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 becomes∆VqC===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===++Thus∆V1= 100.0 V – 21.1 V = 78.9 V and∆q1=C1∆V1= (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.