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

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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 ...
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