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Finding electric flux through a surface, simple problem, whats wrong?

  • Thread starter mr_coffee
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  • #1
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At each point on the surface of the cube shown in Figure 23-27, the electric field is parallel to the z axis. The length of each edge of the cube is 4.0 m. On the top face of the cube E = -40 k N/C, and on the bottom face of the cube E = +29 k N/C. Determine the net charge contained within the cube.
Image: here

Flux = (29E3 N/C)(4.0)^2;
Flux = 464000 k going up
Flux = (-40E3 N/C)(4.0)^2;
Flux = -640000 k going down
Net Flux = -640000 + 464000 = -176000

I submitted it and was inncorrect,
So i tried added them together and i got 1.104E6, thats the flux, now I found an equation that says, Flux = Qtotal/Eo, so I took Flux*Eo = Qtotal and got 9.7704E-6 and it is still wrong!
 
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Answers and Replies

  • #2
Doc Al
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When finding the flux through a surface, think in terms of inward (-) and outward (+).
mr_coffee said:
Flux = (29E3 N/C)(4.0)^2;
Flux = 464000 k going up
That should be negative, since it points inward.
Flux = (-40E3 N/C)(4.0)^2;
Flux = -640000 k going down
This is also inward, so negative.
Net Flux = -640000 + 464000 = -176000
Redo this, since both are negative.

I submitted it and was inncorrect,
So i tried added them together and i got 1.104E6, thats the flux, now I found an equation that says, Flux = Qtotal/Eo, so I took Flux*Eo = Qtotal and got 9.7704E-6 and it is still wrong!
Signs matter! The flux is negative.
 
  • #3
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So i got -1.104E-6 for the Flux. So Q = -1.104E-6*8.85E-12 = -9.7704E-6? I have only one more chance so i have to make sure i'm doing this one right, thank you!
 
  • #4
Doc Al
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Well... does that follow from what I said? And do you think what I said is accurate? :wink:
 
  • #5
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ahh it was wrong, what did i mess up?
 
  • #6
Doc Al
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mr_coffee said:
ahh it was wrong, what did i mess up?
Nothing that I can see. But those on-line systems can be picky. (Units, significant figures, etc.)
 
  • #7
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yeah i hate this system, i'll ask the professor tomarrow whats up, thanks!
 

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