24
Solution
(a) Since the electric field has only an x component, for faces
perpendicular to x direction, the angle between E and DS is
± p/2 Therefore, the flux  f = E DS is separately zero for each face
of the cube except the two shaded ones Now the magnitude of
the electric field at the left face is
EL = ax1/2 = aa1/2
(x = a at the left face)