61. (a) At x= 0.040 m, the net field has a rightward (+x) contribution (computed using Eq. 23-13) from the charge lying between x= –0.050 m and x= 0.040 m, and a leftward (–x) contribution (again computed using Eq. 23-13) from the charge in the region from 0.040 mx=to x= 0.050 m. Thus, since σ= q/A= ρV/A= ∆xin this situation, we have 93122200(0.090m)
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