Gauss' Theorem - Net Flux Out - Comparing two vector Fields

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Homework Statement:
Vector field [itex] F = \begin{pmatrix} x^2 \\ 2y^2 \\ 3z \end{pmatrix} [/itex] has net out flux of [itex] 4 \pi [/itex] for a unit sphere centred at the origin. If we are now given a vector field [itex] F_1 = \begin{pmatrix} x^2 \\ 2y^2 \\ 3(z+1) \end{pmatrix} [/itex], then how does the net out flux compare if we consider a unit sphere centred at (1,0,0) - is it bigger or smaller?
Relevant Equations:
Gauss' Theorem
Hi,

I just have a quick question about a problem involving Gauss' Theorem.

Question: Vector field [itex] F = \begin{pmatrix} x^2 \\ 2y^2 \\ 3z \end{pmatrix} [/itex] has net out flux of [itex] 4 \pi [/itex] for a unit sphere centred at the origin (calculated in earlier part of question). If we are now given a vector field [itex] F_1 = \begin{pmatrix} x^2 \\ 2y^2 \\ 3(z+1) \end{pmatrix} [/itex], then how does the net out flux compare if we consider a unit sphere centred at (1,0,0) - is it bigger or smaller (than the 4 [itex] \pi [/itex] calculated earlier)?

This is the final part of the question and is worth very few marks, suggesting that I shouldn't need to do any calculations, but I am having trouble arriving at an answer intuitively. The whole question has been about Gauss' Law.

Method:
I am stuck and do not know what else to do beyond the realisation that [itex] \nabla \cdot \vec F_1 = \nabla \cdot \vec F [/itex].

Other initial thoughts that I have include:
- This sphere is completely in the +ve regions of the x-axis
- Some points of interest are on the surface of the new sphere: (0,0,0) and (x,y,z=-1)

I would appreciate any help.

Kind regards
 

Answers and Replies

  • #2
Charles Link
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I believe the y and z terms of ## \nabla \cdot F ## are going to give identical contributions in the volume integrals over the respective spheres. Do you agree?
Look next at the x term. For the sphere centered at the origin, this term is symmetric in the volume integral centered at the origin and it will vanish. How does it behave for the sphere centered at (1,0,0)?
It appears they are just looking for a qualitative answer, with no need to quantify it.
Finally, look again at the z term. It looks like that makes for a positive answer, with both cases the same. In fact that's where the ## 4 \pi ## comes from for the first case.
 
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  • #3
Master1022
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Thank you for your reply.

I believe the y and z terms of ## \nabla \cdot F ## are going to give identical contributions in the volume integrals over the respective spheres. Do you agree?
Look next at the x term. For the sphere centered at the origin, this term is symmetric in the volume integral centered at the origin and it will vanish. How does it behave for the sphere centered at (1,0,0)?

Yes, I agree about the y and z terms. For the x- term of the divergence, it is linear in [itex] x [/itex]. So am I correct in thinking that its contribution will be less for the shifted sphere? I think this might be the case because the 'more positive side' of the shifted sphere will give almost twice its original contribution, but the side nearest the origin will now give a -ve contribution as the divergence will oppose the normal surface vector. Hence the sum of these two contributions will be [itex] < 2 \times arbitrary [/itex] [itex] unit [/itex]

In the original case, the x component of [itex] \nabla \cdot \vec F [/itex] was alligned with the normal vector at both 'ends' of the sphere and thus giving [itex] = 2 \times arbitrary [/itex] [itex] unit [/itex]
 
  • #4
Charles Link
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With the ## \nabla \cdot F ##, you are doing a volume integral, (in order to compute a surface flux integral), and you don't have any direction in the volume integral. The first case, centered at the origin gets a minus for negative x. For the shifted case, the x is always positive. I do not agree with your answer that it will be less.
 
  • #5
Master1022
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With the ## \nabla \cdot F ##, you are doing a volume integral, (in order to compute a surface flux integral), and you don't have any direction in the volume integral. The first case, centered at the origin gets a minus for negative x. For the shifted case, the x is always positive. I do not agree with your answer that it will be less.
Oh yes, I accidentally combined aspects of both integrals. Now I think I understand what you were originally saying - so the shifted version is all in the positive x-region, so there it only has a positive contribution to the volume integral. Hence, the total flux out will be greater.

From a surface integral perspective, could I pursue the following logic:
- There the 'more negative' side of the sphere will always have an opposition to the surface normal vector whether it is in the original or shifted sphere and this opposition will be the same amount.
- Moreover, for the shifted sphere, we have increased the 'more positive' side's contribution in proportion to [itex] x^2 [/itex] and thus the net flux out will be greater
- we have increased positive contribution whilst keeping the negative contribution the same.

Thanks for the help.
 
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  • #6
Charles Link
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this opposition will be the same amount
This opposition (e.g. for the x term for the left side) is the same amount in absolute value by symmetry (as the right side) for the unshifted case. For this sentence I don't agree completely with how you stated it, but I think you understand the concept.
 

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