Solution details - Revision history

Abdehait at 14:31, 23 September 2014

2014-09-23T14:31:35Z

← Older revision		Revision as of 14:31, 23 September 2014
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	<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> D_L \frac {\partial^2{}}{\partial{z^2}}</math> -
	<math> W \frac {\partial{}}{\partial{z}}</math> ''n'' = 0		<math> W \frac {\partial{}}{\partial{z}}</math> ''n'' = 0

2013-11-12T05:23:33Z

Solution Analysis

← Older revision		Revision as of 05:23, 12 November 2013
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	<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 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>		for <math> r'= \sqrt {\rho'^2 + z^2}</math>

2013-11-12T05:22:44Z

Solution Analysis

← Older revision		Revision as of 05:22, 12 November 2013
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	<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 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>		for <math> r'= \sqrt {\rho'^2 + z^2}</math>

2013-11-03T19:00:00Z

Solution Analysis

← Older revision		Revision as of 19:00, 3 November 2013
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	<math> n_{monopole} = (\frac{\pi}{2\lambda_L })^{1/2} \frac{ e^{-\lambda_L (r'-z)}}{r'} </math>		<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>		for <math> r'= \sqrt {\rho'^2 + z^2}</math>

2013-11-03T18:58:06Z

Solution Analysis

← Older revision		Revision as of 18:58, 3 November 2013
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	<math> n_{monopole} = (\frac{\pi}{2\lambda_L })^{1/2} \frac{ e^{-\lambda_L (r'-z)}}{r'} </math>		<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>

2013-11-03T18:56:43Z

Solution Analysis

← Older revision		Revision as of 18:56, 3 November 2013
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	<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>		for <math> r'= \sqrt {\rho'^2 + z^2}</math>

2013-11-03T18:54:24Z

Solution Analysis

← Older revision		Revision as of 18:54, 3 November 2013
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	<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}		for <math> r'= \sqrt {\rho'^2 + z^2}</math>








	<references/>		<references/>

2013-11-03T18:51:52Z

Solution Analysis

← Older revision		Revision as of 18:51, 3 November 2013
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	<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 {\ro'^2 + z^2}		for <math> r'= \sqrt {\rho'^2 + z^2}
	<references/>		<references/>

2013-11-03T18:50:47Z

Solution Analysis

← Older revision		Revision as of 18:50, 3 November 2013
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	<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{\partial}{\partial z}~~(\frac{\pi}{2\lambda_L })^{1/2} \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 {\ro'^2 + z^2}
	<references/>		<references/>

2013-11-03T18:47:51Z

Solution Analysis

← Older revision		Revision as of 18:47, 3 November 2013
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	<math> n_{monopole} = (\frac{\pi}{2\lambda_L })^{1/2} \frac{ e^{-\lambda_L (r'-z)}}{r'} </math>		<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{\partial}{\partial z}(\frac{\pi}{2\lambda_L })^{1/2} \frac{ e^{-\lambda_L (r'-z)}}{r'} </math>

	<references/>		<references/>