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

The electric field components in Fig. are E_(x)=ax^(1//2), E_(y)=E_(z)=0 , in which alpha=800N//C m^(1//2) . Calculate the flux through the cube

The electric field in a region of space is given by E = 5i + 2j N/C . The electric flux due to this field through an area 2m^(2) lying in the YZ plane. In S.I. units is

A cube is arranged such that its length, breadth and height are along X,Y and Z directions. One of its corners is situated at the origin Length of each side of the cube is 25 cm. The components of electric field are E_(x) = 400 sqrt(2) N//C, E_(y) = 0 and E_(z) = 0 respectively. The flux coming out of the cube at one end will be