What is the flux through (a) each cube face

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Homework Help Overview

The problem involves calculating the electric flux through the faces of a cube when a point charge is placed at one of its corners. The subject area pertains to electrostatics and the application of Gauss's law.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to model the electric field due to a point charge and evaluate the flux through the cube faces. Some participants question the validity of the initial expression for the electric field and suggest using symmetry arguments and Gauss's law instead. There is also discussion about the total flux and how it relates to the flux through individual faces of the cube.

Discussion Status

Participants have explored various interpretations of the problem, with some suggesting that symmetry can simplify the calculations. There is a recognition of the total flux through the cube and how it can be divided among the faces, although there is no explicit consensus on the final answer.

Contextual Notes

There is mention of the original poster's confusion regarding the correct approach and the need for clarification on the assumptions made about the charge's position and the resulting flux calculations.

danago
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A point particle with charge q is placed at a corner of a cube of edge a . What is the flux through (a) each cube face forming that corner and (b) each of the other cube faces?

I tried modelling the problem by positioning the charge such that it is at point (0,0,0) in the x-y-z space. The cube then extends out in the positive x, y and z directions.

I then defined the unit vectors i, j and k to extend in the x, y and z directions respectively. Now I am not sure if my next step is valid, but it seemed to be a decent thing to do. I said that the electric field due to the charge is given by:

[tex] \overrightarrow E = \frac{{kq}}{{x^2 }}\widehat{\underline i } + \frac{{kq}}{{y^2 }}\widehat{\underline j } + \frac{{kq}}{{z^2 }}\widehat{\underline k }[/tex]

I then evaluated the integral of the dot product of vectors E and dA over one of the cubes faces, and came up with the expression for flux as kq, however this wasnt the correct result.

Anyone able to help with the question?

Thanks,
Dan.
 
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That expression for the field strength is not correct. The field points radially outward from the point charge; its magnitude is kq/r^2.

In any case, you don't need that at all. Instead, put on your thinking cap. :wink:

Picture the field emanating uniformly in all directions. Using symmetry arguments and Gauss's law (and the meaning of flux) will tell you the flux without the need for any calculation.
 
Sorry forgot to mention, i got part (a) correct as 0.

It was part (b) i was having trouble with. Is (b) still solvable using symmetry arguments?
 
danago said:
Sorry forgot to mention, i got part (a) correct as 0.
Good.
It was part (b) i was having trouble with. Is (b) still solvable using symmetry arguments?
Sure!

Hint: What's the total flux through all sides of the cube? Which side has the most flux? :wink:
 
Doc Al said:
Good.

Sure!

Hint: What's the total flux through all sides of the cube? Which side has the most flux? :wink:

Will the total flux be [tex]\Phi=q / \epsilon_0[/tex]? And will the flux through each of the other faces be the same, therefore making the flux through each face just be [tex]\Phi=q / (3\epsilon_0)[/tex]?
 
Last edited:
You got it!
 
Thanks very much :smile: That made it much easier than whatever i was trying to do originally haha.
 
Hmm i tried inputting that answer, but it isn't showing up as correct. Apparently the answer is supposed to be [tex]\Phi=q / (24\epsilon_0)[/tex]. Any idea what's going on there?
 
Yes. Big whoops! (:redface:) [itex]\Phi=q / \epsilon_0[/itex] is the total flux in all directions. How much goes through that cube?
 
  • #10
Doc Al said:
Yes. Big whoops! (:redface:) [itex]\Phi=q / \epsilon_0[/itex] is the total flux in all directions. How much goes through that cube?

Oh i think i see what's going on here. I can picture the charge as being in the center of 8 smaller cubes, so the flux through just the cube will be 1/8th of the total flux; and then dividing by 3 gives the correct answer :smile:

Thanks again Doc Al!
 
  • #11
Exactly. (Sorry about not catching that the first time around. Asleep at the wheel.)
 

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