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# HW03sol - Classical Electrodynamics I PHY 505 Fall 2010...

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1 C:\User\Teaching\505-506\F10\HW03sol.doc PHY 505 Classical Electrodynamics I Fall 2010 Homework 3 with solutions Problem 3.1 (to be graded of 15 points). Calculate the mutual capacitance (per unit length) between two cylindrical conductors, forming a system with the cross-section shown in Fig. on the right, in the limit t << w << R . Hint : You may like to use the “ elliptical ” (not “ellipsoidal”!) coordinates defined by the following equation: ), cosh( i c iy x (*) with the appropriate choice of constant c . In these orthogonal 2D coordinates, the Laplace operator is very simple: 2 2 2 2 2 2 2 2 2 ) cos (cosh 1 c . (**) (This is not quite surprising, because Eq. (*) may be also considered as a conformal map z = c cosh w , where z = x + iy , and w = + i .) Solution: On the [ x, y ] plane, the lines of constant are ellipses with horizontal and vertical semi-axes c cosh and c sinh , respectively. For = 0, the ellipse degenerates into a straight horizontal segment – c < x < + c , while for >> 1, the ellipse is virtually a circle of radius = ( c /2)exp . As a result, if we select the axes x and y as shown in figure, and take c = w /2, the boundary conditions on the conductor surfaces (see Fig.) may be satisfied, at t << w << R , by a potential distribution ( ), with no dependence on : , ) 4 (ln , 0 ) 0 ( V W R (***) where V is the voltage between the conductors. Hence the boundary problem may be satisfied by a function ( ) if it satisfies the 1D Laplace equation following from Eq. (**): . 0 2 2 d d This equation shows that ( ) is just a linear function c 1 + c 2 . Selecting two constants c 1,2 to satisfy boundary conditions (***), we get ) / 4 ln( w R V . To calculate the surface charge of the central conductor, we need to calculate the (normal) electric field on its surface: , sin ) 2 / )( / 4 ln( ) (sinh sin ) 2 / )( / 4 ln( 0 2 / 2 / 0 w w R V w w R V y E w x w y n so that the two-surface charge density R w t

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