chapter25- 022

# chapter25- 022 - 22 Using Equation 25-14 the capacitances...

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22. Using Equation 25-14, the capacitances are 12 2 2 01 1 11 12 2 2 02 2 22 2 2 (8.85 10 C /N m )(0.050 m) 2.53 pF ln( / ) ln(15 mm/5.0 mm) 2 2 (8.85 10 C /N m )(0.090 m) 3.61 pF . ln( / ) ln(10 mm/2.5 mm) L C ba L C ba πε π πε π ×⋅ ×⋅ == = Initially, the total equivalent capacitance is 12 1 2 12 12 1 2 1 2 1 2 1 1 1 (2.53 pF)(3.61 pF) 1.49 pF 2.53 pF 3.61 pF CC C C C CC C C C C C + =+= = = = ++ , and the charge on the positive plate of each one is (1.49 pF )(10 V) = 14.9 pC. Next, capacitor 2 is modified as described in the problem, with the effect that 12 2 2 02 2 22 2 2 (8.85 10 C /N m )(0.090 m) 2.17 pF . ln( / ) ln(25 mm/2.5 mm) L C ba πε π ×⋅ == = The new total equivalent capacitance is 12 12 12 (2.53 pF)(2.17 pF) 1.17 pF 2.53 pF 2.17 pF CC C == = ++ and the new charge on the positive plate of each one is (1.17 pF)(10 V) = 11.7 pC. Thus we see that the charge transferred from the battery (considered in absolute value) as a result of the modification is 14.9 pC – 11.7 pC = 3.2 pC. (a) This charge, divided by e gives the number of electrons that pass point P . Thus, 12 7 19 3.2 10 C 2.0 10 1.6 10 C N × == × × . (b) These electrons move rightwards in the figure (that is, away from the battery) since
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## This note was uploaded on 01/25/2011 for the course PHYS 221 taught by Professor Hk during the Spring '10 term at The Petroleum Institute.

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