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Find the net amount of charged contained in a cube

  • Thread starter mr_coffee
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Hello everyone i'm confused on where I should go next with this problem. It is found experimentally that the electric field in a ceritan region of Earth's atmosphere is directed vertically down. At an altitude of 300m the field has a magnitude 60.0 N/c; at an altitude of 200m, the magnitude is 100 N/c. Find the net amount of charge contained in a cube 100m on edge, with horizontal faces at altitudes of 200 and 300m. I posted a picture of what I think it should look like: http://img331.imageshack.us/img331/143/physicssss8uv.jpg [Broken]
I found the electric flux at 200 and 300m (values are on image): The flux at 300m should be negative. So then I used Eo(Flux) = q_enclosed for both faces and found. At 300m the enclosed charge is 3.186x10^-5, and at 200m I found the enclosed charge to be 5.31x10^-5. But now i'm lost on what i should do. THe answer is 3.54x10^-6 C. Thanks.
 
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mr_coffee said:
I found the electric flux at 200 and 300m (values are on image): The flux at 300m should be negative. So then I used Eo(Flux) = q_enclosed for both faces and found. At 300m the enclosed charge is 3.186x10^-5, and at 200m I found the enclosed charge to be 5.31x10^-5. But now i'm lost on what i should do. THe answer is 3.54x10^-6 C. Thanks.
You need to find the total charge enclosed within the cube. Find the net flux through the cube and apply Gauss's law. (Recheck your calculation for the flux through the top and bottom sides of the cube. What's the area of the sides? What's the field?)
 
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excellent, thanks i got it. For some reason I used Area of cube = 100^2 * 6; Insteed of just 100^2.
 
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but how do you find the net flux? It's equal to q/Eo.. so what do you plug in for q?
 
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flux = EA = Q/ε

You're given the electric field at both surfaces, so you can find E. A is just the area of the two squares.

Then ε is constant, just solve for Q.
 

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