The electric field components are `E_(x)=ax^(-1//2),E_(y)=E_(z)=0` in which a =800 N/C `m^(1//2)` calculate (a) the flux through the cube and (b) the charge within the cube asume that a=0.1 m

Explanation

For any closed surface, all area vectors are to be shown perpendicularly outward.
Here, out of six faces of a cube, each having area `a^2`, electric flux gets associated with faces 1 and 2 only (because remaining four faces are parallel to given electric field along X-axis.
Electric flux linked with face-1,
`phi_(1) = A_(1)E_(1)cos theta_(1) =A_(1) (alphax_(1)^(1//2)) cos theta_(1)`
`=(a^(2)) (alpha a^(1//2))cos (180^(@)) (therefore vecA_(1) || -vecE_(1))`
`therefore phi_(1) = -alpha a^(5//2)`........(1) `(therefore x_(1) =a)`
Electric flux linked with face -2 is:
`phi_(2) = A_(2)E_(2) cos theta_(2)`
`therefore theta_(2) = A_(2)(alphax_(2)^(1//2)) cos theta_(2)`
`phi_(2) = sqrt(2)(alpha a^(5//2))`......... (2)
Net electric flux passing through given cube is,
`phi = phi_(1) + phi_(2)`
`=-alphaa^(5//2) + sqrt(2) alpha a^(5//2)`
`= (800)(0.1)^(5//2) (1.414-1)`
`=((800)(0.414))/(3.162)^(5)`
`phi = 1.048 (Nm^(2))/C`
(Here, `phi` comes out to be positive which indicates that this flux comes out of given cube) Now, if net charge enclosed by given cube is q then according to Gauss.s theorem,
`phi = q/epsilon_(0)`
`therefore q =-phi epsilon_(0) = (1.048) (8.85 xx 10^(-12))`
`therefore q = 9.2748 xx 10^(-12)` C