HW07sol - Classical Electrodynamics I PHY 505 Fall 2010...

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1 C:\User\Teaching\505-506\F10\HW07sol.doc PHY 505 Classical Electrodynamics I Fall 2010 Homework 7 with solutions Problem 7.1 (to be graded of 15 points). Calculate resistance between two large conductors separated with a very thin, plane, insulating partition, with a circular hole of radius R in it – see Fig. on the right. Hint : You may like to use the degenerate ellipsoidal coordinates which have been used in class to find the self-capacitance of a round disk in vacuum – see Sec. 2.3 of the lecture notes Solution : Selecting parameter R of the degenerate ellipsoidal coordinates given by Eq. (2.59) of the lecture notes, to be equal to the hole radius, and placing the origin into its center, we see that the intact part of the partition corresponds to = /2 = const, and at the conductor points slightly above the partition, z R sinh ( /2 - ), so that sinh 1 0 0 R z n R r z R r z . This means that if the potential is a function of alone, the boundary condition of having no current flow through the partition is automatically satisfied. (Due to the problem symmetry, the same conclusion is valid for the lower semi-space as well.) From the disk capacitance problem (Sec. 2.3 of the lecture notes), we already know that with this assumption, the Laplace equation may be also satisfied if . ) sinh arctan( 2 1 c c For the sake of notation simplicity, let us accept (anti)symmetric boundary conditions at infinity: when i.e. , when ), sgn( 2 ) sgn( 2 z V z V (with applied voltage driving current up along axis z ); in this case the solution takes the form ). sinh arctan( V Note that according to this formula, the contribution of distant parts of the conductor (with r >> R ) into the total voltage drop is negligible; this is why the shape of external electrodes is not important, provided that they are not too small or too close to the hole. What remains now is to find total current
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HW07sol - Classical Electrodynamics I PHY 505 Fall 2010...

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