please add procedures ### Find the Flux Through Each SurfaceA closed surface with dimensions \( a = b = 0.400 \, \text{m} \) and \( c = 0.800 \, \text{m} \) is located as shown in the figure. The left edge of the closed surface is located at position \( x = a \). The electric field throughout the region is non-uniform and given by:\[\mathbf{E} = yz^2 \, \mathbf{i} + (xz^2 + 2) \, \mathbf{j} + (2xyz - 1) \, \mathbf{k}\]where \( x, y, z \) are in meters.#### Diagram ExplanationThe diagram depicts a rectangular prism aligned with the coordinate axes (x, y, z). The lengths of the sides aligned with the x and y axes are both \( a = b = 0.400 \, \text{m} \), while the side aligned with the z axis has a length \( c = 0.800 \, \text{m} \).The electric field vector is expressed in three components, relating to the unit vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\):- \( yz^2 \, \mathbf{i} \): Component in the x-direction.- \( (xz^2 + 2) \, \mathbf{j} \): Component in the y-direction.- \( (2xyz - 1) \, \mathbf{k} \): Component in the z-direction. This setup requires calculating the electric flux through each surface of the prism based on the given electric field expression.

Let us consider a closed surface with dimensions as mentioned in question which are written below:- a=0.400 m b=0.400 m c=0.800 m Also, we have given that left edge of the closed surface is located at x=a. Clearly electric field is not uniform throughout the region so it will change at every point.The changing non-uniform electric field is given by the equation as written below:- E=yz2i+(xz2+2)j+(2xyz-1)k--------(1) (Here value of x, y & z are in meter). We can see that electric field is directed along the x-axis according to figure provided. So, j and k components will be zero and we will be calculating net electric field only along i component. Now if we put j=k=0 in (1) we get, E=yz2i----------(2) We can say from (2) electric field will be perpendicular to axis y and axis z or simply it will act along x-axis. so we will be having Enet as net electric field which will be given by E1 and E2 respectively.E1=y=-E(where x=a)-----------------(3) E2=z=+E(where x=a+c)----------------(4) here we are taking y as x=a because it is parallel along front surface of a which is acting along x axis (also seen as height of closed surface). And we are taking z as x=a+c because it is perpendicular along front surface of a which is acting along x axis (also seen as length of closed surface). Enet=E1+E2----------(5) Putting value of E1 & E2 from (3) and (4) in (5) we get, Enet=-E(where x=a)+E(where x=a+c)------(6) putting value of E from (2) in (6) we get, Enet=-yz2i(where x=a)+yz2i(where x=a+c)------------(7) we don't have value of x in above equation so directly putting y=b and z=c in (7) we get, Enet=-bc2+bc2 Enet=0 N/C We clearly know that when electric field is zero for a closed surface than whatever be the area of a closed surface the net flux will be zero. So, the flux through each surface will be zero.