F)(2.25 V) = 22.5C.57. The pairC3andC4are in parallel and consequently equivalent to 30F.Since thisnumerical value is identical to that of the others (with which it is in series, with thebattery), we observe that each has one-third the battery voltage across it.Hence, 3.0 V isacrossC4, producing a chargeq4=C4V4= (15F)(3.0 V) = 45C.58. (a) HereDis not attached to anything, so that the 6Cand 4Ccapacitors are in series(equivalent to 2.4C). This is then in parallel with the 2Ccapacitor, which produces anequivalence of 4.4C. Finally the 4.4Cis in series withCand we obtaineq4.40.820.82(40F)33F4.4CCCCCCwhere we have used the fact thatC= 40F.(b) Now,Bis the point that is not attached to anything, so that the 6Cand 2Ccapacitorsare now in series (equivalent to 1.5C), which is then in parallel with the 4Ccapacitor (andthus equivalent to 5.5C). The 5.5Cis then in series with theCcapacitor; consequently,eq5.50.8534F .5.5CCCCCC59. The pairC1andC2are in parallel, as are the pairC3andC4; they reduce to equivalentvalues 6.0F and 3.0F, respectively.These are now in series and reduce to 2.0F,across which we have the battery voltage. Consequently, the charge on the 2.0Fequivalence is (2.0F)(12 V) = 24C.This charge on the 3.0F equivalence (ofC3andC4) has a voltage ofV=qC=24C3F= 8.0 V.