ch23_p24 - P . Using Eq. 23-12, we have E net = E 1 + E 2 =...

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24. We reason that point P (the point on the x axis where the net electric field is zero) cannot be between the lines of charge (since their charges have opposite sign). We reason further that P is not to the left of “line 1” since its magnitude of charge (per unit length) exceeds that of “line 2”; thus, we look in the region to the right of “line 2” for
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Unformatted text preview: P . Using Eq. 23-12, we have E net = E 1 + E 2 = λ 1 2 πε o ( x + L/2 ) + λ 2 2 πε o ( x − L/2 ) . Setting this equal to zero and solving for x we find x = λ 1 − λ 2 λ 1 + λ 2 L 2 which, for the values given in the problem, yields x = 8.0 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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