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$ \begin{equation*} \nabla\cdot\bar{D}=\rho \longrightarrow \oint\bar{D}\cdot d\bar{S}=\int\rho dV=Q_{enc}, \qquad \begin{aligned} d\bar{l}&=d\rho\hat{\rho}+\rho d\phi\hat{\phi}+dz\hat{z}\\ d\bar{S}_{\rho}&=\rho d\phi dz\hat{\rho} \end{aligned} \end{equation*} \begin{align*} \text{\underline{For $\rho<a$}:}& \qquad Q_{enc}=0 \longrightarrow \boxed{\bar{E}=0} \\ \text{\underline{For $a<\rho<b$}:}& \qquad %\begin{aligned} \int_0^L\int_0^{2\pi}(\epsilon_r\epsilon_0E_{\rho})\rho d\phi dz=\int_0^L\int_0^{2\pi}\rho_{sa}(ad\phi dz)\\ &\qquad \qquad \quad \, (\epsilon_r\epsilon_0E_{\rho})(\bcancel{2\pi}\rho)\bcancel{L}=\rho_{sa}(\bcancel{2\pi}a)\bcancel{L}) \\ %\end{aligned}\\ & \qquad \boxed{\bar{E}=\frac{\rho_{sa}a}{\epsilon_r\epsilon_0\rho}\hat{\rho}}\\ \text{\underline{For $b<\rho<c$}:}&\qquad \rho EC \quad \longrightarrow \quad \boxed{\bar{E}=0}\\ \text{\underline{For $c<\rho$}:}& \qquad \int_0^L\int_0^{2\pi}(\epsilon_r\epsilon_0E_{\rho})\rho d\phi dz=\rho_{sa}(2\pi a)L \\ &\qquad \boxed{\bar{E}=\frac{\rho_{sa}a}{\epsilon_r\epsilon_0\rho}\hat{\rho}} \end{align*} Then \begin{equation*} \boxed{ \bar{E}=\begin{cases} 0&\rho<a\\ \frac{\rho_{sa}a}{\epsilon_r\epsilon_0\rho}\hat{\rho}&a<\rho<b\\ 0&b<\rho<c\\ \frac{\rho_{sa}a}{\epsilon_r\epsilon_0\rho}\hat{\rho}&c<\rho \end{cases}} \left(\frac{V}{m}\right) \end{equation*} $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang