# How Gauss’ Law is applied to cylinders

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• ynbo0813
ynbo0813
If I understand correctly, Gauss’ Law is (roughly) derived as follows:

Part A
1. Electric Flux = EA
2. E = q / (∈4πr^2)
3. A of the surface of a sphere is 4πr^2
4. They cancel out and therefore EA =q/∈
Line 4 seems to only apply to a sphere, as it is based on line 3.

Now, Gauss’ Law is applied to cylinders as follows:

Part B
1. A of a cylinder is 2πrL
2. EA of a cylinder = E2πrL
3. EA is also = q/∈ (from 4 in Part A)
4. q/∈ = E2πrL (from 2 and 3 in Part B)
5. Solve for E = q/∈2πrL
6. q = λL
7. E = λ / ∈2πr
What I’m confused about is line 3 in Part B. I don’t see how we can use the result of line 4 (Part A) in line 3 (Part B), as line 4 (Part A) was derived from the area of a sphere, not that of a cylinder. In fact, if we were to derive Gauss’ Law for cylinders in the same way we did for spheres we would get as follows (similar to lines Part A):

Part C
1. Electric Flux = EA
2. E = q / (∈4πr^2)
3. A of the surface of a sphere is
4. EA = q2πrL / (∈4πr^2)
5. EA = qL / ∈2r (or λL^2 / ∈2r)
I know I must have made a misstep, misunderstanding, or false assumption somewhere. I’m just trying to figure out what it is.

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You can apply Gauss's Law in this way only if you have very symmetric conditions. One should note that the really fundamental laws of Maxwell theory are the differential laws, i.e., Gauss's theorem in the form
$$\vec{\nabla} \cdot \vec{E}=\frac{\rho}{\epsilon_0},$$
where ##\rho## is the charge density.

Using now Gauss's integral theorem you can derive the integral form
$$\int_V \mathrm{d}^3 x \vec{\nabla} \cdot \vec{E}=\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E} = \int_{V} \mathrm{d}^3 x \frac{\rho}{\epsilon_0}=\frac{Q_V}{\epsilon_0}.$$
Here ##V## is an arbitrary volume, ##\partial V## it's boundary surface with the surface-normal vectors pointing out of the volume, and ##Q_V## the total charge contained in this volume.

Now you can use this integral form if you have very symmetric special cases. E.g., for an infinitely long charged straight wire (charge per length ##\lambda##) the electric field is, because of the symmetry under translations along the direction of the wire and rotations around the wire (written in cylinder coordinates, ##(R,\varphi,z)##)
$$\vec{E}=E_R(R) \vec{e}_R.$$
To get the field, take an arbitrary cylinder of radius ##a## around the wire and arbitrary height ##h## you find
$$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}=\int_0^h \mathrm{d} z \int_0^{2 \pi} \mathrm{d} \varphi a E_R(a) = 2 \pi a h E_R=\frac{Q_V}{\epsilon_0} = \frac{\lambda h}{\epsilon_0},$$
and from this
$$E_R(a)=\frac{\lambda}{2 \pi a}.$$
So writing ##R## again instead of ##a## you have found the electric field
$$\vec{E}=E_R(R) \vec{e}_R=\frac{\lambda}{2 \pi R} \vec{e}_R.$$

Delta2, etotheipi and ynbo0813
ynbo0813
You can apply Gauss's Law in this way only if you have very symmetric conditions.
Are you saying that the way I derived Gauss' Law (using algebra) can only be done in very symmetric conditions, but for a more general derivation I should have used the integral form?

Are you saying that the way I derived Gauss' Law (using algebra) can only be done in very symmetric conditions, but for a more general derivation I should have used the integral form?

That seems backwards, why do you keep talking about 'deriving Gauss' law'? If you have spherical, cylindrical or plane symmetry, then you can exploit that to easily determine the fields from the charge distribution, using Gauss' law [e.g. like what @vanhees71 showed for the infinite charged cylinder]. But ##\nabla \cdot \vec{E} = \rho/\varepsilon_0## is still always true whether you have symmetry or not though, sure.

vanhees71 and ynbo0813
ynbo0813
why do you keep talking about 'deriving Gauss' law'?
Doesn't EA =q/∈ (i.e. Gauss' law) come from somewhere? Of course, you can start with Gauss' Law and derive Coulomb's law from it (or something similar such as Part A Line 2 in the original post), but you can also start with Coulomb's Law and derive Gauss' law from it.

Delta2
Gauss' law is one of the four Maxwell equations, from which everything else in electromagnetism can be derived. In that sense, it's "more fundamental" than Coulomb's law for the electric field of a point charge.

ynbo0813
ynbo0813
But ##\nabla \cdot \vec{E} = \rho/\varepsilon_0## is still always true whether you have symmetry or not though, sure.
Then why can you only use it with very symmetrical cases, such as cylinders?

Then why can you only use it with very symmetrical cases, such as cylinders?

Who says that? Symmetry only makes the computation easier. For arbitrary surfaces, it is a case of performing a surface integral.

ynbo0813
ynbo0813
Who says that? Symmetry only makes the computation easier. For arbitrary surfaces, it is a case of performing a surface integral.
Ah ok.

Maybe it would be helpful to point out where I went wrong in Part C of the original post. I think it might give more clarity.

I don't understand what you tried to do in part C, sorry.

Delta2
ynbo0813
I don't understand what you tried to do in part C, sorry.
In Part C I did the same thing I did in Part A, just with cylinders. Part A is the general derivation of Gauss' Law from Coulomb's Law that I came across. It uses the area of the surface of a sphere. In Part C I did the same thing, just using the area of the surface a cylinder.

Well why do you still use the factor of ##4\pi r^2##? That has nothing to do with a cylinder!

ynbo0813
Well why do you still use the factor of ##4\pi r^2##? That has nothing to do with a cylinder!
From Coulomb's Law

From Coulomb's Law

Coulomb's law applies to point particles, or charge distributions with spherical symmetry, not cylinders!

[HINT: @vanhees71 already wrote the solution!]

ynbo0813
ynbo0813
Coulomb's law applies to point particles, or charge distributions with spherical symmetry, not cylinders!

[HINT: @vanhees71 already wrote the solution!]
But if Gauss' law can be derived from Coulomb's Law (which only applies to point particles or charge distributions with spherical symmetry) how can it be applied to cylinders?

Gauss' law isn't derived from Coulomb's law, it's one of the fundamental Maxwell equations. It's pretty easy to show that Coulomb's law is equivalent to Gauss' law in the case of a point particle or a spherically symmetric charge distribution.

Gauss' law, either in the form ##\nabla \cdot \vec{E} = \rho/\varepsilon_0##, or in integral form ##\int_{\Sigma} \vec{E} \cdot d\vec{S} = q/\varepsilon_0## [true for any closed surface ##\Sigma##] always holds.

I recommend an introductory electromagnetism textbook, e.g. Griffiths, to gain comfort with such concepts.

[As an aside, in some simple examples when you work with a continuous charge distribution, you might choose to integrate over all charge elements ##dq## which you can take to be approximately 'point-like', and e.g. find the electric field at some arbitrary point ##\mathcal{P}## via a summation of Coulomb's law terms.]

alan123hk and ynbo0813
alan123hk
If I understand correctly, Gauss’ Law is (roughly) derived as follows:

The integral form of Gauss's law is applicable to all closed surface shapes, so it cannot be derived only by verifying spherical and cylindrical shapes, nor by verifying it in spherical shapes and then applying the concept to cylindrical shapes, etc. You should try to prove that it works for all shapes.

ynbo0813 and etotheipi
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One should note that the really fundamental laws of Maxwell theory are the differential laws,
Why you are saying this? The way I know it, integral and differential forms are equivalent (as long as the fields are differentiable) and also if I am not mistaken integral form preceded historically the differential form. So , why you saying that the differential form is more fundamental?

About the equivalence of Coulomb's law and Gauss's law I found the following wikipedia section quite good:
https://en.wikipedia.org/wiki/Gauss's_law#Relation_to_Coulomb's_law

The summary of the above link: For stationary charge distributions they are equivalent, if we have moving charges gauss's law still holds but coulomb's law becomes an approximation.

Last edited:
ynbo0813
ynbo0813
About the equivalence of Coulomb's law and Gauss's law I found the following wikipedia section quite good:
https://en.wikipedia.org/wiki/Gauss's_law#Relation_to_Coulomb's_law

The summary of the above link: For stationary charge distributions they are equivalent, if we have moving charges gauss's law still holds but coulomb's law becomes an approximation.
Another way to describe the difference between Coulomb's Law and Gauss' Law may be this (and please correct me if I'm wrong):

Coulomb's Law can be used to find the electric field at a particular point in space. However, where the charge distribution is more complex (like a cylinder) Coulomb's Law is unhelpful.

Gauss' Law can be used to describe the electric flux through any closed surface, no matter the shape of the surface or the charge distribution. However, it can only be used to find the electric field at a particular point in space in special symmetrical cases such as cylinders.

In other words, a cylinder is too complex to simply use Coulomb's Law to find the electric field, but it is symmetrical enough to use Gauss' Law to find the electric field.

Homework Helper
Gold Member
Another way to describe the difference between Coulomb's Law and Gauss' Law may be this (and please correct me if I'm wrong):

Coulomb's Law can be used to find the electric field at a particular point in space. However, where the charge distribution is more complex (like a cylinder) Coulomb's Law is unhelpful.

Gauss' Law can be used to describe the electric flux through any closed surface, no matter the shape of the surface or the charge distribution. However, it can only be used to find the electric field at a particular point in space in special symmetrical cases such as cylinders.

In other words, a cylinder is too complex to simply use Coulomb's Law to find the electric field, but it is symmetrical enough to use Gauss' Law to find the electric field.
You are correct here , however you are referring to cases where we should choose which law is more easy to apply. Of course in some cases it might be easier to use coulomb's law and in some others gauss's law.

My wikipedia link does not refer to that but to when these two laws are equivalent i.e when we can derive the one from the other, or simply put when both laws hold or when one law holds but the other does not hold.

ynbo0813
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2022 Award
Doesn't EA =q/∈ (i.e. Gauss' law) come from somewhere? Of course, you can start with Gauss' Law and derive Coulomb's law from it (or something similar such as Part A Line 2 in the original post), but you can also start with Coulomb's Law and derive Gauss' law from it.
Sure! On the fundamental level the differential Maxwell equations (+ the Lorentz-force law or equivalently the conservation laws due to Noether's theorem applied to Poincare symmetry of special relativity) are the most comprehensive description of electromagnetical phenomena within the realm of classical physics.

Of course you can always transform the differential equations to integral equations. From Gauss's Law in differential form,
$$\vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho,$$
you get the integral form by just choosing a volume ##V## with boundary ##\partial V##, with the surface-normal vectors pointing out of the volume by convention, and take the volume integral of Gauss's Law and then apply Gauss's integral theorem:
$$\int_{V} \mathrm{d}^3 x \vec{\nabla} \cdot \vec{E} = \int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E} = \frac{1}{\epsilon_0} \int_{V} \mathrm{d}^3 x \rho=\frac{1}{\epsilon_0} Q_V.$$
That tells you that the flux of the electric field through a closed (!) surface equals the charge contained in the enclosed volume (times a conversion factor ##1/\epsilon_0## due to the choice of SI units). This holds for any ##V## and arbitrary fields fulfilling Gauss's Law.

To derive, however, ##\vec{E}## for the given source ##\rho## from Gauss's Law in integral form alone is usually only possible for charge distributions of high symmetry.

ynbo0813, etotheipi and Delta2