A cube has sides of length `L`. It is placed with one corner at the origin ( refer to Fig.2.119). The electric field is uniform and given by `vec€ = - B hat (i) + C hat (j) - D hat(k)` , where `B , C , and D` are positive constants.
The total flux passing through the cube is

Options

A. `( B + C + D)L^(2)`
B. `2 ( B + C + D) L^(2)`
C. `6 ( B + C + D) L^(2)`
D. zero

Answer

D

Explanation

Given that `vec(E ) = - B hat (i) + C hat (j) - D hat (k) , Phi = vec ( E). Vec(A)` , edge length `L`, and
`vec(n )_(S_(1)) = - hat (j) rArr Phi_(1) = vec(E ). A hat(n)_(S_(1)) = - CL^(2)`
`vec(n )_(S_(2)) = + hat (k) rArr Phi_(2) = vec(E ). A hat(n)_(S_(2)) = - DL^(2)`
`vec(n )_(S_(3)) = + hat (j) rArr Phi_(3) = vec(E ). A hat(n)_(S_(3)) = + CL^(2)`
`vec(n )_(S_(4)) = - hat (k) rArr Phi_(4) = vec(E ). A hat(n)_(S_(4)) = + DL^(2)`
`vec(n )_(S_(5)) = + hat (i) rArr Phi_(5) = vec(E ). A hat(n)_(S_(5)) = - BL^(2)`
`vec(n )_(S_(6)) = - hat (i) rArr Phi_(6) = vec(E ). A hat(n)_(S_(6)) = + BL^(2)`
Total flux is `Sigma_(i = 1)^(6) Phi_(i) = 0`