18. We determine each capacitance from the slope of the appropriate line in the graph. Thus, C1= (12 µC)/(2.0 V) = 6.0 µF. Similarly, C2= 4.0 µF and C3= 2.0 µF. The total equivalent capacitance is given by 123123123123111()CCCCCCCCCC++=+=++, or 123123(6.0 F)(4.0 F 2.0 F)36F3.0F6.0F4.0F2.0F 12CC CCµµµµµµ++====++. This implies that the charge on capacitor 1 is 1q=(3.0 µF)(6.0 V) = 18
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