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Electric Flux Through A Box

  • Thread starter eMac
  • Start date
  • #1
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Homework Statement


http://i885.photobucket.com/albums/ac52/HummusNFalafel/showmepl.jpg
A empty cardboard box is placed in an electric field pointing upwards with a strength of 1200 N/C as shown above. The bottom of the box is tilted up by an angle Θ = 13° from the horizontal. Side A of the box faces out of the page and is 11 cm high and 16 cm long. Sides B and C are 11 cm high and 11 cm deep. Sides D and E are 16 cm long and 11 cm deep.


b) What is the flux through side B of the box?

Homework Equations



Electric Flux = EAcos(theta)


The Attempt at a Solution



I do not know what angle to use for any of the sides. I tried 90-13 to get 77 but that doesnt work as well. Im not sure if I'm using the correct area as well. Im multiplying base x height but im not sure if that is the correct equation to use. Any help will be much appreciated, thanks in advance.
 

Answers and Replies

  • #2
gneill
Mentor
20,793
2,773
It boils down to determining the cross sectional area of the side in question that is presented to the field. So in this case the field is running directly upwards (in the positive z-axis direction. If you were to look down from above the xy plane (from some height up the z axis), what cross sectional area would side B occupy?
 
  • #3
17
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It boils down to determining the cross sectional area of the side in question that is presented to the field. So in this case the field is running directly upwards (in the positive z-axis direction. If you were to look down from above the xy plane (from some height up the z axis), what cross sectional area would side B occupy?
I'm not sure if I understand, are you referring to the field in 3-D.
 
  • #4
1,137
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find the area vector of side B and use the flux formula
flux = E.S (dot product)
 
  • #5
gneill
Mentor
20,793
2,773
I'm not sure if I understand, are you referring to the field in 3-D.
Certainly. It's a 3D box and you are looking for the flux through one particular side. So what cross sectional area does the side present to the field? Think of it as determining the area of a shadow projection of the side onto the xy plane (since the field is the "light" moving strictly vertically in this case).
 
  • #6
17
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Certainly. It's a 3D box and you are looking for the flux through one particular side. So what cross sectional area does the side present to the field? Think of it as determining the area of a shadow projection of the side onto the xy plane (since the field is the "light" moving strictly vertically in this case).
I'm really not following your explanation. Would the area just be base x height?
 
  • #7
gneill
Mentor
20,793
2,773
I'm really not following your explanation. Would the area just be base x height?
That's the area of the side that you would see if you looked at it straight-on. The field is not "looking" at the side straight-on; It's viewing it from an angle. You want to find the cross sectional area that the side presents to the field.

Picture the electric field as a stream of particles. In the midst of this stream is the side of the box that you're interested in, a rectangle of some area A. The amount of the stream that is intercepted by the side depends upon the angle that the side makes with the direction of flow of the stream. If the side is perpendicular to the stream, presenting its whole face to it, then the area of the stream that's intercepted is A. If the side is at some angle to the flow, then it presents a smaller cross section and proportionately less of the particles of the stream are intercepted.

You want to find the cross sectional area of that the side presents in the direction of the "stream" that is the electric field, and multiply that area by the field strength. This will be the total flux passing through the side.
 
  • #8
17
0
That's the area of the side that you would see if you looked at it straight-on. The field is not "looking" at the side straight-on; It's viewing it from an angle. You want to find the cross sectional area that the side presents to the field.

Picture the electric field as a stream of particles. In the midst of this stream is the side of the box that you're interested in, a rectangle of some area A. The amount of the stream that is intercepted by the side depends upon the angle that the side makes with the direction of flow of the stream. If the side is perpendicular to the stream, presenting its whole face to it, then the area of the stream that's intercepted is A. If the side is at some angle to the flow, then it presents a smaller cross section and proportionately less of the particles of the stream are intercepted.

You want to find the cross sectional area of that the side presents in the direction of the "stream" that is the electric field, and multiply that area by the field strength. This will be the total flux passing through the side.
That's my main problem, I'm not sure what angle the side makes with the stream.
 
  • #9
16
0
That's my main problem, I'm not sure what angle the side makes with the stream.
So you've been given the angle theta which is 13 degrees.

jmUKi.jpg


You'll need to find the cross sectional area which will be given by the distance x multiplied by the height of the box which I believe is given by the question.

The angle that you need to work out the length of x is also in the problem. Look carefully at the diagram and try to picture the box rotating and the angles being made in those two triangles.
 

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