P24_064 - 64(a At x = 0.040 m the net field has a...

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: 64. (a) At x = 0.040 m, the net field has a rightward (+x) contribution (computed using Eq. 24-13) from the charge lying between x = −0.050 m and x = 0.040 m, and a leftward (−x) contribution (again computed using Eq. 24-13) from the charge in the region from x = 0.040 m to x = 0.050 m. Thus, since σ = q/A = ρV /A = ρ∆x in this situation, we have E= ρ(0.090 m) ρ(0.010 m) − = 5. 4 N / C . 2 ε0 2 ε0 (b) In this case, the field contributions from all layers of charge point rightward, and we obtain E= ρ(0.100 m) = 6. 8 N / C . 2 ε0 ...
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online