# Difference across its plates we convert the given

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difference across its plates. We convert the given value of the energy to Joules. Since1 J1 Ws,=we multiply by (103W/kW)(3600 s/h) to obtain710 kWh3.610J=×. Thus,CUV==×=22 3610100072272.JVF.chbg
32. LetV= 1.00 m3. Using Eq. 25-25, the energy stored is()()22212380211C8.8510150V m1.00m9.9610J.22N mUuEε===×=×VV
33. The energy per unit volume isuEerer==FHGIKJ=121243202022204εεεεππ02.08.8510C /N mε=×, we have22....
()()2221111110.0F50.0V1.25 10J.22UCVµ===×(g) Again, from part (a),422.50 10Cq=×.(h)V2= 50.0 V, as shown in (a).(i) The energy stored inC2is()()223222115.00F50.0V6.25 10J.22UC Vµ===×34. (a) The potential difference acrossC1(the same as acrossC2) is given by()()31212315.0F100V50.0V.10.0F+5.00F+15.0FC VVVCCCµµµµ====++Also,V3=V – V1=V – V2= 100 V – 50.0 V = 50.0 V. Thus,()()()()4111422244431210.0F50.0V5.00 10C5.00F50.0V2.50 10C5.00 10C2.50 10C=7.50 10C.qCVqC Vqqqµµ===×===×=+=×+××(b) The potential differenceV3was found in the course of solving for the charges in part(a). Its value isV3= 50.0 V.(c) The energy stored inC3is()()222333/ 215.0F50.0V/ 21.88 10J.UC Vµ===×(d) From part (a), we have415.00 10Cq=×, and(e)V1= 50.0 V, as shown in (a).(f) The energy stored inC1is
35. (a) Letqbe the charge on the positive plate. Since the capacitance of a parallel-platecapacitor is given by0iA dε, the charge is0iiqCVAVdε==. After the plates arepulled apart, their separation isfdand the potential difference isVf. Then02ffqAVdε=and000.ffffiiiidddAVqVVAA ddεεε===With33.00 10mid=×,6.00 ViV=and38.00 10mfd=×, we have16.0 VfV=.(b) The initial energy stored in the capacitor is21222422211031(8.8510C/Nm)(8.5010m )(6.00 V)4.5110J.222(3.0010m)iiiiAVUCVdε××====××(c) The final energy stored is22200011.22fffiffiiffiiiidddAAAVUVVUddddddεεε====With/8.00/3.00fidd=, we have101.2010J.fU=×(d) The work done to pull the plates apart is the difference in the energy:W = UfUi=117.5210J.×
(d) From the figure,4121212(10.0F)(5.00F)(100 V)3.33 10C.10.0F5.00FC C VqqCCµµµµ====×++(e)V1=q1/C1= 3.33×10–4C/10.0µF = 33.3 V.(f)23111125.55 10JUCV==×.(g) From part (d), we have4213.33 10C.qq==×(h)V2=V – V1= 100 V – 33.3 V = 66.7 V.(i)22122221.11 10JUC V==×.36. (a) The chargeq3in the figure is433(4.00F)(100 V)4.0010CqC Vµ===×.(b)V3=V= 100 V.(c) UsingUCViii=122, we have22133322.00 10JUC V==×.
37. (a) They each store the same charge, so the maximum voltage is across the smallestcapacitor. With 100 V across 10µF, then the voltage across the 20µF capacitor is 50 Vand the voltage across the 25µF capacitor is 40 V. Therefore, the voltage across thearrangement is 190 V.(b) Using Eq. 25-21 or Eq. 25-22, we sum the energies on the capacitors and obtainUtotal= 0.095 J.
38. (a) We calculate the charged surface area of the cylindrical volume as follows:Arhr=+=+=220 252πππ(0.20π(0.20m)(0.10 m)m)m22.

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