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| \left ( \frac{q}{k} \right )^2 d^{\prime 2} | | \left ( \frac{q}{k} \right )^2 d^{\prime 2} |
| & = \begin{alignat}{2} [ | | & = \begin{alignat}{2} [ |
− | & x^2 \cos^2 \phi_g - x \cos \phi_g \sin \phi_g ( v_{2y} ) + \sin^2 \phi_g ( v_{2y} )^2 \\ | + | & x^2 + ( v_{2y} )^2 \\ |
− | & + x^2 \sin^2 \phi_g + x \sin \phi_g \cos \phi_g ( v_{2y} ) - d^{\prime} x \sin \phi_g \\ | + | & - 2 d^{\prime} x \sin \phi_g \\ |
− | & + x \sin \phi_g \cos \phi_g ( v_{2y} ) + \cos^2 \phi_g ( v_{2y} )^2 - d^{\prime} \cos \phi_g ( v_{2y} ) \\ | + | & + x \sin \phi_g \cos \phi_g ( v_{2y} ) - d^{\prime} \cos \phi_g ( v_{2y} ) \\ |
− | & - d^{\prime} x \sin \phi_g - d^{\prime} \cos \phi_g ( v_{2y} ) + d^{\prime 2} \\ | + | & - d^{\prime} \cos \phi_g ( v_{2y} ) + d^{\prime 2} \\ |
| & + d^2 \sin^2 \theta_g + d z \sin \theta_g \cos \theta_g + z^2 \cos^2 \theta_g ] \end{alignat} \\ | | & + d^2 \sin^2 \theta_g + d z \sin \theta_g \cos \theta_g + z^2 \cos^2 \theta_g ] \end{alignat} \\ |
| | | |
| + | & = \begin{alignat}{2} [ |
| + | & x^2 + ( d^2 \cos^2 \theta_g -dz \cos \theta_g \sin\theta_g + z^2 \sin^2 \theta_g ) \\ |
| + | & - 2 d^{\prime} x \sin \phi_g \\ |
| + | & + (x \sin \phi_g - d^{\prime} - d^{\prime}) \cos \phi_g ( v_{2y} ) \\ |
| + | & + d^{\prime 2} + d^2 \sin^2 \theta_g + d z \sin \theta_g \cos \theta_g + z^2 \cos^2 \theta_g ] \end{alignat} \\ |
| | | |
| + | & = \begin{alignat}{2} [ |
| + | & d^{\prime 2} + x^2 + d^2 + z^2 \\ |
| + | & - 2 d^{\prime} x \sin \phi_g \\ |
| + | & + (x \sin \phi_g - 2 d^{\prime}) \cos \phi_g ( v_{2y} ) ] \end{alignat} \\ |
| + | |
| + | & = 2 d^{\prime 2} - 2 d^{\prime} x \sin \phi_g + (x \sin \phi_g - 2 d^{\prime}) \cos \phi_g ( v_{2y} ) \\ |
| | | |
| & = ? \\ | | & = ? \\ |
Revision as of 15:25, 13 January 2016
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We define:
And calculate:
Grouping and rearranging: