Difference between revisions of "ECE PhD QE FO 2012 Problem1.1" - Rhea

Latest revision as of 22:24, 25 April 2017

$ (1) \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{{For $\rho<a$}:}& \qquad Q_{enc}=0 \longrightarrow \boxed{\bar{E}=0} \\ \text{{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{{For $b<\rho<c$}:}&\qquad \rho EC \quad \longrightarrow \quad \boxed{\bar{E}=0}\\ \text{{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*} $

(2)

$ \begin{align*} C&=\frac{Q}{V} & V_2-V_1&=-\int_1^2\bar{E}\cdot d\bar{l} & Q&=\int_0^1\int_0^{2\pi}\rho_{sa}(ad\phi dz) \\ & & &=-\int_b^a\frac{\rho_{sa}a}{\epsilon_r\epsilon_0\rho}d\rho & &=\rho_{sa}(2\pi a)\\ & & &=\frac{\rho_{sa}a}{\epsilon_r\epsilon_0}ln(\frac{b}{a}) & & \end{align*} $ $ \begin{equation*} C=\frac{\bcancel{\rho_{sa}}(2\pi\bcancel{a})}{\frac{\bcancel{\rho_{sa}}\bcancel{a}}{\epsilon_r\epsilon_0}ln(\frac{b}{a})} =\boxed{\frac{2\pi \epsilon_r\epsilon_0}{ln(\frac{b}{a})} \left(\frac{F}{m}\right)} \end{equation*} $

(3)

$ \begin{equation*} \nabla \times \bar{H}=\bar{J} \longrightarrow \oint \bar{H}\cdot d\bar{l}=\int_S\bar{J}\cdot d\bar{S}=I_{enc} \end{equation*} $ $ \begin{align*} \int_0^{2\pi}H_{\phi}(\rho d\phi)&=I &\text{(assuming $I$ is out of the page)}\\ H_{\phi}(2\pi \rho)&= I &\\ \end{align*} $ $ \begin{equation*} \boxed{\bar{H}=\frac{I}{2\pi\rho}\hat{\phi}} \left(\frac{A}{m}\right) \qquad \qquad a<\rho<b \end{equation*} $

(4)

$ \begin{align*} L=\frac{\Phi}{I} \qquad \text{\underline{where}:}& & \Phi&=\int \bar{H}\cdot d\bar{S} & dS_{\phi}&=(drdz)\hat{\phi}\\ & & \Phi&=\int_0^1\int_a^b\frac{\mu_0 I}{2\pi\rho}dr dz & &\\ & & &=\frac{\mu_0 I}{2\pi}ln(\frac{b}{a}) & &\\ L=\frac{\frac{\mu_0 I}{2\pi}ln(\frac{b}{a})}{I}& & & \boxed{L={\mu_0 ln(\frac{b}{a})}{2\pi}}\left(\frac{H}{m}\right) \end{align*} $

(5)

$ \begin{align*} G&=\frac{1}{R} & R&=\frac{V}{I}=\frac{-\int_1^2\bar{E}\cdot d\bar{l}}{\int_S\bar{J}\cdot d\bar{S}} \end{align*} $ $ \begin{equation*} \bar{J}=\sigma\bar{E}=\left(\frac{\sigma\rho_{sa}a}{\epsilon_r\epsilon_0\rho}\right)\hat{\rho} \end{equation*} $ $ \begin{align*} I&=\int_0^1\int_0^{2\pi}\left(\frac{\sigma\rho_{sa}a}{\epsilon_r\epsilon_0\bcancel{\rho}}\right)\bcancel{\rho}d\phi dz & V&=\frac{\rho_{sa}a}{\epsilon_r\epsilon_0}ln(\frac{b}{a})\\ &=\frac{2\pi\sigma\rho_{sa}a}{\epsilon_r\epsilon_0} & & \end{align*} $ $ \begin{equation} G=\frac{I}{V}=\frac{\frac{2\pi\sigma\bcancel{\rho_{sa}}\bcancel{a}}{\bcancel{\epsilon_r\epsilon_0}} } {\frac{\bcancel{\rho_{sa}}\bcancel{a}ln(\frac{b}{a})}{\bcancel{\epsilon_r\epsilon_0}}} = \boxed{\frac{2\pi\sigma}{ln(\frac{b}{a})}} \left(\frac{S}{m}\right) \end{equation} $