ch23_p49 - 49. At all points where there is an electric...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
problem, use a Gaussian surface in the form of a sphere that is concentric with the sphere of charge and passes through the point where the electric field is to be found. The field is uniform on the surface, so 2 4 EdA rE ⋅= π ³ G G v , where r is the radius of the Gaussian surface. For r < a , the charge enclosed by the Gaussian surface is q 1 ( r / a ) 3 . Gauss’ law yields 3 2 11 3 00 4 . 4 qq r r E aa π επ ε §· = ¡ = ¨¸¨¸ ©¹ (a) For r = 0, the above equation implies E = 0. (b) For r = a /2, we have 92 2 1 5 2 1 32 2 0 (/ 2 ) (8.99 10 N m /C )(5.00 10 C) 5.62 10 N/C. 4 2(2.00 10 m) qa E a πε ×⋅ × == = × × (c) For r = a , we have 2 1 5 1 22 2 0 (8.99 10 N m /C )(5.00 10 C) 0.112 N/C. 4 (2.00 10 m) q E a × = × In the case where a < r < b , the charge enclosed by the Gaussian surface is q 1 , so Gauss’ law leads to 2 2 4. 4 E r = ¡ = (d) For r = 1.50 a , we have 2 1 5 1 2 0 (8.99 10 N m /C )(5.00 10 C) 0.0499 N/C. 4 (1.50 2.00 10 m) q E r × = ×× (e) In the region b < r < c , since the shell is conducting, the electric field is zero. Thus, for r = 2.30 a , we have
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.

Page1 / 2

ch23_p49 - 49. At all points where there is an electric...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online