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# Opt2sol - Classical Electrodynamics I PHY 505 Fall 2010...

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1 C:\User\Teaching\505-506\F10\Opt2sol.doc PHY 505 Classical Electrodynamics I Fall 2010 Optional problems Set 2 with solutions Problem O.7. Solve the problem shown in Fig. 2.19 of the lecture notes (reproduced on the right); in particular, find the distribution of the electrostatic potential at the cylinder’s axis. Solution : Due to the problem’s 2 h -periodicity along axis z , its particular solutions should be proportional to linear combinations of sin kz and cos kz with 2 kh = 2 m , where m are integers. Moreover, if the origin of axis z is selected exactly at the gap between the rings, the solution should be an odd function of z , so that for the internal problem ( R , where R is the rings’ radius) we may write 1 0 sin , m m h mz h m I c z . The modified Bessel function of the first kind, used in this expression, has zero index due to the evident axial symmetry of the problem (so that in Eq. (2.128) of the lecture notes we have to take F = const). The argument of this function follows from the discreteness of the variable separation constant k : = k = m /2 h . Finally, we had to drop the modified Bessel functions of the second kind, K 0 ( ), from our solution, because they diverge at 0 – see Fig. 2.20 and thus cannot be used to represent the finite potential inside the rings. 1 Coefficients c m should be found from the boundary condition on the conducting rings; since the proper periodicity is already incorporated in our solution, it is sufficient to require that . 0 for , 2 sin , 1 0 h z V h mz h mR I c z R m m Multiplying both sides of this expression by sin( m’z /2 h ), and integrating the result from 0 to h , we get even. for , 0 odd, for , / 1 1 0 m m m h mR I V c m . The plot below shows the resulting electrostatic potential at the axis of the system, (0, z ), for several values of ratio R / h . At R > h , the result is dominated by the first term of the series (with m = 1), which is proportional to 1/ I 0 ( R /2 h ), so that at R >> h , according to Eq. (2.158), the potential variation amplitude equals 2 V ( R /2 h ) 1/2 exp{- R /2 h }. On the contrary, at R << h , the potential follows that of the closest ring electrode at most of the axis, besides narrow ( z ~ R ) intervals centered to the gap between the electrodes. 1 Note that these function have to be used in the solution of the external problem ( R ), which is absolutely similar to that discussed above, besides the replacement of I 0 ( m /2 h ) with K 0 ( m /2 h ). 2 / V 2 / V 2 / V z h t

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