Compute the charge enclosed by a cube of 2m each edge centered at the origin and with the edges parallel to the axes. Given D = 10y3 /3 j

To compute the charge enclosed by the cube using the given electric flux density ( mathbf{D} = frac{10y^3}{3} hat{mathbf{j}} ) C/m(^2), we need to apply Gauss’s Law in integral form, which relates the electric flux through a closed surface to the charge enclosed by that surface. However, because we’re dealing with the electric flux density (mathbf{D}) directly, we can integrate (mathbf{D}) over the surface of the cube to find the total charge enclosed without explicitly invoking Gauss’s Law.

Given that the cube has its edges parallel to the axes and is centered at the origin with edge length 2 m, it extends from (-1) m to (1) m along the x, y, and z axes.

Since (mathbf{D}) only has a y-component ((frac{10y^3}{3} hat{mathbf{j}})), the flux through the cube will only occur through the faces perpendicular to the y-axis, i.e., the faces at (-1) m and (1) m in the y-direction. The areas of the faces through which (mathbf{D}) passes are parallel to the xz-plane.

The total charge enclosed ((Q_{text{enc}})) by the cube can be obtained by integrating the normal component of (math