ch23_p33 - These three conditions are sufficient for...

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33. In the region between sheets 1 and 2, the net field is E 1 E 2 + E 3 = 2.0 × 10 5 N/C . In the region between sheets 2 and 3, the net field is at its greatest value: E 1 + E 2 + E 3 = 6.0 × 10 5 N/C . The net field vanishes in the region to the right of sheet 3, where E 1 + E 2 = E 3 . We note the implication that σ 3 is negative (and is the largest surface-density, in magnitude).
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Unformatted text preview: These three conditions are sufficient for finding the fields: E 1 = 1.0 × 10 5 N/C , E 2 = 2.0 × 10 5 N/C , E 3 = 3.0 × 10 5 N/C . From Eq. 23-13, we infer (from these values of E ) | σ 3 | | σ 2 | = 3.0 x 10 5 N/C 2.0 x 10 5 N/C = 1.5 . Recalling our observation, above, about σ 3 , we conclude σ 3 σ 2 = –1.5 ....
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.

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