Unformatted text preview: 64. (a) At x = 0.040 m, the net ﬁeld has a rightward (+x) contribution (computed using Eq. 2413) from
the charge lying between x = −0.050 m and x = 0.040 m, and a leftward (−x) contribution (again
computed using Eq. 2413) 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 ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Charge

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