Compute the Gauss law for D= 10ρ3 /4 i, in cylindrical coordinates with ρ= 4m, z=0 and z=5

To compute Gauss’s law using the given electric flux density, (D = frac{10rho^3}{4} hat{i}) in cylindrical coordinates, we follow the integral form of Gauss’s law for electric fields, which states:

[

oint_S mathbf{D} cdot dmathbf{A} = Q_{text{enc}}

]

Where:

– (oint_S mathbf{D} cdot dmathbf{A}) is the electric flux through a closed surface (S),

– (Q_{text{enc}}) is the total electric charge enclosed by the surface.

Given that (D) is in the direction of (hat{i}), which implies it’s purely radial in cylindrical coordinates, this simplifies the problem since we only need to consider the component of (mathbf{D}) that is normal to the surface (S). For a cylindrical surface with radius (rho) and height from (z=0) to (z=5), the relevant surface area for the flux calculation is the curved surface area, because the electric flux density is radially outward and will not penetrate the top and bottom surfaces normally.

The curved surface area of a cylinder is (A = 2pirho h), where (rho) is the radius of the cylinder and (h) is the height. For (rho = 4