hw03 - PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS...

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

View Full Document Right Arrow Icon

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

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

Unformatted text preview: PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS CLASS 3. 23.08, 23.11, 23.53, 23.67; Q25.02, Q25.06, Q25.07 23.08 a) Call the length, width, and height of the room l, w, and h . Then, from Gauss law, the charge enclosed is given by: ) ( 2 ) ( lh wh lw E A d E q + +- = = , where E is the magnitude of the electric field at each surface, and the negative sign is because the field points inward. Then, the charge per volume in the room is: + +- = + +- = = w l h E lwh lh wh lw E V q 1 1 1 2 ) ( 2 + + - =- m . 2 1 m . 3 1 m 5 . 2 1 ) N/C 600 )( F/m 10 85 . 8 ( 2 12 = 1.3 10-8 C/m b) C/electron 10 6 . 1 C/m 10 3 . 1 19 3 8-- - - = = e n n = 8.1 10 10 electrons/m 23.11. The cube is sketched below. a) The flux through the three sides that touch the charge is zero since the electric field is parallel to each of those faces....
View Full Document

This note was uploaded on 04/07/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

Page1 / 2

hw03 - PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS...

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