# Solving Problem: Help Needed from Experts

• Jim Newt
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.

#### Jim Newt

Hi all,

I hope you can see the attachment. Could someone give me a more detailed explanantion of solving this problem? Any help would be greatly appreciated. Thanks,

Jim

#### Attachments

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

I'm having problems with the whole concept. For instance, why does the "right side" have a value of 18 dot 9, and the bottom have a value of -4 dot -9? Thanks,

Jim

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 $$dA=\hat{n}\cdot dA$$, more intuitive to me at least . . .

top: the normal vector $$\hat{n}$$ points in the positive direction, so the flux
is $$\int \vec{E} \cdot \hat{n}dA= \int +E_{z} dA = -4\int dA = -36$$

Bottom: the normal vector $$\hat{n}$$ to the bottom points in the -z direction, so the flux is is
$$\int \vec{E} \cdot \hat{n}dA= \int -E_{z} dA = 4\int dA = +36$$

Right facing side: the normal vector $$\hat{n}$$ to the right points in the +j direction, so the flux is
$$\int \vec{E} \cdot \hat{n}dA= \int E_{y} dA = \int 2y^2 dA = \int 2y^2 dx dz =2y^2_{y=3} \int dxdz = 18 * 9 = 162$$

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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