Find the flux of the electric field through each of the five surfaces of the inclined plane as shown in figure. What is the total flux through the entire closed surface?

Explanation

Note that flux through the face ABF, CDE, and BCEF is zero. Area vector of face ABF points in the positive z- directions area vector of CDF points in the negative z-direction, and area vector of BCEF points in the negative y- direction. In all the three cases, field E is normal to area vector.
Flux through face ABCD
Magnitude of area vector of face ABCD = ab
`phi_(E) = vec E * vecA = E(ab)cos(90^@-theta) = Eab sin theta`

Flux through face ADEF :
Magnitude of area vector of face
ADFE = `(a sin theta)b = ab sintheta`
`phi_E = E cos 180^@'(ab sin theta) = -Eab sin theta`
Thus, we obtain
`(phi_E)_(ABF) = 0, (phi_E)_(CDE) = 0, (phi_E)_(BCEF) = 0`
`(phi_E)_(ABCD) = +E ab sin theta and (phi_E)_(ADEF) = -Eabsintheta. `
Flux is a scalar quantity, therefore the total flux is algebraic
sum of flux through each surface.
`phi_(t otal) = (phi_E)_(ABF) + (phi_E)_(CDE) + (phi_E)_(BCEF)`
`+(phi_E)_(ABCD) + (phi_E)_(ADEF)`
`= 0+0+0+Eabsin theta - E ab sin theta = 0` .