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# A cube is located with one corner at the origin of anx, y, z, coordinate system. One of the cube'sfaces lies in the x, y plane, another in they, z pla

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A cube is located with one corner at the origin of anx, y, z, coordinate system. One of the cube'sfaces lies in the x, y plane, another in they, z plane, and another in the x, zplane. In other words, the cube is in the first octant of thecoordinate system. The edges of the cube are 0.18 m long. A uniform electric field is parallelto the x, y plane and points in the direction of the+y axis. The magnitude of the field is 1500 N/C. Find the electric flux through each ofthe six faces of the cube. bottom face
Nm2/C
top face
Nm2/C
each of the other four faces
Nm2/C

2020-12-01

The electric flux through each face of the cube is given by $$\displaystyle\phi_{{{E}}}={\left({E}{\cos{{\left(\theta\right)}}}\right)}{A}$$
where E is the magnitude of the electricfeild at the face, A is the area of the face, and Ø is theangle between the electric field and the outward nornal of theface. We can use this expression to calculate the electric flux
$$\displaystyle\theta_{{{E}}}$$
through each of the six faces of the cube. a) on the bottom face of the cube, the outward normal pointsparallel to the -y axis, in the opposite direction to the electricfield, and $$\displaystyle\theta={180}$$
Therefore,
$$\displaystyle{\left(\phi_{{{E}}}\right)}={\left({1500}{\frac{{{N}}}{{{C}}}}\right)}{\left({\cos{{180}}}\right)}{\left({.2}^{{{2}}}\right)}=-{6.0}\times{10}^{{{1}}}{N}{)}\times\frac{{m}^{{{2}}}}{{C}}$$ (in your case instead ofthe .2 use your .18m i used the number from the book.
On the top face of the cube, the outward normal pointsparallel to the +y axis, and the flux=0.0°. The electric fluxis, therefore,
$$\theta _{E}=(1500)(cos(0)(.2)^{2}=+6.0\times 10^{1} N•m²/C$$
On each of the other four faces, the outward normals areperpendicular to the direction of the electric field, so
$$\displaystyle\phi\ne{q}{90}$$
so for each of the four sides faces,
$$(\theta _{s}=(1500)(cos(90)(.2)^{2}= 0N•m²/C$$
b) The total flux through the cube is:
$$\displaystyle{\left(\theta_{{{r}}}\right)}=\theta_{{{B}}}+\theta_{{{r}}}+\theta_{{{S}{1}}}\theta_{{{S}{2}}}+\theta_{{{S}{3}}}+\theta_{{{S}{4}}}$$
you should end up with a final answer of 0N•m²/C . Because each side is 0, and the bottom isa negative and the top is a postive of each amounts of flux. sothey cancel.