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Electric Flux through a Cube

  • Thread starter 123yt
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  • #1
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Homework Statement


A surface in the shape of a cube with edge lengths a = 2.7 m is oriented in an x-y-z coordinate system so that the top and bottom faces are parallel to the x-y plane, the left and right faces are parallel to the x-z plane, and the front and back faces are parallel to the y-z plane (see diagram), with the center of the cube at the origin. A charge Q = -0.3 nC is located at the origin.

Now imagine the same charge moves to position (0 m, -2.7 m,0 m).

Calculate the electric flux through the left face of the cube.

[URL]http://homework.phyast.pitt.edu/res/pitt/PHYS0175/Python%20Assignments/PHYS%200175%20Lab%203%20Figure%201.JPG[/URL]

Homework Equations


Flux = Q/e0
E = G*q / r^2


The Attempt at a Solution


Tried integrating the electric field and area, but got two variables that needed to be integrated.
 
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Answers and Replies

  • #2
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please can you provide an image?
 
  • #3
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The flux formula that you have listed is the total flux throughout a given shape. Since only the perpendicular component to the counts towards the total flux.
The shape or size of the object does not change the amount of flux given the same charge enclosed.
Drawing a picture would help. And you shouldn't need the second formula. Just a hint you can imagine the cube in question as one of six cubes and divide the total flux equally among all perpendicular faces.
 
  • #4
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If the charge was just inside the cube, I'd just take the charge, divide by e0, and divide that by 6.

The problem is that the charge is outside the cube, meaning that the cube itself doesn't enclose any charge. The net flux is 0, but I have to find the flux through a single face.
 
  • #5
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The reason I suggest drawing a picture is because it sounds like the charge is moved to a position that is parallel to the left side so the flux through this side should be zero but I wasn't sure if I was envisioning it correctly...
The idea that you are taking 123yt only works with the charge being placed at the center of the cube because the perpendicular component is the same for all faces. That is why I suggest constructing a shape that allows for the charge to be in the center again.
 
  • #6
SammyS
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If the charge was just inside the cube, I'd just take the charge, divide by e0, and divide that by 6.
Yes, this IS correct.
The problem is that the charge is outside the cube, meaning that the cube itself doesn't enclose any charge. The net flux is 0, but I have to find the flux through a single face.
You are only asked to find the flux through one side of the cube!

Yes, it's true that the flux through the entire cube is zero.
 
  • #7
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So you mean draw a Gaussian cube around the charge, with one of the cube sides replaced by the side that I'm trying to find the flux of?

Hmm, I just realized something incredibly obvious. If I already have the flux of one of the cube faces while the charge was in the center, and then the charge moved to the opposite side but otherwise still the exact same position from the face, then the flux would just be the reverse of the flux through the face while it was inside the cube.

Thanks, I got it figured out now.
 

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