
#1
Jan2508, 11:36 PM

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A cube of side l55cm is placed in a uniform field E= 7.30 X10^3 N/C with edges parallel to the field lines.
a. what is the net flux through the cube? b. what is the flux through each of its six faces? Face 1 perpendicular to E (field lines enter the cube) Face 2 perpendicular to E (field lines exit the cube) Face 3 parallel to E Face 4 parallel to E Face 5 parallel to E Face 6 parallel to E Now...it might just be throwing me off...but I THOUGHT (which i could totally be wrong) that the parallel sides will have a flux value of zero because E is not entering or exiting those areas. However..I could be completely wrong because it seems I am teaching this problem to myself at the moment...but I am still lost on how to find the net flux and when it enters and exits... OH...I have an equation that the electric flux = EA cos (theta) I substituted the values in...figuring area as 3025 cm^2 and then converting to m^2 and subbing everything in the equation but I am doing something completely wrong.... HELP Please! 



#2
Jan2608, 05:52 AM

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You are correct in thinking faces parallel to the field will have zero flux. Remember that these things are vectors as well. The convention is to take outward flux as positive and inward flux as negative.




#3
Jan2608, 03:08 PM

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I took the part of the question "with edges parallel to the field lines" to mean that the angle theta would be 0 degrees... I took the 55cm and figured the area to be 3025 cm^2 and then converted to .3025 m^2 and E is given. EAcos[tex]\Theta[/tex] = (7.30 x 10^3 N/C)(.3025m^2)(cos 0) = 2208.5 which would enter the cube so it would be positive... yes??? 



#4
Jan2608, 03:41 PM

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[SOLVED] Help needed with Gauss's Law and Net Flux
Yes that is the formula one would use. More generally it is:
[tex]\oint_{S} \mathbf{E} \cdot d\mathbf{S}[/tex] If you read what I said before if the answer is positive then the flux is leaving the cube. There was a reason why I said you must consider this vectorially. 



#5
Jan2608, 03:52 PM

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I have read and read and I cant seem to figure out anymore than I have... 



#6
Jan2608, 03:57 PM

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You have one side where the field lines exit the cube now what about the side where they enter?




#7
Jan2608, 03:58 PM

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It seems to me you're making this more difficult than it is  you've already said everything you need to get the answer.
Can you tell us exactly how you're seeing the angle theta, first for face 1, then for face 2, then for faces 3  6? I suspect you're off by 90 degrees. 



#8
Jan2608, 04:07 PM

P: 11

for face 3 through 6 i didnt use any angle because the electric flux entered through one side and exited through another so only face 1 and face 2 (both perpendiculer to E) would have a value... so for the angle for face 1 would be 180 and for face 2 would be zero... so flux through face 1 would be (7.2 x 10^3)(.3025m^2)(cos180) = 2208.25 N m^2/C and then flux through face 2 would be (7.2 x 10^3)(.3025m^2)(cos0) = 2208.25 N m^2/C 



#9
Jan2608, 04:16 PM

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#10
Jan2608, 04:37 PM

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#11
Jan2608, 04:50 PM

P: 662

Bingo!
And by the way  you're right that you can neglect any contribution from sides 3  6, but even if you didn't, the math still works. In those cases, the angle is 90 deg (remember, it's the angle between the field vector and the normal to the surface), so the cos gives you zero, as you'd expect. 



#12
Jan2608, 04:56 PM

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