This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 6.13. Perfectlyconducting concentric spheres have radii of 2 and 6 cm. The region 2 < r < 3 cm is filled with a solid conducting material for which = 100 S/m, while the portion for which 3 < r < 6 cm has = 25 S/m. The inner sphere is held at 1 V while the outer is at V = 0. a. Find E and J everywhere: From symmetry, E and J will be radiallydirected, and we note the fact that the current, I , must be constant at any crosssection; i.e., through any spherical surface at radius r between the spheres. Thus we require that in both regions, J = I 4 r 2 a r The fields will thus be E 1 = I 4 1 r 2 a r (2 < r < 3) and E 2 = I 4 2 r 2 a r (3 < r < 6) where 1 = 100 S/m and 2 = 25 S/m. Since we know the voltage between spheres (1V), we can find the value of I through: 1V = Z . 03 . 06 I 4 2 r 2 dr Z . 02 . 03 I 4 1 r 2 dr = I . 24 1 1 + 1 2 and so I = . 24 (1 / 1 + 1 / 2 ) = 15 . 08A Then finally, with I = 15 . 08 A substituted into the field expressions above, we find...
View
Full
Document
This note was uploaded on 10/02/2011 for the course EE 1 taught by Professor Joshi during the Fall '08 term at UCLA.
 Fall '08
 Joshi
 Electrical Engineering

Click to edit the document details