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