20. We note that the total equivalent capacitance is C123 = [(C3)−1+ (C1 + C2)−1]−1= 6 µF. (a) Thus, the charge that passed point ais C123 Vbatt= (6 µF)(12 V) = 72 µC. Dividing this by the value e= 1.60×10−19 C gives the number of electrons: 4.5×1014, which travel to the left – towards the positive terminal of the battery. (b) The equivalent capacitance of the parallel pair is C12 = C1 + C2= 12 µF. Thus, the voltage across the pair (which is the same as the voltage across C1and C2 individually) is 72 µC12 µF= 6 V . Thus, the charge on C1 is (4 µF)(6 V) = 24 µC, and dividing this by egives the number of electrons (1.5×1014) which have passed (upward) though point b. (c) Similarly, the charge on C2 is (8 µF)(6 V) = 48 µ
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