ch24_p79 - × 10 − 6 m and e = 1.60 × 10 − 19 C we...

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7 9. The net potential at point P (the place where we are to place the third electron) due to the fixed charges is computed using Eq. 24-27 (which assumes V 0 as r ): V P = 1 4 πε 0 - e d + 1 4 πε 0 -e d = - e 2 πε 0 d . Thus, with d = 2.00
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Unformatted text preview: × 10 − 6 m and e = 1.60 × 10 − 19 C, we find V P = − 1.438 × 10 − 3 V. Then the required “applied” work is, by Eq. 24-14, W app = ( − e ) V P = 2.30 × 10 − 22 J ....
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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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