Question 4.8: Given that D = zρ cos^2φ az C/m^2, calculate the charge dens...

\rho _{v}=\nabla\cdot D=\frac{\partial D_{z}}{\partial z}=\rho \cos^{2}\phi

At \left(1,\pi/4,3\right),  \rho_{v}=1\cdot \cos^{2} \left(\pi/4\right)=0.5C/m^{3}. The total charge enclosed by the cylinder can be found in two different ways.

Method 1: This method is based directly on the definition of the total volume charge.

Q=\int_{v}\rho_{v}dv=\int_{v}\rho\cos^{2}\phi\rho d\phi d\rho dz

=\int_{z=-2}^{2}dz\int_{\phi=0}^{2\pi}\cos^{2}\phi d\phi\int_{\rho=0}^{1} \rho^{2} d\rho=4\left(\pi\right)\left(1/3\right)=\frac{4\pi}{3}C

Method 2: Alternatively, we can use Gauss’s law

Q=\Psi=\oint_{S}D\cdot dS=\left[\int_{S}+\int_{t}+\int_{b}\right]D\cdot dS

=\Psi_{s}+ \Psi_{t} +\Psi_{b}

where \Psi_{s}, \Psi_{t}, and \Psi_{b} are the flux through the sides (curved surface), the top surface, and the bottom surface of the cylinder, respectively (see Figure 3.18). Since D does not have component along a_{\rho}, \Psi_{s}=0, for \Psi_{t}, dS=\rho d\phi d\rho a_{z} so

\Psi_{t}=\int_{\rho=0}^{1}\int_{\phi=0}^{2\pi}z\rho \cos^{2}\phi\rho d\phi d\rho |_{z=2}

=2\int_{0}^{1}\rho ^{2}d\rho\int_{0}^{2\pi}\cos^{2}\phi d\phi =2\left(\frac{1}{3}\right)\pi=\frac{2\pi}{3}

and for \Psi_{b},  dS=-\rho d\phi d\rho a_{z} , so

\Psi_{b}=-\int_{\rho=0}^{1}\int_{\phi=0}^{2\pi}z\rho \cos^{2}\phi\rho d\phi d\rho |_{z=-2}

=2\int_{0}^{1}\rho ^{2}d\rho\int_{0}^{2\pi}\cos^{2}\phi d\phi=\frac{2\pi}{3}

Thus

Q=\Psi=0+\frac{2\pi}{3}+\frac{2\pi}{3}=\frac{4\pi}{3}C

as obtained earlier.