ch25_p58 - (a This charge divided by e gives the number of...

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58. Equation 25-14 leads to C 1 = 2.53 pF and C 1 = 2.17 pF. Initially, the total equivalent capacitance is C 12 = [ ( C 1 ) 1 + ( C 2 ) 1 ] 1 = 1.488 pF, and the charge on the positive plate of each one is (1.488 pF )(10 V) = 14.88 pC. Next, capacitor 2 is modified as described in the problem, with the effect that the new value of C 2 is 2.17 pF (again using Eq. 25-14). The new total equivalent capacitance is C 12 = [ ( C 1 ) 1 + ( C 2 ) 1 ] 1 = 1.170 pF, and the new charge on the positive plate of each one is (1.170 pF)(10 V) = 11.70 pC. Thus we see that the charge transferred from the battery (considered in absolute value) as a result of the modification is 14.88 pC – 11.70 pC = 3.18 pC.
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Unformatted text preview: (a) This charge, divided by e gives the number of electrons that pass point P . Thus, 3.18 × 10-12 1.6 × 10-19 = 2.0 × 10 7 . (b) These electrons move rightwards in the figure (that is, away from the battery) since the positive plates (the ones closest to point P ) of the capacitors have suffered a decease in their positive charges. The usual reason for a metal plate to be positive is that it has more protons than electrons. Thus, in this problem some electrons have “returned” to the positive plates (making them less positive)....
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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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