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A cube, placed in a region of electric field, is oriented such that one of its corners is at the origin and its edges are parallel to the x, y, and z axes as shown to the left. The length of each of its edges is 1.3 m.

(a) What is the flux through the cube if the electric field is given by E = (3.00y)j ?

Φ = ? Nm^2/C

(b) What is the flux through the cube if the electric field is given by (-4.00)i + (6.00+3.00y)j ?

Φ= ? Nm^2/C

(c) For part (b), how much charge must be enclosed by the cube?

q = ? C

Φ = int(E * dA)

I get the feeling this should be really easy but so far its not. So for part a the field is going through the top and bottom surfaces and the formula I believe should just shrink down from E dot dA the just E*A so if we do it for just one surface it should be (3e-6)(1.3^2) = 5.07e-6 but since it goes through 2 faces parralel with the area vectors summing those two should make it equal to zero correct?

For part b it looks like the electric field vector doesn't even pass through the cube (4 to the left and 9 up from the origin?) So shouldn't the answer be zero?

For part c I assume that when I do get the right answer I can just set it equal to Qencl/ε and solve for Q right?

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# Homework Help: Flux Through a Cube

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