Doubts in applying gauss' law

In summary, the conversation revolved around solving a problem involving flux and charge density. The speaker had no issues with part a, where they used triple integration to find the flux crossing the surface at two different values of r. However, they were unsure about part b, where they applied Gauss' law and took D (electric displacement vector) outside the integral, assuming it was constant even though the charge density was not uniform over the volume. The speaker and the book were both wondering how this was possible.
  • #1
FOIWATER
Gold Member
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I actually have solved the problem and received the answers that the book provided.

However I am second guessing what I did,

This is the problem:

let rho v be 10/r^2 mC/m^3 between 1<r<4
let rho v be zero elsewhere
a)find the net flux crossing the surface at r=2 m, r= 6m
b)determine D at r=1, r=5

a) I have no issues with, I can simply say that the Flux crossing the surface equals the charge enclosed and solve for the charge enclosed by integrating the volume charge density over the volume. I used triple integration in spherical coordinates with r from 1 to 2, theta from 0 to pi, and phi from 0 to 2pi. I got 40pi mC

Same goes for r=6m, I just integrated the same except r went from 1 to 4 (since there is no more charge after r = 4, or rather, the charge density was given as zero) (but the flux will remain the same at 6m as it was at 4m) 120pi

My issue is with what I did in b), applying gauss' law.

I know gauss' law holds, and for r=1, since there is no charge inside a spherical gaussian surface, there is no flux density on the surface so D=0 but for r=5m, I assumed a gaussian surface (a sphere) of radius 5. I know the flux at 5m is the same as in part b, 120pi. so:
flux = Qenc = surface integral D dot ds, D and ds are both vectors.
How could I, and the book, assume D is constant and take it outside the integral sign?
The charge density inside the sphere was given as 10/r^2, which isn't uniform over the volume?

How could a sphere be perpindicular to the flux lines contain within when the charge density is not uniform?
 
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  • #2
At first glace, it looks like conceptual issue is that the charge density isn't uniform, but is IS still functionally radial. With no angular dependence or other weird things going, the flux lines are still perpindicular to the sphere.
 
  • #3
OK that was what I was thinking could be going on.

Thanks
 
  • #4
Sure thing!
 
  • #5


First of all, it is completely normal to have doubts about applying Gauss' law, even for experienced scientists. It is a complex concept and it is always good to double check your calculations and assumptions.

In this particular problem, it seems like you have correctly applied Gauss' law in part a) by considering the charge enclosed within the Gaussian surface. However, in part b), you are correct in realizing that the charge density is not uniform within the Gaussian surface. This means that the electric field (represented by D) is not constant over the surface, and therefore cannot be taken outside the integral.

In order to properly apply Gauss' law in this case, you would need to consider the differential charge elements within the spherical Gaussian surface and integrate over the entire surface to find the total flux. This would involve using the charge density function and taking into account the varying distance from the center of the sphere.

It is also worth noting that Gauss' law is a mathematical representation of the physical concept of electric flux, and it is not always applicable in every situation. In cases where the charge density is not uniform, it may be more appropriate to use other methods, such as direct integration or the divergence theorem.

In summary, it is important to carefully consider the assumptions and limitations of Gauss' law when applying it to a problem, and to always double check your calculations and results. If you have any further doubts or concerns, it may be helpful to consult with a colleague or mentor for guidance.
 

What is Gauss' law?

Gauss' law, named after mathematician and physicist Carl Friedrich Gauss, is a fundamental law of electromagnetism that relates the distribution of electric charges to the resulting electric field. It states that the electric flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space.

How do you apply Gauss' law?

To apply Gauss' law, you must first identify the closed surface or Gaussian surface that encloses the charge distribution. This surface can be any shape, as long as it is a closed surface. Then, you must calculate the net charge enclosed by the surface and determine the electric field at each point on the surface. Finally, you can use the formula Q/ε0 = ∫E⃗ · dA⃗ to solve for the electric field or charge distribution.

What is the significance of Gauss' law?

Gauss' law is significant because it allows for the calculation of electric fields in complex situations, such as those involving non-uniform charge distributions. It also provides a direct relationship between the electric field and the charge distribution, making it a powerful tool in the study of electromagnetism.

What are some common doubts in applying Gauss' law?

Some common doubts in applying Gauss' law include confusion about the choice of Gaussian surface, difficulty in calculating the net charge enclosed by the surface, and understanding the relationship between the electric field and the charge distribution. It is important to carefully consider the situation and choose an appropriate Gaussian surface, as well as to have a good understanding of the mathematical principles behind the law.

Are there any limitations to Gauss' law?

Yes, there are limitations to Gauss' law. It only applies to static electric fields and does not take into account any time-varying effects. It also assumes that the permittivity of free space is constant, which may not always be the case. Additionally, it does not account for any quantum mechanical effects, making it less accurate at very small scales.

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