Gauss's law and nonuniform electric field

In summary, the conversation discusses the concept of flux and how it relates to a Gaussian surface in the shape of a cube. The electric field is given and it is mentioned that a constant vector does not contribute to the flux. The participants also discuss the calculation of flux for individual faces and the importance of the outward normal component. There is a mention of nonzero flux from the constant vector on one side, with the same but opposite sign on the opposite side.
  • #1
boredbluejay
6
0

Homework Statement


A Gaussian surface is in the shape of a cube with edge length 1.40m. The electric field is E=[-4i+(6+3y)j]N/C.

I got an answer, but the solution manual stated that we treat the electric field as E=3yj+E0, where E0=-4i+6j, which does not contribute to the flux. Why is this? Please help!
 
Physics news on Phys.org
  • #2
The flux of a constant vector is zero as the contribution from one side of the cube is positive and negative from the opposite side.

ehild
 
  • #3
Ah, okay. I didn't know that. But when calculating the flux of individual faces, does the constant vector contribute?
 
  • #4
You know how to get a flux of a vector field B(r) at an surface area A? you integrate the normal component of the vector B : Φ=∫BndA.
The normal component refers to the outward normal. Although there is nonzero flux from the constant vector on on side, there is the same with opposite sign on the opposite side.

ehild
 
  • #5


I can provide some clarification on this situation. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. In simpler terms, the electric flux is a measure of the electric field passing through a given surface.

In this problem, we have a nonuniform electric field given by E=[-4i+(6+3y)j]N/C. This means that the strength and direction of the electric field is changing as we move through the space. In order to calculate the electric flux passing through a closed surface, we need to consider the electric field at every point on that surface.

Now, let's consider the Gaussian surface in the shape of a cube with edge length 1.40m. This means that the surface has six sides, and each side has an area of 1.40m^2. In order to calculate the electric flux through this surface, we need to calculate the electric field passing through each side and then add them all together.

However, in this problem, we have an electric field that can be broken down into two components: E=3yj+E0, where E0=-4i+6j. The first component, 3yj, is dependent on the y-coordinate and does not contribute to the flux because the sides of the cube have no y-component. Therefore, we only need to consider the second component, E0, which is constant and does not change as we move through the surface. This means that we can simply multiply E0 by the area of each side and add them together, without having to consider the changing y-component.

In summary, we treat the electric field as E=3yj+E0 because the y-component does not contribute to the electric flux through the cube's sides. This simplification allows us to calculate the electric flux more easily and accurately. I hope this helps clarify the solution manual's approach.
 

FAQ: Gauss's law and nonuniform electric field

What is Gauss's Law and how is it related to electric fields?

Gauss's Law is a fundamental law of electromagnetism that relates the electric flux through a closed surface to the total charge enclosed by that surface. In other words, it describes the relationship between the distribution of electric charges and the resulting electric field in a given region.

How is Gauss's Law applied to nonuniform electric fields?

In the case of nonuniform electric fields, the electric flux through a closed surface can still be calculated using Gauss's Law. However, the total charge enclosed by the surface must be found by taking into account the nonuniformity of the electric field. This can be done by considering small areas of the surface and approximating the electric field as being uniform over each area.

Can Gauss's Law be used to determine the electric field at a specific point?

Yes, Gauss's Law can be used to determine the electric field at a specific point if the electric flux through a closed surface that encloses that point is known. This is because the electric flux is directly proportional to the total charge enclosed, which in turn is related to the electric field through Gauss's Law.

What is the relationship between Gauss's Law and the divergence of the electric field?

Gauss's Law can also be expressed in differential form, known as the divergence form, which relates the divergence of the electric field to the charge density at a given point. In this form, Gauss's Law can be used to calculate the electric field at a specific point by taking the divergence of the electric field at that point.

Are there any limitations to the application of Gauss's Law?

One limitation of Gauss's Law is that it only applies to static electric fields, meaning that the charges and electric field do not change with time. It also assumes that the medium in which the electric field exists is linear and homogeneous. Additionally, Gauss's Law is only applicable to electrostatics and cannot be used to analyze the behavior of electric fields in dynamic systems or in the presence of changing magnetic fields.

Back
Top