Solution details - Revision history

Abdehait at 14:31, 23 September 2014

2014-09-23T14:31:35Z

2013-11-12T05:23:33Z

Solution Analysis

2013-11-12T05:22:44Z

Solution Analysis

2013-11-03T19:00:00Z

Solution Analysis

2013-11-03T18:58:06Z

Solution Analysis

2013-11-03T18:56:43Z

Solution Analysis

2013-11-03T18:54:24Z

Solution Analysis

2013-11-03T18:51:52Z

Solution Analysis

2013-11-03T18:50:47Z

Solution Analysis

2013-11-03T18:47:51Z

Solution Analysis

@@ Line 3: / Line 3: @@
-<math>  (\frac {\partial^2{}}{\partial{x^2}} +\frac {\partial^2{}}{\partial{x^2}})</math>''n'' +
+<math>  D (\frac {\partial^2{}}{\partial{x^2}} +\frac {\partial^2{}}{\partial{x^2}})</math>''n'' +
 <math> D_L \frac {\partial^2{}}{\partial{z^2}}</math> -
 <math> W \frac {\partial{}}{\partial{z}}</math> ''n'' = 0

@@ Line 102: / Line 102: @@
 <math> n_{dipole}  = (\frac{\pi}{2\lambda_L})^{1/2} \frac{z}{r'} (1+\frac{1}{\lambda_L r'})\frac{ e^{-\lambda_L (r'-z)}}{r'} = </math>
-<math> -\frac{\partial}{\partial r'}n_{monopole}  = -(\frac{\pi}{2\lambda_L })^{1/2} \frac{\partial}{\partial z}\frac{ e^{-\lambda_L (r'-z)}}{r'}   </math>
+<math> -\frac{\partial}{\partial r'}n_{monopole}  = -(\frac{\pi}{2\lambda_L })^{1/2} \frac{\partial}{\partial r'}\frac{ e^{-\lambda_L (r'-z)}}{r'}   </math>
 for <math> r'= \sqrt {\rho'^2 + z^2}</math>

@@ Line 102: / Line 102: @@
 <math> n_{dipole}  = (\frac{\pi}{2\lambda_L})^{1/2} \frac{z}{r'} (1+\frac{1}{\lambda_L r'})\frac{ e^{-\lambda_L (r'-z)}}{r'} = </math>
-<math> -\frac{\partial}{\partial z}n_{monopole}  = -(\frac{\pi}{2\lambda_L })^{1/2} \frac{\partial}{\partial z}\frac{ e^{-\lambda_L (r'-z)}}{r'}   </math>
+<math> -\frac{\partial}{\partial r'}n_{monopole}  = -(\frac{\pi}{2\lambda_L })^{1/2} \frac{\partial}{\partial z}\frac{ e^{-\lambda_L (r'-z)}}{r'}   </math>
 for <math> r'= \sqrt {\rho'^2 + z^2}</math>

@@ Line 100: / Line 100: @@
 <math> n_{monopole}  = (\frac{\pi}{2\lambda_L })^{1/2} \frac{ e^{-\lambda_L (r'-z)}}{r'}   </math>
-<math> n_{dipole}  = -(\frac{\pi}{2\lambda_L})^{1/2} \frac{z}{r'} (1+\frac{1}{\lambda_L r'})\frac{ e^{-\lambda_L (r'-z)}}{r'} = </math>
+<math> n_{dipole}  = (\frac{\pi}{2\lambda_L})^{1/2} \frac{z}{r'} (1+\frac{1}{\lambda_L r'})\frac{ e^{-\lambda_L (r'-z)}}{r'} = </math>
-<math> -\frac{\partial}{\partial z}n_{monopole}  = (\frac{\pi}{2\lambda_L })^{1/2} -\frac{\partial}{\partial z}\frac{ e^{-\lambda_L (r'-z)}}{r'}   </math>
+<math> -\frac{\partial}{\partial z}n_{monopole}  = -(\frac{\pi}{2\lambda_L })^{1/2} \frac{\partial}{\partial z}\frac{ e^{-\lambda_L (r'-z)}}{r'}   </math>
 for <math> r'= \sqrt {\rho'^2 + z^2}</math>

@@ Line 100: / Line 100: @@
 <math> n_{monopole}  = (\frac{\pi}{2\lambda_L })^{1/2} \frac{ e^{-\lambda_L (r'-z)}}{r'}   </math>
-<math> n_{dipole}  = (\frac{\pi}{2\lambda_L})^{1/2} \frac{z}{r'} (1+\frac{1}{\lambda_L r'})\frac{ e^{-\lambda_L (r'-z)}}{r'} = </math>
+<math> n_{dipole}  = -(\frac{\pi}{2\lambda_L})^{1/2} \frac{z}{r'} (1+\frac{1}{\lambda_L r'})\frac{ e^{-\lambda_L (r'-z)}}{r'} = </math>
 <math> -\frac{\partial}{\partial z}n_{monopole}  = (\frac{\pi}{2\lambda_L })^{1/2} -\frac{\partial}{\partial z}\frac{ e^{-\lambda_L (r'-z)}}{r'}   </math>

@@ Line 104: / Line 104: @@
 <math> \frac{\partial}{\partial z}n_{monopole}  = (\frac{\pi}{2\lambda_L })^{1/2} \frac{\partial}{\partial z}\frac{ e^{-\lambda_L (r'-z)}}{r'}   </math>
-for <math> r'= \sqrt {\rho'^2 + z^2}
+for <math> r'= \sqrt {\rho'^2 + z^2}</math>
 <references/>