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- Thread starter Jim Newt
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In summary, the conversation revolves around solving a problem involving finding the flux through a box with six faces. Jim is struggling with understanding the concept and asks for a more detailed explanation. The other person explains that the flux is calculated using the surface integral, with the normal vector pointing in the direction of the face. They provide examples for the top, bottom, right, and left faces, and suggest Jim try solving for the front and back faces.

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The answer looks right to me, from a quick glance. What are you having problems with?

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Jim

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Well, we have some field that is pointing through the surfaces of the box. So you are asked to find the flux through the entire box. The box obviously has six faces, I'll go through a few of them. . .

It's easier to write [tex]dA=\hat{n}\cdot dA[/tex], more intuitive to me at least . . .

top: the normal vector [tex]\hat{n}[/tex] points in the positive direction, so the flux

is [tex]\int \vec{E} \cdot \hat{n}dA= \int +E_{z} dA = -4\int dA = -36 [/tex]

Bottom: the normal vector [tex]\hat{n}[/tex] to the bottom points in the -z direction, so the flux is is

[tex]\int \vec{E} \cdot \hat{n}dA= \int -E_{z} dA = 4\int dA = +36 [/tex]

Right facing side: the normal vector [tex]\hat{n}[/tex] to the right points in the +j direction, so the flux is

[tex]\int \vec{E} \cdot \hat{n}dA= \int E_{y} dA = \int 2y^2 dA = \int 2y^2 dx dz

=2y^2_{y=3} \int dxdz = 18 * 9 = 162 [/tex]

left facing side: Same stuff here, but y=0 on the left side so the flux=0

I'll let you do the front and back sides. They are similar to the right/left faces. Hope that helps. . .

It's easier to write [tex]dA=\hat{n}\cdot dA[/tex], more intuitive to me at least . . .

top: the normal vector [tex]\hat{n}[/tex] points in the positive direction, so the flux

is [tex]\int \vec{E} \cdot \hat{n}dA= \int +E_{z} dA = -4\int dA = -36 [/tex]

Bottom: the normal vector [tex]\hat{n}[/tex] to the bottom points in the -z direction, so the flux is is

[tex]\int \vec{E} \cdot \hat{n}dA= \int -E_{z} dA = 4\int dA = +36 [/tex]

Right facing side: the normal vector [tex]\hat{n}[/tex] to the right points in the +j direction, so the flux is

[tex]\int \vec{E} \cdot \hat{n}dA= \int E_{y} dA = \int 2y^2 dA = \int 2y^2 dx dz

=2y^2_{y=3} \int dxdz = 18 * 9 = 162 [/tex]

left facing side: Same stuff here, but y=0 on the left side so the flux=0

I'll let you do the front and back sides. They are similar to the right/left faces. Hope that helps. . .

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