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.