Question

Given that :
\[\large \color{green}{D = 30 e^{\frac{-\rho}{b}}\hat{a_{\rho}} - 2\frac{z}{b} \hat{a_z}} \; \; \; \; C/m^2\]
in Cylindrical Coordinates, find the \(\color{red}{\textsf{Outward Flux}}\) crossing the right circular cylinder described by \(\color{blue}{\rho = 2b\; , \; z = 0 \; \; and \; \; z = 5b.}\)

Answer 1

\(a_{\rho}\) and \(a_z\) are unit vectors along radial direction and along z axis of Cylinder.

Answer 2

I have to try it on my own.. :)

Answer 3

We can find the Outward Flux as :

\[\psi = \oint D.ds\]

Answer 4

\[\psi = \int\limits\limits_0^{2 \pi} \int\limits\limits_0^{5b}(30 e^{\frac{-\rho}{b}}\hat{a_{\rho}}) \cdot (\rho \cdot d \phi \cdot dz \hat{a_{\rho}})|_{\rho = 2b} - 2\int\limits\limits_0^{2 \pi} \int\limits\limits_0^{2b}(\frac{z}{b} \hat{a_z}) \cdot (\rho \cdot d \rho \cdot d \phi \; \hat{a_z})|_{z = 5b}\]

Answer 5

\[\psi = 30 e^{-2} \cdot(2b) \cdot (5b) \cdot (2 \pi) - 2 \cdot(2 \pi) \cdot (5) \int\limits_0^{2b} \rho d \rho\]
\[\psi = 254.9 b^2 - 62.8 \times \frac{1}{2}[4b^2] \implies \psi = 254.9 b^2 - 125.6b^2 \implies \color{green}{\psi = 129.3b^2}\]

Answer 6

\[\color{blue}{\textsf{Outward Flux} \; (\psi) \approx 129b^2} \; \; \; \; \: \: \: \: \color{green}{\textsf{Hooray...!!}}\]