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


by DieCommie
Tags: charge, corner, cube, electric, flux, point
DieCommie
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#1
Sep1-06, 04:09 PM
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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 Im off track all together! Thx for any help
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Andrew Mason
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Sep1-06, 05:00 PM
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Quote Quote by DieCommie
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 Im 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
DieCommie
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#3
Sep1-06, 05:14 PM
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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.

nrqed
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Sep1-06, 05:29 PM
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Find Electric Flux through cube's side, point charge on corner


Quote Quote by DieCommie
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
Andrew Mason
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Sep1-06, 07:50 PM
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Quote Quote by DieCommie
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
DieCommie
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#6
Sep1-06, 10:01 PM
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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.
itheo92
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#7
Jun18-10, 11:22 AM
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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?
Doc Al
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Jun18-10, 11:26 AM
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Quote Quote by itheo92 View Post
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.
sa_max
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#9
Feb4-12, 05:49 PM
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Quote Quote by DieCommie View Post
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.
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.
Doc Al
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Feb4-12, 06:02 PM
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Quote Quote by sa_max View Post
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.
sa_max
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Feb4-12, 06:42 PM
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Quote Quote by Doc Al View Post
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.


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