(Created page with "<math> \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\ha...") |
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\end{aligned} | \end{aligned} | ||
\end{equation*} | \end{equation*} | ||
| + | </math> | ||
| + | <math> | ||
\begin{align*} | \begin{align*} | ||
| − | \text{ | + | \text{{For $\rho<a$}:}& \qquad Q_{enc}=0 \longrightarrow \boxed{\bar{E}=0} \\ |
| − | \text{ | + | \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)\\ | \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}) \\ | &\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}}\\ | & \qquad \boxed{\bar{E}=\frac{\rho_{sa}a}{\epsilon_r\epsilon_0\rho}\hat{\rho}}\\ | ||
| − | \text{ | + | \text{{For $b<\rho<c$}:}&\qquad \rho EC \quad \longrightarrow \quad \boxed{\bar{E}=0}\\ |
| − | \text{ | + | \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}} | &\qquad \boxed{\bar{E}=\frac{\rho_{sa}a}{\epsilon_r\epsilon_0\rho}\hat{\rho}} | ||
\end{align*} | \end{align*} | ||
| + | </math> | ||
Then | Then | ||
| + | |||
| + | <math> | ||
\begin{equation*} | \begin{equation*} | ||
\boxed{ | \boxed{ | ||
Revision as of 21:04, 24 April 2017
$ \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*} $
