Gauss's Law: Understanding the Role of E in Electric Field Calculation

[FS98]

Gauss’s law is stated as follows

What exactly does E describe. If you use a Gaussian cylinder for example, where does the value of E describe the electric field?

 

[Orodruin]

On the cylinder surface. Note that you have an integral, which depends on the value of the integrand at all points on the integration domain.

 

[FS98]

On the cylinder surface. Note that you have an integral, which depends on the value of the integrand at all points on the integration domain.

What about the circles on the end of the cylinder? I’ve seen some examples where the value for E only depends on the radius of the cylinder. It doesn’t make sense for a point on the center of one of the circular ends of a cylinder to have the same electric field as the top for a situation like a long rod.

 

[Orodruin]

Of course it doesn't, the electric field is generally not constant on the surface.

 

[FS98]

Of course it doesn't, the electric field is generally not constant on the surface.

But the value of E solved for with Gauss’s law in some examples I’ve seen was some constants multiplied by r. If E represented the electric field at on the surface, and was only dependent on r, then wouldn’t the electric field have to be the same on every point on a cylinder of a specified radius, including the ends?

 

[Orodruin]

But the value of E solved for with Gauss’s law in some examples I’ve seen was some constants multiplied by r.

In order to do this you need to be able to argue that the electric field component in the normal direction of the surface is constant on the surface due to symmetries. It is not a generally applicable method. The symmetries have to do with the source and not anything to do with the surface itself. In many cases you can choose a surface cleverly such that you can utilise symmetry arguments.

 

[jtbell]

including the ends

Consider the significance of the dot product $\mathbf{E} \cdot d \mathbf{A}$, particularly in the special cases where $\mathbf{E}$ is perpendicular and parallel to the surface at the location of $d \mathbf{A}$. Which case applies at the ends of the cylindrical Gaussian surface, in the examples that you're looking at?

It would help if you describe a specific example that confuses you, or point us to a page that describes it, preferably with a diagram. Otherwise we have to guess, and we might guess the wrong example, which would lead to confusion.

 

[FS98]

Consider the significance of the dot product $\mathbf{E} \cdot d \mathbf{A}$, particularly in the special cases where $\mathbf{E}$ is perpendicular and parallel to the surface at the location of $d \mathbf{A}$. Which case applies at the ends of the cylindrical Gaussian surface, in the examples that you're looking at?

It would help if you describe a specific example that confuses you, or point us to a page that describes it, preferably with a diagram. Otherwise we have to guess, and we might guess the wrong example, which would lead to confusion.

The example I’m thinking of is from this video.



A Gaussian cylinder is used to find the electric field from an infinitely long charged rod. I’ll upload a picture of finished problem from the video. The answer that the uploaded gets is E = λ/(2πε0r). If E is the electric field at any point on the surface, I’m not sure this makes sense because of the ends of the cylinders. I would think that the electric field at the center of one of the circular ends would be 0 for example rather than the value of E calculated.

I understand that in the example the flux on the ends is 0, but I don’t see how that would imply that the electric field calculated doesn’t apply at those points.

 

[Orodruin]

I understand that in the example the flux on the ends is 0, but I don’t see how that would imply that the electric field calculated doesn’t apply at those points.

It does apply. But the contribution to the integral is zero due to the field being radial and therefore parallel to the surface. The reason you do a cylinder in the first place is the symmetry of the problem. You can argue that the field is radial everywhere and therefore $\vec E \cdot d\vec S = E\, dS$ on the mantle with $E$ constant. The contribution from the mantle is therefore $EA$, where $A$ is the mantle area. The contribution to the integral from the end caps is zero because on the end caps $\vec E \cdot d\vec S = 0$. The total integral is therefore $EA$, which needs to be compared with the enclosed charge.

Edit: Also note that the guy in your video is being rather sloppy in his notation at 3:40. What he is writing down is the integral over the mantle (after arguing that the integral over the end caps are zero) and as such it is not a closed surface integral and should not have the ring.

 

[jtbell]

But the contribution to the integral [on the end caps] is zero due to the field being radial and therefore parallel to the surface.

...and perpendicular to $d\mathbf{A}$, which is what matters for the dot product $\mathbf{E} \cdot d\mathbf{A}$. Note the diagrams in these examples. They use $\Delta \mathbf{A}$ instead of $d\mathbf{A}$, but the idea is the same.

 

[FS98]

It does apply. But the contribution to the integral is zero due to the field being radial and therefore parallel to the surface.

I think my problem is the math. I’m familiar with calc 1 integrals but the math here seems to be different than what I’m used to.

Is the following correct?

The value E gives the electric field at all parts of the cylinder.

The dot product area integral for the cylinder is equal to the dot product area integral for the end caps plus the dot product area integral for the curved part of the cylinder.

The dot product area integral for the end caps is equal to 0 because of the dot product.

Our new dot product area integral contains the value E which now only describes the electric field along the curved area of the cylinder.

 

[jtbell]

Yes, you’ve described it correctly.

The flux integrals that you’ll see in this course will probably all be similar to this one, in that you’ll be able to split them into parts that are either zero like the end caps here, or a simple product of the area and some constant field magnitude.

If you continue on to Calculus III and more advanced physics courses, you’ll get examples that you actually have to solve as “real integrals.” The important thing in this course is to grasp them conceptually.