Importance/Use of Gauss' Law and Ampere's Circuital Law

[Aurelius120]

Homework Statement

What is the use of Gauss' Law and Ampere's Circuital Law? When to use them over Coulomb's Law and Biot Savart Law?

Relevant Equations

Ampere's Law $$\int \vec B.d\vec l=\mu _{○}I_{enclosed}$$
Gauss' Law$$\int \vec E.d\vec A=\frac{Q_{enclosed}}{\epsilon_{○}}$$

Abbreviations
ACL- Ampere's Circuital Law
BSL - Biot Savart Law
GL- Gauss Law
CL- Coulomb's Law

So the question is like I mentioned. I know how to use GL and ACL in place of CL and BSL. I also know that they make the calculations simpler.

The problem is whenever GL and ACL are used in place of CL or BSL, they are used in the following way:
Gauss Law:$$\int{ E. dA}=E\int{dA}=\frac{Q_{in}}{\epsilon_{○}} \implies E=\frac{Q_{in}}{\epsilon_{○}A}$$
Amperes Circuital Law:
$$\int{ B. dl}=B\int{dl}=\mu_{○} I_{in} \implies B=\frac{\mu_{○}I_{in}}{L}$$

In the uses above, we drag the $E$ and $B$ vectors out of the integral under the assumption that they are uniform for the selected surface or loop.
What is the basis of these assumptions?

My teacher said that it will work that way for the questions and examples we will be given and I need not worry about that.
Allegedly, there are certain symmteries or conditions that they use to verify which questions can be solved that way.

So what are those conditions or symmetries, that we are not taught?

What is the importance of such laws to say scientists or engineers ? It seems that using those laws are as difficult as using CL or BSL. I mean it would be as tedious to check whether certain conditions are followed in a given situation to drag $B$ or $E$ out of the integral as using CL or BSL.
What then is the use of GL or ACL? Why teach them without teaching those conditions?

 

[PhDeezNutz]

For Gauss’ Law

Spherical, cylindrical, and planar symmetry

For Ampere’s Law

Cylindrical, and planar

 

[PhDeezNutz]

Please see this insight article by well known member @Orodruin

https://www.physicsforums.com/insights/a-physics-misconception-with-gauss-law/

 

[kuruman]

What is the basis of these assumptions?

The basis of these assumptions is symmetry. For Gauss's law, in order to drag, as you say, $E$ out of the integral, two conditions must be satisfied
1. Vector $\mathbf{E}$ must be perpendicular to the Gaussian surface everywhere. If that is the case, then the vectors are parallel everywhere and you can write $\mathbf{E}\cdot d\mathbf{A}=E~dA.$
2. The magnitude $E$ is constant everywhere on the Gaussian surface.

These conditions are satisfied if you place a point charge at the center of a sphere. They are not satisfied if you place a point charge away from the center or if you place a point charge at the center of a prolate (football-shaped) ellipsoid of revolution.

Similar considerations apply with Amperian loops
1. Vector $\mathbf{B}$ must be tangent to the Amperian loop everywhere. If that is the case, then the vectors are parallel everywhere and you can write $\mathbf{B}\cdot d\mathbf{l}=B~dl.$
2. The magnitude $B$ is constant everywhere on the Amperian loop.

These conditions are satisfied if you place a very long current-carrying perpendicular to the plane of a circle passing through the center of the circle. They are not satisfied if you place the current away from the center or if you place it at the center of an ellipse.

 

[Aurelius120]

The basis of these assumptions is symmetry. For Gauss's law, in order to drag, as you say, $E$ out of the integral, two conditions must be satisfied
1. Vector $\mathbf{E}$ must be perpendicular to the Gaussian surface everywhere. If that is the case, then the vectors are parallel everywhere and you can write $\mathbf{E}\cdot d\mathbf{A}=E~dA.$
2. The magnitude $E$ is constant everywhere on the Gaussian surface.

If I am given a non-conventional charge distribution, how do I check for these conditions?

Similar considerations apply with Amperian loops
1. Vector $\mathbf{B}$ must be tangent to the Amperian loop everywhere. If that is the case, then the vectors are parallel everywhere and you can write $\mathbf{B}\cdot d\mathbf{l}=B~dl.$
2. The magnitude $B$ is constant everywhere on the Amperian loop.

If I am given an uncommon current distribution, how do I check?

Is it pure mathematics to draw the field lines for checking these conditions?
How do we know the pattern of field lines for say a line charge distribution or that cylinder is the Gaussian surface of choice for it? We assume that field lines are radial for point charge so point charge at center of sphere is rather intuitive

 

[Orodruin]

Symmetry. You will not get anywhere with those methods without symmetry arguments.

We assume that field lines are radial for point charge so point charge at center of sphere is rather intuitive

We do not need to assume this. We can deduce that it must be so from the spherical symmetry. Same thing for the field pointing in the radial direction relative to an infinite line charge - that situation has cylinder symmetry.

 

[kuruman]

If I am given a non-conventional charge distribution, how do I check for these conditions?

If I am given an uncommon current distribution, how do I check?

You need to say more about your distributions. Non-conventional and uncommon can mean anything.

Is it pure mathematics to draw the field lines for checking these conditions?

It's common sense and the rules for drawing field lines. If you have not been taught these rules, I am sure you can find them on the web.

We assume that field lines are radial for point charge so point charge at center of sphere is rather intuitive

That the field lines due to a point charge are radial is not assumption; it is the case. Here is a little test of your intuition. Shown below are 3 positive point charges as red circles with electric field lines pointing away from them. Each charge is enclosed by a closed Gaussian surface depicted in green. In (A) it is a sphere, in (B) it is a cube and in (C) it is a piece of a sphere with a flat circular top.

Question 1
For each of the three surfaces state whether Gauss's law can be used to find the electric field on the surface and justify your answer.

Question 2
For each of the three surfaces state whether Gauss's law valid and justify your answer.

I hope that answering these questions will sharpen your understanding of Gauss's law.

 

[Orodruin]

It's common sense and the rules for drawing field lines. If you have not been taught these rules, I am sure you can find them on the web.

It is symmetry arguments that are very accurately described by mathematics. Drawing field lines may help or hinder, but is generally difficult to do for a charge distribution you do not already know the field for. Also, I for one am not certain that it is easy to find common sense on the internet …

 

[Aurelius120]

That the field lines due to a point charge are radial is not assumption; it is the case. Here is a little test of your intuition. Shown below are 3 positive point charges as red circles with electric field lines pointing away from them. Each charge is enclosed by a closed Gaussian surface depicted in green. In (A) it is a sphere, in (B) it is a cube and in (C) it is a piece of a sphere with a flat circular top

Yep sorry. Had been using that so often I forgot how that was actually derived.

Shown below are 3 positive point charges as red circles with electric field lines pointing away from them. Each charge is enclosed by a closed Gaussian surface depicted in green. In (A) it is a sphere, in (B) it is a cube and in (C) it is a piece of a sphere with a flat circular top.

Question 1
For each of the three surfaces state whether Gauss's law can be used to find the electric field on the surface and justify your answer.

Question 2
For each of the three surfaces state whether Gauss's law valid and justify your answer.
View attachment 337294
I hope that answering these questions will sharpen your understanding of Gauss's law.

1) Gauss Law can't be used because
A) Field is not uniform for Gaussian surface
B)Field is not orthogonal to Gaussian surface at every point.
C) Field is not uniform and not perpendicular to Gaussian Surface
Therefore E can't be pulled out of the integral
2) Gauss Law is valid because it is a fundamental law and $$\int \vec E.d\vec A=\frac{Q_{in}}{\epsilon_{○}}$$

 

[Aurelius120]

Symmetry. You will not get anywhere with those methods without symmetry arguments.

We can deduce that it must be so from the spherical symmetry. Same thing for the field pointing in the radial direction relative to an infinite line charge - that situation has cylinder symmetry.

Spherical symmetry for point charge is easier to visualise because of coulombs law for point charge.

Is there a mathematical proof of why cylindrical symmetry implies that field is uniform and perpendicular to curved surface of cylinder ?
From this site cylindrical symmetry means
A charge distribution has cylindrical symmetry if the charge density depends only upon the distance from the axis of the cylinder and does not vary along the axis or with the direction about the axis.
Now why does this automatically translate to uniform and orthogonal field on curved surface of chosen cyinder?

https://www.jove.com/science-education/13719/gauss-s-law-cylindrical-symmetry

 

[Aurelius120]

You need to say more about your distributions. Non-conventional and uncommon can mean anything.

Something weird but existing/useful in real life.
Let's see something like a uniformly charged Earth shaped object(geoid/oblate spheroid)

Or current flowing in Earth's core(imperfect sphere) causing magnetic fields

 

[Orodruin]

Spherical symmetry for point charge is easier to visualise because of coulombs law for point charge.

Is there a mathematical proof of why cylindrical symmetry implies that field is uniform and perpendicular to curved surface of cylinder ?
From this site cylindrical symmetry means
A charge distribution has cylindrical symmetry if the charge density depends only upon the distance from the axis of the cylinder and does not vary along the axis or with the direction about the axis.
Now why does this automatically translate to uniform and orthogonal field on curved surface of chosen cyinder?

https://www.jove.com/science-education/13719/gauss-s-law-cylindrical-symmetry

Cylinder symmetry consists of arbitrary rotations along the cylinder axis, arbitrary translations around the cylinder axis, reflections in any plane perpendicular to the cylinder axis, and reflections in any plane containing the cylinder axis.

A charge distribution has a particular symmetry if it does not change when you perform a symmetry transformation. Generally, since the field for a given charge distribution is unique, this means that the field cannot change under the symmetry transformations either.

To the cylinder symmetry. The reflection symmetries imply that any component orthogonal to the reflection surface change sign. For something changing sign not to change, it must be zero (ie, x = -x implies x=0). From this we can conclude that the angular and longitudinal components of the electric field must be zero. Furthermore, the translations and rotations can be used to transform any point at distance r from the cylinder axis to any other point at the same distance r from the cylinder axis. Therefore, the magnitude of the field can only depend on the distance r.

The same argument for the magnetic field can be found here:
https://www.physicsforums.com/insights/symmetry-arguments-and-the-infinite-wire-with-a-current/
The result is different because the magnetic field is a pseudovector and not a vector and therefore transforms differently.

 

[kuruman]

Yep sorry. Had been using that so often I forgot how that was actually derived.

1) Gauss Law can't be used because
A) Field is not uniform for Gaussian surface
B)Field is not orthogonal to Gaussian surface at every point.
C) Field is not uniform and not perpendicular to Gaussian Surface
Therefore E can't be pulled out of the integral
2) Gauss Law is valid because it is a fundamental law and $$\int \vec E.d\vec A=\frac{Q_{in}}{\epsilon_{○}}$$

Great. The MIT site below has a comprehensive introductory treatment of Gauss's law and includes the symmetries and surfaces that go together, solved examples, how to use Gauss's law in general to show things like why free charges reside on the surface of a conductor and other goodies.

https://web.mit.edu/8.02t/www/802TEAL3D/visualizations/coursenotes/modules/guide04.pdf

 

[PhDeezNutz]

At some point in this thread OP asked about “more general charge and current distributions that lack symmetry” (or some thing to that effect). Other tools besides Gauss’ and Ampere’s law must be used.In such cases the “multi-pole expansion” must be used.

Scalar multipole expansion for electric potential (take the gradient of this then you have electric field)

Vector multipole expansion for magnetic potential (take the curl of this you have magnetic field)

Edit: I think OP is very perceptive to ask such questions at an early stage because the limitations of those laws are a stumbling block for many many students later on down the road.

 

[Orodruin]

In such cases the “multi-pole expansion” must be used.

”Must be used” is a pretty strong wording. You can use any approach that will allow you to get the potential or field, including both multipole expansions as well as other things like Green’s function methods.

 

[PhDeezNutz]

Fair, I was actually going to edit my post to say something like

“When given prescribed boundary conditions things like Method of Images and Greens Functions can also be used”

But I got lazy. But you are absolutely right.

 

[Aurelius120]

Cylinder symmetry consists of arbitrary rotations along the cylinder axis, arbitrary translations around the cylinder axis, reflections in any plane perpendicular to the cylinder axis, and reflections in any plane containing the cylinder axis.

A charge distribution has a particular symmetry if it does not change when you perform a symmetry transformation. Generally, since the field for a given charge distribution is unique, this means that the field cannot change under the symmetry transformations either.

To the cylinder symmetry. The reflection symmetries imply that any component orthogonal to the reflection surface change sign. For something changing sign not to change, it must be zero (ie, x = -x implies x=0). From this we can conclude that the angular and longitudinal components of the electric field must be zero. Furthermore, the translations and rotations can be used to transform any point at distance r from the cylinder axis to any other point at the same distance r from the cylinder axis. Therefore, the magnitude of the field can only depend on the distance r.

This.
Nowhere on the internet did I find this. I guess they leave it for us to figure out.

Generally, since the field for a given charge distribution is unique, this means that the field cannot change under the symmetry transformations either

Since you use 'Generally', does it mean there are/maybe cases where we might not pull $E$ out of the integral despite symmetries being followed?

 

[PhDeezNutz]

Since you use 'Generally', does it mean there are/maybe cases where we might not pull $E$ out of the integral despite symmetries being followed?

On the contrary E can be pulled out whenever symmetry is obeyed. When symmetry is not obeyed you must resort to other methods such as the

Multi-pole expansion.

Method of Images.

Greens Functions.

Which are entire topics in and of the themselves.

Multipole expansions and Method of Images are a topic for upper level undergrad classes.

Greens functions you might not encounter until grad school. I view them as “integrated method of images” but that may be wrong.

 

[Aurelius120]

I encounter questions like this
A point charge $q$ is placed at the corner of a cube with side $a$. Find flux through entire surface and flux through each face of cube.

Since Gaussian surface for point charge is a sphere, it certainly is an approximation.

 

[Orodruin]

On the contrary E can be pulled out whenever symmetry is obeyed.

This comes with caveats. See the Insight you quoted in #3.

 

[PhDeezNutz]

This comes with caveats. See the Insight you quoted in #3.

Symmetry must be obeyed with the field in conjunction with the chosen surface?

 

[PhDeezNutz]

I encounter questions like this
A point charge $q$ is placed at the corner of a cube with side $a$. Find flux through entire surface and flux through each face of cube.
View attachment 337322

View attachment 337321
Since Gaussian surface for point charge is a sphere, it certainly is an approximation.

The flux through the entire surface is 0.25q/episolon

The flux through the three faces that are coplanar with the charge are 0.

The flux through the remaining three faces are identical and not 0.

 

[Orodruin]

Greens functions you might not encounter until grad school. I view them as “integrated method of images” but that may be wrong.

Green's function methods in themselves a priori have little to do with the method of images. Rather it is a way of constructing the solution to an inhomogeneous linear differential equation from special cases of the same differential equation, but with the inhomogeneity replaced by a delta distribution. Integrating these solutions over the actual inhomogeneity builds up the actual inhomogeneity and thereby the solution. These methods are not restricted to PDEs but can also be used to solve ODEs. Most people have already come into contact with them simply by solving for a potential or field from an extended charge by integrating the point particle potential/field over that charge. The point particle potential just so happens to be a Green's function for the Poisson equation.

However, in some cases you can use the method of images to find the appropriate Green's function. I have a chapter dedicated to Green's function methods in my book if you have access to it.

Symmetry must be obeyed with the field in conjunction with the chosen surface?

Take the infinite line charge for example. We will typically use a finite cylinder as our Gaussian surface. By the symmetry of the problem we know that the field is radial and therefore the end cap integrals vanish (the field being parallel to the end caps). The integral that the field magnitude can be pulled out of is the mantle area integral, since it is at constant distance from the line charge - but we cannot forget to argue that the end cap integrals vanish. The Gaussian surface itself only obeys the rotational symmetry, the symmetry of reflections in planes containing the line charge, and one reflection in a plane orthogonal to the line charge (the plane through the middle of the surface). This is a sub-group of the full symmetry group of the line charge.

 

[PhDeezNutz]

I will definitely nab a copy of your book with my next paycheck. Been meaning to do it for awhile.

 

[nasu]

The charge inside the cube it's 1/8 of the total charge, isn't it? @PhDeezNutz

 

[Orodruin]

Since Gaussian surface for point charge is a sphere, it certainly is an approximation.

There is no approximation here. A particular charge configuration does not have a single Gaussian surface associated to it. You can use any Gaussian surface and Gauss' law will hold. However, in order to use Gauss' law to determine the electric field (which is not what the question asks you to do), then you better make sure to choose your surface wisely.

The charge inside the cube it's 1/8 of the total charge, isn't it? @PhDeezNutz

Technically, the charge is not inside the cube, it is a point charge at one of the corners. This makes the entire thing a bit precarious, but it is easy to argue for 1/8 of the flux entering the cube.

 

[Orodruin]

Nowhere on the internet did I find this.

Well ... technically ...

I guess they leave it for us to figure out.

Many times, symmetry arguments are more hand-waved and not made very explicit. However, they do rest on a solid mathematical foundation that can be made quite formal.

 

[PhDeezNutz]

The charge inside the cube it's 1/8 of the total charge, isn't it? @PhDeezNutz

The flux throughout the entire corner cube is 1/4th

Since only 3 of those 6 surfaces are relevant in regards to flux…..the flux through those surfaces is 1/8

Maybe I wasn’t clear. Hopefully it is now.

Either way I’ve had way too much to drink this Saturday night.

 

[nasu]

No, the flux through the whole cube is 1/8. There are 8 cubes sharing that corner.
Of course, if you consider a really point-like charge the point is neither inside nor outside and the problem is ambiguous. Already mentioned by @Orodruin. But we can imagine a small sphere instead of a mathematical point. 1/8 of the sphere is inside. In 3D space we have 8 octants and not 4 quadrants as in 2D.

 

[Aurelius120]

There is no approximation here.

The flux through the remaining three faces are identical and not 0.

There was one question involving a similar case where the flux through each face wasn't identical as per my teacher but the answer demanded we treat them as such.Maybe I mixed up

Perhaps that was the same question but with the charge at midpoint of an edge of the cube (instead of corner)

 

[PhDeezNutz]

@nasu you are right

1/8th the flux through the entire cube

And since only 3/6 of the faces have actual nonzero flux through them

That 1/8th gets divided by 3

1/24th through each of the facesSorry for being absent minded and apologies to OP for not being more measured with my responses.