Flux due to a charge located at the corner of a cube

In summary: Also, the symmetry argument only applies to surfaces which are spherically, cylindrical, or planar. This problem does not have those restrictions, so the symmetry argument does not apply. Additionally, since the electric field is not being integrated, it is not possible to pull out the electric field vector.
  • #1
SmartAries
4
0
Homework Statement
A particle with charge 5.0 micro-coulombs is placed at the corner of a cube. The total electric flux through all sides of the cube is:

A) 0
B) 7.1 * 10^4
C) 9.4 * 10^4
D) 1.4 * 10^5
E) 5.6 * 10^5
Relevant Equations
flux = q / epsilon
The correct answer is B, but I am not sure why.

I have a few confusions regarding this problem. First of all, I had thought that we cannot use Gauss' Law to determine the flux through a SIDE of a cube since Gauss' Law only works for SURFACES. How can we determine how an electric field pierces a 2D-line?

Next, if I interpret the word "sides" as synonymous with "faces," I get answer choice E.

flux = charge / epsilon
flux = (5.0 * 10^-6) / (8.85 * 10^-12) = 5.6 * 10^5 (which is answer choice E)

Additionally, if the charge is located at a corner of the cube, wouldn't that mean that there would only be flux through 3 faces of the cube? I think this because wouldn't the electric field lines from the charge be parallel to the three faces that are adjacent to the corner where the charge is located? If I am correct in this interpretation, then the total flux through all sides of the cube would be calculated using

flux = charge / (2 * epsilon)

because we would divide total flux by 2 since only 3 / 6 faces are being "pierced" by the electric field. If I follow through with this calculation, I do not get answer choice B.

I have attached an image of the original problem as well as a solution that I found online that I am having trouble deciphering. I would be incredibly grateful for any help; I've been stuck on this problem for hours!
 

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  • #2
Gauss Law is applicable on surface which bound volume(This should be clear from derivation). First your solution is wrong since there is no symmetry argument that you can apply to lone cube. This is essential since you can't take out vector ##\vec E## from integration as it behaves differently at each surface. Coming to your doubt, note that you can imagine a big cube with centre as origin and section it in 8 equal pieces along all 3 axis from centre. Your cube is one of those 8 pieces. By placing charge at the centre of your big cube, you have made your system in which charge is placed at corner of all 8 small cubes. From here onward, symmetry arguments are valid and answer comes as you have shown in figure.
 
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  • #3
SmartAries said:
interpret the word "sides" as synonymous with "faces,"
In the context of a solid body, side means face, not edge.
 
  • #4
Abhishek11235 said:
Gauss Law is applicable on surface which bound volume(This should be clear from derivation). First your solution is wrong since there is no symmetry argument that you can apply to lone cube. This is essential since you can't take out vector ##\vec E## from integration as it behaves differently at each surface. Coming to your doubt, note that you can imagine a big cube with centre as origin and section it in 8 equal pieces along all 3 axis from centre. Your cube is one of those 8 pieces. By placing charge at the centre of your big cube, you have made your system in which charge is placed at corner of all 8 small cubes. From here onward, symmetry arguments are valid and answer comes as you have shown in figure.

Thanks for your solution, I understand it now! I have one question regarding symmetry. I apologize as I'm still studying Guass' Law so I hope to not come across as clueless. I thought the symmetry argument only applied to spherical, cylindrical, and planar surfaces. In the case of a large cube centered at the origin, wouldn't we still not be able to apply the symmetry argument as the surface is cubical?Additionally, why was my initial solution incorrect if I was not dealing with the integration of the electric field (which would require symmetry if we wanted to pull E out of the intergral)? I was merely calculating flux by dividing total charge by epsilon.

I understand the correct answer, but I was hoping you could clarify a little bit on the applications of the symmetry argument to this specific homework problem/question.
 
  • #5
SmartAries said:
Thanks for your solution, I understand it now! I have one question regarding symmetry. I apologize as I'm still studying Guass' Law so I hope to not come across as clueless. I thought the symmetry argument only applied to spherical, cylindrical, and planar surfaces. In the case of a large cube centered at the origin, wouldn't we still not be able to apply the symmetry argument as the surface is cubical?Additionally, why was my initial solution incorrect if I was not dealing with the integration of the electric field (which would require symmetry if we wanted to pull E out of the intergral)? I was merely calculating flux by dividing total charge by epsilon.

Gauss law is not a valid tool here simply because there is charge at the boundary (Introduction to Electrodynamics, Griffith). The field is not smooth on the surface. There is discontinuity! Hence, Gauss theorem is repealed and so does your method! By modifying the problem, Gauss law is made true as charge is enclosed within volume!
 
  • #6
SmartAries said:
Thanks for your solution, I understand it now! I have one question regarding symmetry. I apologize as I'm still studying Guass' Law so I hope to not come across as clueless. I thought the symmetry argument only applied to spherical, cylindrical, and planar surfaces. In the case of a large cube centered at the origin, wouldn't we still not be able to apply the symmetry argument as the surface is cubical?Additionally, why was my initial solution incorrect if I was not dealing with the integration of the electric field (which would require symmetry if we wanted to pull E out of the intergral)? I was merely calculating flux by dividing total charge by epsilon.

I understand the correct answer, but I was hoping you could clarify a little bit on the applications of the symmetry argument to this specific homework problem/question.
The flux through a cube of 8 times the size where the charge is at the center would be answer E. Your cube is 1/8 of that (with the other 7 identical). Hence answer B.
I don't think the exercise composer want you to worry like @Abhishek11235 in his/her last post: the faces with the charge in them simply have flux zero.
 
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  • #7
A common trick when using Guass's law is to "force" a symmetrical interpretation when you're given a situation where the charge lies on what you would otherwise be a boundary of the Gaussian surface. As others have suggested, make it so that the charge lies at the center of a larger structure that would make the overall symmetry easy to analyze.

If the charge were to be surrounded by eight identical cubical spaces, then by symmetry the one cube that you're interested in would pass 1/8 the total flux produced by the charge:

1582073455745.png
 

Related to Flux due to a charge located at the corner of a cube

What is flux?

Flux is a measure of the flow of a physical quantity through a given surface. In the case of electric flux, it measures the flow of electric field lines through a surface.

What is a charge located at the corner of a cube?

A charge located at the corner of a cube refers to a point charge that is positioned at one of the eight corners of a cube-shaped object. This charge creates an electric field that extends outward from the corner.

How is flux due to a charge located at the corner of a cube calculated?

Flux due to a charge located at the corner of a cube can be calculated using the formula Φ = q/ε₀, where Φ is the electric flux, q is the magnitude of the charge, and ε₀ is the permittivity of free space.

What is the direction of flux due to a charge located at the corner of a cube?

The direction of flux due to a charge located at the corner of a cube is perpendicular to the surface of the cube and points away from the charge if it is positive, or towards the charge if it is negative.

How does the distance from the charge affect the flux at a point on the surface of the cube?

The flux at a point on the surface of the cube is inversely proportional to the square of the distance from the charge. This means that as the distance increases, the flux decreases, and vice versa.

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