An open surface
For an open surface it doesn't make sense to talk about the charge enclosed by this surface. Gauss's law is not applicable because this law holds only for closed surfaces.
In case of a homogenous electric field and a plane (i.e. open) surface patch the definition of flux from above simpifies to $$\Phi=\mathbf{E}\cdot\mathbf{S}=ES\cos(\theta).$$ But you cannot relate this flux to any charge.

2. A closed surface
For a closed surface Gauss's law holds true. $$\Phi=\oint \mathbf{E}(\mathbf{r})\cdot d\mathbf{S} =Q_\text{enclosed}/\epsilon_0$$ The remarkable thing of this law is, that the flux depends only on the enclosed charge, but not on the detailed geometric shape of the surface. So this is a law with physical content, connecting two physical quantities (charge $$Q$$ and electric field $$\mathbf{E}$$).

• So can I say that "Gauss's law" doesnt depend on the geometry but flux in general can be calculated with other parameters like E and S ? Nov 22, 2021 at 4:45

Nope! The electric flux depends on only the charge enclosed, regardless of the geometry of your Gaussian surface.

Here is Gaussβs Law in integral form: $$\oint \textbf{E} \cdot d\textbf{A} = \frac{Q_{enc}}{\epsilon_0}$$

Altering the surface would make computing the integral more difficult, but the net flux would still be the same regardless because some of the area vectors will be parallel with the electric field vectors and others may end up being perpendicular (causing the dot product to go to zero).

• Altering the geometry of the doesn't make computing the integral more difficult since the integral always equals $Q/\epsilon$ regardless of the geometry. Nov 21, 2021 at 18:52
• @BobD Youβre completely correct, I should have clarified that I meant actually computing EdA via brute force. Nov 21, 2021 at 19:00

$$\int \nabla \cdot E dv = Q enclosed / \epsilon_0$$

This is an equivalent form of gauss law. This states that the flux is dependant on the divergence of E about my volume enclosed by my gaussian surface. suppose I have a charge Q located at the center, the divergence of E is zero everywhere apart from the center at which It is $$\rho / \epsilon_0$$

If I were to now change my surface to be a different surface that still encloses the single point of divergence of E. would the flux change?

Clearly not as the integral is dependant on the sum of the divergences about the volume. The sum hasn't changed since the divergence of the points in the new volume is 0 apart from where any charge is located

• So can I say that "Gauss's law" doesnt depend on the geometry but flux in general can be calculated with other parameters like E and S ? Nov 22, 2021 at 4:46
• unclear what you mean. gauss law describes a closed surface flux for the E field which states that the flux about any closed surface is equal to the charge enclosed by that surface. flux as a concept is (field strength × area) in its simplest form or how many field lines flow through an area ish. Nov 22, 2021 at 5:03

Electric flux density does indeed depend on the shape and location of the Gaussian surface. However, this is not what Gauss's law is describing. Gauss's law says that the total electric flux of a surface that completely encloses a charge depends only on the total amount of charge that is enclosed, and is the summation of all the differential fluxes through each small differential area of the Gaussian surface, as described mathematically by VaxTensor. The "nice" thing about Gauss's law is that the total amount of electric flux exiting or entering a Gaussian surface is the same for ALL Gaussian surfaces, which allows selection of the most convenient shape and location of the Gaussian surface. Thus, the integration of the flux equation can often be a trivial mathematical operation if the proper Gaussian surface is specified (e.g., a sphere for a point charge and a cylinder for a line charge).

• So can I say that "Gauss's law" doesnt depend on the geometry but flux in general can be calculated with other parameters like E and S ? Nov 22, 2021 at 4:46
• @HarshDarji, you could say that. But note - for a point charge, you would put the charge at the center of a Gaussian surface. At that point, the electric field strength is constant over the whole Gaussian surface, allowing you to "poll" the "E" across the integral. In addition, since you know the surface area of a sphere (e.g., $4 \pi r^2$), you don't have to integrate "dA". Nov 22, 2021 at 16:34