# On which objects can we apply Gauss' Law to find the electric field?

• zenterix
In summary: In some cases (like for the cylinder), you can just apply the law and get the flux. In other cases (like for the ring), you need to go beyond the flux to get the field.
zenterix
Homework Statement
The title of this section is "Charge Distributions with Enough Symmetry for Gauss's Law"

The problem is stated as follows:

On which of the objects shown in the figure above can you apply Gauss's Lw to find the Electric Field from the given charge density? (Choose all that apply, see picture below)

1. Infinite line of charge with a uniform charge density ##\lambda_0##

2. Infinite cylinder of charge with charge density ##\rho=\rho_0\frac{\cos{\theta}}{r}##

3. A slab of charge, infinite in ##y## an ##z## with a constant charge density ##\rho##

4. A ring of charge with uniform charge density ##\lambda_0##

5. A sphere with uniform charge density ##\rho_0##
Relevant Equations
As far as I can tell, you could in theory apply Gauss's Law to any of the objects in the figure below. This answer, according to the automated answer from MIT Open Library, is incorrect.

Calculating the flux of the electric field generated by a charge distribution can be made easier or harder depending on the Gaussian surface chosen. However, as long as we can calculate the flux of the electric field through a closed surface, we can apply Gauss's Law. Is this true?

In the case of the infinite line and cylinder, we use as Gaussian surface a cylinder with end-caps perpendicular to the line or cylinder.

For the infinite slab, we can use a cylinder with the shaft of the cylinder perpendicular to the slab.

For the ring, we could use a piece of a torus around a small piece of the ring.

For the solid sphere we use an enclosing sphere.

What am I missing?

I also don't get the title of the section: "Charge distributions with enough symmetry for Gauss's Law".

I thought Gauss's Law was valid for any closed surface enclosing a charge. I don't understand what "enough symmetry" means in the title above. I get that with symmetry it's easier to solve the integrals involved by hand.

Gauss' law is always true. But, unless there is enough symmetry, you won't be able to use Gauss' law to find the electric field.

"Enough symmetry"usually means that you can choose a Gaussian surface such that ##\oint \mathbf E \cdot \mathbf {dA} =EA##

or you can choose a Gaussian surface such that ##\int \mathbf E \cdot \mathbf {dA} = EA## for some parts of the surface and ##\int \mathbf E \cdot \mathbf {dA} =0## for the other parts of the surface.

zenterix, PeroK and Steve4Physics
So just to confirm: we can apply Gauss's Law to all of the charges in the pictures in my original post?

If so, then the course is actually asking something slightly different; but I think it doesn't matter because in that case the question is simply formulated in a way that is not clear enough.

zenterix said:
So just to confirm: we can apply Gauss's Law to all of the charges in the pictures in my original post?
In SI units, Gauss' law is ##\oint \mathbf E \cdot \mathbf{dA} = \frac{Q_{\rm enclosed}}{\epsilon_0}##. This applies to any closed surface in any situation.

However, whether or not you can apply Gauss' law to find the electric field at some point depends on whether or not there is enough symmetry in the charge distribution.

zenterix said:
If so, then the course is actually asking something slightly different; but I think it doesn't matter because in that case the question is simply formulated in a way that is not clear enough.
For each picture, you should be able to state whether or not it is possible to find E by application of Gauss' law.

TSny said:
For each picture, you should be able to state whether or not it is possible to find E by application of Gauss' law.
And I have stated my conclusion: yes, I can find E by applying it to all of the objects. I've actually done all calculations except for the ring and sphere.

Infinite Line

$$\vec{E}=\frac{\lambda}{2\pi r\epsilon_0}\hat{r}$$

Infinite Cylinder, uniform charge density

Inside the cylinder
$$\vec{E}=\frac{\rho r}{2\epsilon_0}\hat{r},\ 0<r \leq a$$

Outside the cylinder
$$\vec{E}=\frac{\rho a^2}{2r\epsilon_0},\ r \geq a$$

The cylinder in the pictures has some weird expression for charge density. The latter is not uniform, and given the ##\cos{\theta}## it appears there are regions of positive and negative charge on the cylinder. Doing the calculations it appears that electric flux through a cylindrical Gaussian surface is zero for this case.

I haven't done the sphere yet but I see it is an example in my textbook so it can be done.

The ring I haven't done nor intend to. I made two attempts at this question in my self-study course: one said yes we can do the ring and the other said no we can't do the ring; both were wrong.

I have thought about the question. This is one of these questions in between lecture videos. I would like to know the answer at this point, but it isn't available.

I’ll add a few words to what @TSny ny has said.

zenterix said:
I thought Gauss's Law was valid for any closed s
Yes, Gauss’s Law (GL) is valid. But that doesn't mean it can always be used to find electric field.

GL only gives the total flux passing through a Gaussian surface (S) in terms of the enclosed charge.

GL can’t give you the field at each point on S except in special cases - where the magnitude of the field is known to be constant (or zero) over each part of S.

For example, consider the ring of charge (radius R). In the plane of the ring, the magnitude of the field at R-ΔR (from the centre) is not the same as the magnitude of the field at R+ΔRE.

In fact, inside the ring, you should be able to see that the field at the centre is zero. But outside the ring, you have to go to infinity before the field drops to zero.

So:
“For the ring, we could use a piece of a torus around a small piece of the ring. “
won’t work because we have no way to deal with ##\oint E \cdot dA## since ##E## is not constant.

zenterix and TSny
In your first post you wrote
zenterix said:
Calculating the flux of the electric field generated by a charge distribution can be made easier or harder depending on the Gaussian surface chosen.
Yes, that's right. But keep in mind that there's a difference between calculating the flux and calculating the field E.

zenterix said:
However, as long as we can calculate the flux of the electric field through a closed surface, we can apply Gauss's Law. Is this true?
It depends on what you mean by "apply". In this problem, you want to apply Gauss' law to find the electric field produced by the charge distribution. Unless there is enough symmetry, it will be impossible to apply Gauss' law to find E.

zenterix said:
In the case of the infinite line and cylinder, we use as Gaussian surface a cylinder with end-caps perpendicular to the line or cylinder.
Yes, but it is essential that the linear charge density is uniform. If the charge density is nonuniform, then it would generally be impossible to apply Gauss' law to find E.

zenterix said:
For the infinite slab, we can use a cylinder with the shaft of the cylinder perpendicular to the slab.
Yes. But if the slab has a nonuniform charge density ##\rho##, it will generally be impossible to find E by application of Gauss' law.

zenterix said:
For the ring, we could use a piece of a torus around a small piece of the ring.
Have you been able to do this? What expression did you get for E?

zenterix
TSny said:
Have you been able to do this? What expression did you get for E?
I tried to do it just now. Seems pretty intractable for me from the point of view of calculating the integrals, as far as I did it.

My strategy was:

- Use Coulomb's law to find the electric field generated by the ring at any point ##P##. If I have this then I have an expression to use in a calculation of flux through a Gaussian surface around a piece of the ring.
- First I used Coulomb's law for a ##dq## on the ring.

$$\vec{r}_{dq,p}=\vec{r}_{o,p}-\vec{r}_{o,dq}$$

$$\vec{r}_{o,p}=r_p \hat{r_p} + \rho \cos{\phi_p}\hat{k} = r_p(\cos{\theta_p} \hat{i} + \sin{\theta_p} \hat{j}) + \rho \cos{\phi_p}\hat{k}$$

$$\vec{r}_{o,dq}=R\hat{r}_{dq} = R(\cos{\theta_{dq}}\hat{i}+\sin{\theta_{dq}}\hat{j})$$

$$\vec{r}_{dq,p}=r_p \hat{r_p} + \rho \cos{\phi_p}\hat{k} - R\hat{r}_{dq}$$
$$=(r\cos{\theta_p}-R\cos{\theta_{dq}})\hat{i}+(r\sin{\theta_p}-R\sin{\theta_{dq}})\hat{j}+\rho\cos{\phi_p}\hat{k}$$

Note that ##\vec{r}_{o,p}## doesn't change, but ##\vec{r}_{o,dq}## changes as we change ##\theta_{dq}## from ##0## to ##2\pi##, ie ##\hat{r}_{dq}## changes.

$$dq=\lambda ds = \lambda R d\theta_{dq}$$

$$dE_{dq,p}=\frac{k_edq}{|\vec{r}_{dq,p}|^3}\vec{r}_{dq,p}=\frac{k_e\lambda R d\theta_{dq}}{|\vec{r}_{dq,p}|^3}\vec{r}_{dq,p}$$

If we now sub in ##\vec{r}_{dq,p}## and ##|\vec{r}_{dq,p}|^3## and take the limit of a Riemann sum, we get an integral in ##d\theta_{dq}##, from ##\theta_p## to ##\theta_p+2\pi##. A quick try on Maple of one of the integrals shows a crazy result.

Before I dive in and try to figure out if there is a way to solve the integral, maybe you guys have more experience and can tell me what you think so far, if there is a coordinate system in which the integral might be easier, etc.

At this point, I feel like I should've placed point P above one of the axes ##x## or ##y##. This would make all the ##\theta_p## expressions simpler. But actually it only simplifies things slightly.

Last edited:
zenterix said:
My strategy was:

- Use Coulomb's law to find the electric field generated by the ring at any point ##P##. If I have this then I have an expression to use in a calculation of flux through a Gaussian surface around a piece of the ring.
You are asked to decide if you can use Gauss' law to determine the electric field at any point ##P## (without using Coulomb's law).

Is there a way to choose a closed surface that passes through the point ##P## such that you can set up the equation ##\oint \vec E \cdot \vec {dA} = \frac{Q_{enc}}{\epsilon_0}## for this surface and then solve this equation for ##E## at point ##P##?

Steve4Physics and zenterix
zenterix said:
I tried to do it just now. Seems pretty intractable for me from the point of view of calculating the integrals, as far as I did it.
I think you are missing the point – applying GL simply can not provide you with E as a function of ρ and θ.

The integral from θₚ to θₚ+2π is independent of θₚ. You must get the same answer whatever value of θₚ you use because of Gauss’s Law! - the total flux through the Gaussian surface is the same whatever angle you start/end at when evaluating the integral.

The integration will remove θₚ from the result – so you will end up with no information about the dependence of E on θ.

GL is valid here but simply can’t be used to find E as a function of ρ and θ.

On a different (but related) note, you might find this video on finding the field inside a charged ring – a tricky problem - of interest.

zenterix
Ok, looks like I did miss the point. I was trying to use Coulomb's law to calculate the electric field at an arbitrary point P, to then use Gauss's law. But for what? The point of GL is to have a way to calculate the electric field at any point using a surface integral. If you already have an expression for the electric field, then using Gauss law would only be useful if you somehow didn't know the enclosed charge somewhere. You could use GL to solve for that enclosed charge.

Steve4Physics said:
The integration will remove θₚ from the result – so you will end up with no information about the dependence of E on θ.
Indeed.
TSny said:
Is there a way to choose a closed surface that passes through the point P such that you can set up the equation
Looks like the answer is no, because the surface integral contains the expression for some non-constant electric field generated by the ring. I guess this is the point of the Gaussian surface being symmetric enough: symmetric enough that the electric field is constant on the surface you are integrating on.

Steve4Physics and TSny

## 1. What is Gauss' Law and how is it used to find the electric field?

Gauss' Law is a fundamental law in electromagnetism that relates the electric field to the distribution of electric charges. It states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space. This law can be used to find the electric field in situations with high symmetry, such as a point charge or a charged spherical shell.

## 2. Can Gauss' Law be applied to any object to find the electric field?

No, Gauss' Law can only be applied to objects with high symmetry, such as spheres, cylinders, and planes. This is because the electric field must have a constant magnitude and direction at every point on the surface in order for the law to be applicable.

## 3. How does the shape of an object affect the application of Gauss' Law?

The shape of an object determines the symmetry of the electric field around it. The more symmetric the object, the easier it is to apply Gauss' Law. For example, a sphere has perfect symmetry and the electric field will be the same at every point on its surface, making it ideal for applying the law.

## 4. Can Gauss' Law be used to find the electric field inside an object?

Yes, Gauss' Law can also be used to find the electric field inside an object as long as the charge distribution is known. This is because the law relates the electric field to the enclosed charge, regardless of whether it is inside or outside the object.

## 5. Are there any limitations to using Gauss' Law to find the electric field?

Yes, there are limitations to using Gauss' Law. Besides the requirement of high symmetry, the law also assumes that the electric field is constant on the surface of the object being considered. This may not be the case for objects with varying charge densities or non-uniform electric fields.

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