http://img58.imageshack.us/my.php?image=physics24yu7.jpg I\'m somewhat clueless

Question

http://img58.imageshack.us/my.php?image=physics24yu7.jpg I'm somewhat clueless how voltage is carried throughoutcapacitors like this without knowing any charge. Perhaps thisproblem is easier then I thought, and I'm just overthinking it. Itis an online homework, so I just took a screenshot. Thanks inadvance for the help!! http://img58.imageshack.us/my.php?image=physics24yu7.jpg I'm somewhat clueless how voltage is carried throughoutcapacitors like this without knowing any charge. Perhaps thisproblem is easier then I thought, and I'm just overthinking it. Itis an online homework, so I just took a screenshot. Thanks inadvance for the help!!

Explanation / Answer

The capacitors in the figure are C1=11F,C2= 22 F,C3= 22 F andC4= 11 F.The battery connected to the capacitorsis E = 126 V.Initially,the switch S1 is closed andswitch S2 is left open.When switch S1 isclosed,then the two capacitors C1 and C3 arecharged.The capacitors C1 and C3 areconnected in series,therefore,the charge flowing through the twocapacitors is same.The equivalent capacitance of the two capacitorsC1 and C3 is (1/Ceq) = (1/C1) +(1/C3) or (1/Ceq) = (1/11) + (1/22) = (2 + 1/22) =(3/22) or Ceq= (22/3) F The charge flowing through the two capacitors is Q = Ceq* E = (22/3) * 10-6 * 126 = 924 *10-6 C The potential difference across the two capacitors is E1= (Q/C1) or E1= (924 * 10-6/11 *10-6) or E1= 84 V and E3= (Q/C3) or E3= (924 * 10-6/22 *10-6) or E3= 42 V Therefore,the potential difference between points a and bis Vab= (E1 - E3) = (84 - 42) V= 42 V The potential difference between the two points a and b ispositive. When switch S2 is closed,then the capacitorsC2 and C4 are in series,therefore,the chargeflowing through the two capacitors is same.The equivalentcapacitance of the two capacitors C2 and C4is (1/Ceq') = (1/C2) +(1/C4) or (1/Ceq') = (1/22) + (1/11) = (1 + 2/22) =(3/22) or Ceq'= (22/3) F The capacitors Ceq and Ceq' are inparallel,therefore,the equivalent capacitance is (1/C) = (1/Ceq) + (1/Ceq') or (1/C) = (3/22) + (3/22) = (3 + 3/22) = (6/22) F or C = (22/6) F When the capacitors Ceq and Ceq' are inparallel,then the potential difference across the two capacitors issame,that is, Vab= 42 V. The magnitude of charge that flowed when switch S2is closed is Q = Ceq * Vab= (22/3) *10-6 * 42 or Q = 308 * 10-6 C = 308 C or E3= (924 * 10-6/22 *10-6) or E3= 42 V Therefore,the potential difference between points a and bis Vab= (E1 - E3) = (84 - 42) V= 42 V The potential difference between the two points a and b ispositive. When switch S2 is closed,then the capacitorsC2 and C4 are in series,therefore,the chargeflowing through the two capacitors is same.The equivalentcapacitance of the two capacitors C2 and C4is (1/Ceq') = (1/C2) +(1/C4) or (1/Ceq') = (1/22) + (1/11) = (1 + 2/22) =(3/22) or Ceq'= (22/3) F The capacitors Ceq and Ceq' are inparallel,therefore,the equivalent capacitance is (1/C) = (1/Ceq) + (1/Ceq') or (1/C) = (3/22) + (3/22) = (3 + 3/22) = (6/22) F or C = (22/6) F When the capacitors Ceq and Ceq' are inparallel,then the potential difference across the two capacitors issame,that is, Vab= 42 V. The magnitude of charge that flowed when switch S2is closed is Q = Ceq * Vab= (22/3) *10-6 * 42 or Q = 308 * 10-6 C = 308 C (1/Ceq') = (1/C2) +(1/C4) or (1/Ceq') = (1/22) + (1/11) = (1 + 2/22) =(3/22) or Ceq'= (22/3) F The capacitors Ceq and Ceq' are inparallel,therefore,the equivalent capacitance is (1/C) = (1/Ceq) + (1/Ceq') or (1/C) = (3/22) + (3/22) = (3 + 3/22) = (6/22) F or C = (22/6) F When the capacitors Ceq and Ceq' are inparallel,then the potential difference across the two capacitors issame,that is, Vab= 42 V. The magnitude of charge that flowed when switch S2is closed is Q = Ceq * Vab= (22/3) *10-6 * 42 or Q = 308 * 10-6 C = 308 C Vab= 42 V. The magnitude of charge that flowed when switch S2is closed is Q = Ceq * Vab= (22/3) *10-6 * 42 or Q = 308 * 10-6 C = 308 C

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