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# ch25_p20 - 20 We note that the total equivalent capacitance...

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20. We note that the total equivalent capacitance is C 123 = [ ( C 3 ) 1 + ( C 1 + C 2 ) 1 ] 1 = 6 µ F. (a) Thus, the charge that passed point a is C 123 V batt = (6 µ F)(12 V) = 72 µ C. Dividing this by the value e = 1.60 × 10 19 C gives the number of electrons: 4.5 × 10 14 , which travel to the left – towards the positive terminal of the battery. (b) The equivalent capacitance of the parallel pair is C 12 = C 1 + C 2 = 12 µ F. Thus, the voltage across the pair (which is the same as the voltage across C 1 and C 2 individually) is 72 µ C 12 µ F = 6 V . Thus, the charge on C 1 is (4 µ F)(6 V) = 24 µ C, and dividing this by e gives the number of electrons (1.5 × 10 14 ) which have passed (upward) though point b . (c) Similarly, the charge on C 2 is (8 µ F)(6 V) = 48 µ
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