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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. 2313, 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 09/07/2010 for the course PHYS 247 taught by Professor Zv during the Fall '10 term at The Petroleum Institute.
 Fall '10
 ZV

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