Find Electric Flux through cube's side, point charge on corner

In summary, the conversation discusses finding the electric flux through the top side of a cube with one corner at the origin and a charge located at the origin. The solution involves using a double integral, with one integration in the y direction (theta) and one in the x direction (phi). The component of the electric field perpendicular to the surface is used to set up the integral, and the second integral is set up using Gauss' law and symmetry. The total flux through the cube is found to be 1/24th of the total flux through a gaussian sphere centered on the origin. There is also a reference to using the symmetry method instead of integrals to solve the problem.
  • #1
DieCommie
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0
My problem is this,

Find the electrix flux through the top side of a cube. The cube's corner is on the origin, and is 'a' units on length. The charge 'q' is located at the origin, with the corner of the cube.

I am thinking I need a double integral. One to swipe the box in the y direction (theta), and one to swipe in the x direction (phi).

So, I know the component of the electric field perpendicular to the surface is (k*q*cos(theta))/(a*cos*theta)^2 . So I think I need to integrate this from 0 to (pi/4), then integrate again.

But how to set up the second integral is where I am confused...

Or maybe I am off track all together! Thx for any help
 
Last edited:
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  • #2
DieCommie said:
My problem is this,

Find the electrix flux through the top side of a cube. The cube is centerd on the origin, and is 'a' units on length. The charge 'q' is located at the origin.

I am thinking I need a double integral. One to swipe the box in the y direction (theta), and one to swipe in the x direction (phi).

So, I know the component of the electric field perpendicular to the surface is (k*q*cos(theta))/(a*cos*theta)^2 . So I think I need to integrate this from 0 to (pi/4), then integrate again.

But how to set up the second integral is where I am confused...

Or maybe I am off track all together! Thx for any help
Use Gauss' law and symmetry to show that the flux through one side of the cube is 1/6th of the total flux through a gaussian sphere centred on the origin.

AM
 
  • #3
Thx for the reply.

Im sorry, I mis stated my problem I will edit it.

The cube is not centered on the origin, its corner is at the origin with the charge.

So the flux through one side will not necessarly be 1/6th of the flux.
 
  • #4
DieCommie said:
Thx for the reply.

Im sorry, I mis stated my problem I will edit it.

The cube is not centered on the origin, its corner is at the origin with the charge.

So the flux through one side will not necessarly be 1/6th of the flux.
It's not 1/6 but you still can do it by symmetry. Hint: if you would surround the charge entirely by cubes, how many would you have to put there (so that the charge is completely surrounded)? By symmetry, the flux in each of those cubes will be the same and equal to the total flux over the number of cubes.

It would be quite a pain to do with an integral
 
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  • #5
DieCommie said:
Thx for the reply.

Im sorry, I mis stated my problem I will edit it.

The cube is not centered on the origin, its corner is at the origin with the charge.

So the flux through one side will not necessarly be 1/6th of the flux.
You have to find the solid angle subtended by the cube face. Use Nrged's suggestion. The three faces that intersect the charge have no flux since their areas are perpendicular to the field. ([itex]E\cdot dA = 0[/itex]). It takes 8 cubes to cover a complete sphere centred at the origin of radius a, so each cube subtends 1/8th of the sphere. So each surface has 1/24th of the total flux through a gaussian sphere centred on the origin.

AM
 
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  • #6
I see the solution perfectaly now. Thank you all, you answered my question well.


I did infact get the double integral too, wow what a mess. But getting the double integral was satisfying. :biggrin:
 
  • #7
It takes 8 cubes to cover a complete sphere centred at the origin of radius a, so each cube subtends 1/8th of the sphere

can u please explain it a bit. I can't understand this thing. How would it take 8 cubes to cover a complete sphere?
 
  • #8
itheo92 said:
can u please explain it a bit. I can't understand this thing. How would it take 8 cubes to cover a complete sphere?
Note that the cubes are arranged so that their corners are at the origin.
 
  • #9
DieCommie said:
I see the solution perfectaly now. Thank you all, you answered my question well.


I did infact get the double integral too, wow what a mess. But getting the double integral was satisfying. :biggrin:

I can do this using the gauss's law and the symmetry. Can you explain how did you do this using the integrals. I can't seems to get the correct answer when I'm using the integrals.
 
  • #10
sa_max said:
I can do this using the gauss's law and the symmetry. Can you explain how did you do this using the integrals. I can't seems to get the correct answer when I'm using the integrals.
Realize you are replying to a post made over 5 years ago.
 
  • #11
Doc Al said:
Realize you are replying to a post made over 5 years ago.

Yes I know. I was curious, and can't figure out what I'm doing wrong.I couldn't find any online resource that explains this problem using integrals. so it was worth a try.
 

1. What is electric flux?

Electric flux is a measure of the amount of electric field passing through a given surface. It is represented by the symbol Φ and is measured in units of volt-meters (V*m).

2. How is electric flux calculated?

Electric flux can be calculated by multiplying the strength of the electric field by the area of the surface the field passes through. The formula for electric flux is Φ = E * A * cos(θ), where E is the electric field strength, A is the area of the surface, and θ is the angle between the electric field and the normal vector of the surface.

3. What is a point charge?

A point charge is a hypothetical electric charge that occupies a single point in space. It is often used in physics calculations as a simplified model for an object with a finite size and charge distribution.

4. How can the electric flux through a cube's side be found?

To find the electric flux through a cube's side, we can use the formula Φ = Q/ε0, where Q is the charge of the point charge on the corner of the cube and ε0 is the permittivity of free space. This formula assumes that the cube's side is perpendicular to the electric field created by the point charge.

5. Why is it important to calculate electric flux?

Electric flux is an important concept in understanding the behavior of electric fields. It allows us to quantify the amount of electric field passing through a given surface, which can help us predict the movement of charges and the behavior of electric circuits. Additionally, electric flux is directly related to the strength of the electric field, which is a fundamental property of electric forces.

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