HW01sol - Classical Electrodynamics I PHY 505 Fall 2010...

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1 C:\User\Teaching\505-506\F10\HW01sol.doc PHY 505 Classical Electrodynamics I Fall 2010 Homework 1 with solutions Problem 1.1 (to be graded of 10 points). Calculate the electric field created by a thin, long, straight filament, electrically charged with a constant linear density , using two approaches: (i) directly from Eq. (8), and (ii) using the Gauss law. Solutions : (i) From the translational and axial symmetries of the problem, it is clear that E ( r ) = n E ( ), where is the distance from the filament. 1 Let us select the plane of drawing so that it contains both the filament and the observation point, and take the line of filament for axis z (see Fig. below). Then E ( ) may be calculated as        z z z z z z z dE dE dE E 2 / 1 2 2 cos , where dE is the magnitude of the elementary contribution to the field, created by a small segment dz of the filament, with electric charge dz . According to Eq. (1.3) of the lecture notes, 2 2 0 4 1 z dz dE  , so that the total field  0 2 / 3 2 0 2 / 3 2 2 0 2 1 4 4   d z dz E . (*) (For the last transition, I have used the well-known table integral – see, e.g., MA Eq. (6.5b).) (ii) Taking a round cylinder of radius and length L , with its axis on the filament, for the Gaussian volume, we ensure that the electric field E is the same, and perpendicular to the volume boundary on its side walls, while the field flux through cylinder’s “lids” is zero. As a result, Eq. (1.16) yields 1 I regret using the same letter ( ) as for the charge density per unit volume (which is not used in this problem), but both notations are traditional. Let me hope this will not result in any confusion. z dz 0 E d dE 2 / 1 2 2 ) ( z
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2 C:\User\Teaching\505-506\F10\HW01sol.doc 0 2  L LE , immediately giving the same result (*). 2 We see that for this, highly symmetric problem both solution ways are readily doable, but the Gauss method is still easier. Problem 1.2 (10 points). Use any two (different) approaches you like to calculate the distribution of electrostatic potential and electric field E created, in otherwise free space, by a plane layer of thickness t , with a uniformly distributed charge of density (see Fig. on the right). Solutions : (i) Using as the Gauss volume a pillbox similar to that discussed in the lecture notes (Fig. 1.4), for z > t /2 we get the same field as for the thin charged plane - see Eq. (24), where now = t , so that 2 for ), sgn( 2 , 2 0 0 t z z t E t E z , where z is the Cartesian coordinate perpendicular to the layer, with z = 0 in its middle. . On the other
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This note was uploaded on 09/10/2011 for the course PHY 505 taught by Professor Stephens,p during the Fall '08 term at SUNY Stony Brook.

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HW01sol - Classical Electrodynamics I PHY 505 Fall 2010...

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