# CH-25 - Chapter 25 1. Charge flows until the potential...

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Chapter 251. Charge flows until the potential difference across the capacitor is the same as thepotential difference across the battery. The charge on the capacitor is thenq = CV, andthis is the same as the total charge that has passed through the battery. Thus,q= (25×10–6F)(120 V) = 3.0×10–3C.2. (a) The capacitance of the system isCqV===702035pCVpF..(b) The capacitance is independent ofq; it is still 3.5 pF.(c) The potential difference becomesVqC===2003557pCpFV..3. (a) The capacitance of a parallel-plate capacitor is given byC=ε0A/d, whereAis thearea of each plate anddis the plate separation. Since the plates are circular, the plate areaisA =πR2, whereRis the radius of a plate. Thus,(29(292122210038.85 10F m8.2 10m1.44 10F144pF.1.3 10mRCdπε π----====(b) The charge on the positive plate is given byq = CV, whereVis the potentialdifference across the plates. Thus,q= (1.44×10–10F)(120 V) = 1.73×10–8C = 17.3 nC.4. We useC = Aε0/d.(a) Thus,dAC==×=×--ε0121210088510100885102....mFm.2CN m2chdi1039
CHAPTER 25(b) Sincedis much less than the size of an atom (10–10m), this capacitor cannot beconstructed.5. Assuming conservation of volume, we find the radius of the combined spheres, thenuseC= 4πε0Rto find the capacitance. When the drops combine, the volume is doubled.It is thenV= 2(4π/3)R3. The new radiusR'is given by(293344233RR=ππ′ =RR21 3.The new capacitance is1 30004425.04.CRRRεεε===πππWithR= 2.00 mm, we obtain(29(29123135.048.85 10F m2.00 10m2.80 10FCπ---=%=.6. (a) We use Eq. 25-17:Cabba=-=×-=440 038 08 991040 038 084 509πε......mmmmmmmmpF.N mC22bgbgdibg(b) Let the area required beA. ThenC=ε0A/(b – a), or(29(29(29(2922212C0N m84.5pF40.0mm38.0mm191cm .8.8510C baAε---===×7. The equivalent capacitance is given byCeq=q/V, whereqis the total charge on all thecapacitors andVis the potential difference across any one of them. ForNidenticalcapacitors in parallel,Ceq=NC, whereCis the capacitance of one of them. Thus,/NCq V=and(29(29361 00C9 09 10110V1 00 10Fq.N..VC.-===8. The equivalent capacitance isCCC CCCeqFFFFFF.=++=++=312124 0010 050010 05007 33......μμμμμμbgbg9. The equivalent capacitance is16
(29(29(29123eq12310.0F5.00F4.00F3.16F.10.0F5.00F4.00FCCCCCCCμμμμμμμ++===++++10. The charge that passes through meterAisqC VCV====eqFVC.33 25042000 315..μbgbg11. (a) and (b) The original potential differenceV1acrossC1is(29(29eq1123.16F100.0V21.1V.10.0F5.00FC VVCCμμμ===++ThusV1= 100.0 V – 21.1 V = 78.9 V andq1=C1V1= (10.0μF)(78.9 V) = 7.89×10–4C.12. (a) The potential difference acrossC1isV1= 10.0 V. Thus,q1=C1V1= (10.0μF)(10.0 V) = 1.00×10–4C.(b) LetC= 10.0μF. We first consider the three-capacitor combination consisting ofC2and its two closest neighbors, each of capacitanceC. The equivalent capacitance of thiscombination is2eq21 50C CCC.C.

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Term
Fall
Professor
WILKE
Tags
Potential difference, Electric charge, Ceq
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