ch23_p46 - Consequently, the point should be in the region...

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46. The point where the individual fields cancel cannot be in the region between the shells since the shells have opposite-signed charges. It cannot be inside the radius R of one of the shells since there is only one field contribution there (which would not be canceled by another field contribution and thus would not lead to zero net field). We note shell 2 has greater magnitude of charge (| σ 2 | A 2 ) than shell 1, which implies the point is not to the right of shell 2 (any such point would always be closer to the larger charge and thus no possibility for cancellation of equal-magnitude fields could occur).
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Unformatted text preview: Consequently, the point should be in the region to the left of shell 1 (at a distance r > R 1 from its center); this is where the condition E 1 = E 2 | q 1 | 4 o r 2 = | q 2 | 4 o ( r + L) 2 or 1 A 1 4 o r 2 = | 2 | A 2 4 o ( r + L) 2 . Using the fact that the area of a sphere is A = 4 R 2 , this condition simplifies to r = L ( R 2 / R 1 ) | 2 |/ 1 1 = 3.3 cm . We note that this value satisfies the requirement r > R 1 . The answer, then, is that the net field vanishes at x = r = 3.3 cm....
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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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