# PHY2049 Chapter 25 Solutions - Chapter 25 1. (a) The...

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1011Chapter 251. (a) The capacitance of the system isCqV===Δ702035pCVpF..(b) The capacitance is independent ofq; it is still 3.5 pF.(c) The potential difference becomesΔVqC===2003557pCpFV..2. 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.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,()()2122210038.85 10F m8.2 10m1.44 10F144 pF.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. (a) We use Eq. 25-17:()()()()2209N mC40.0mm38.0mm484.5 pF.8.99 1040.0mm38.0mmabCbaπε===×(b) Let the area required beA. ThenC=ε0A/(b – a), or
CHAPTER 251012()()()()21222084.5pF40.0mm38.0mm191cm .8.8510C /N mC baAε===×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. Itis thenV= 2(4π/3)R3. The new radiusR'is given by()3344233RR=pp′ =RR21 3.The new capacitance is1 30004425.04.CRRRεεε===pppWithR= 2.00 mm, we obtain()()123135.048.85 10F m2.00 10m2.80 10FCπ=××=×.6. We useC = Aε0/d.(a) The distance between the plates is()()212221201.00m8.8510C /N m8.8510m.1.00FAdCε×===×(b) Sincedis much less than the size of an atom (10–10m), this capacitor cannot beconstructed.7. For a given potential differenceV, the charge on the surface of the plate is()qNenAd e==wheredis the depth from which the electrons come in the plate, andnis the density ofconduction electrons. The charge collected on the plate is related to the capacitance andthe potential difference byqCV=(Eq. 25-1). Combining the two expressions leads toCdneAV=.With14//5.010m/Vssd VdV==×and2838.4910/ mn=×(see, for example, SampleProblem — “Charging the plates in a parallel-plate capacitor”), we obtain2831942(8.4910/ m)(1.610C)(5.01014 m/V)6.7910F/mCA=×××=×.
10138. 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()()361 00C9 09 10110V1 00 10Fq.N..VC.===××9. The charge that passes through meterAisqC VCV====eqFVC.33 25042000 315..μbgbg10. The equivalent capacitance isCCC CCCeqFFFFFF.=++=++=312124 0010 050010 05007 33.

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Term
Summer
Professor
hardies
Tags
Electric charge, Dielectric, Ceq
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