Solutions to Problem Set 11. The potential on the axiszisΦ(z) =14π0ZSσdSr=14π0Za02πρdρpz2+ρ2=σ20hpz2+a2- |z|iBy symmetry,Epoints along the axis away from the disc (forσ >0),E=Ezez, thusEz=-∂Φ(z)∂z=±σ201-|z|√z2+a2where the plus sign corresponds toz >0 while the minus sign forz <0.2. The conductor and the metal plate have the same potential when they are contacted.Afterthe first contact,q/C1= (Q-q)/C2.After the final contact, there is no charge transfer, soqf/C1=Q/C2. The ultimate charge isqf=qQQ-q.3. We start with the Poisson equation,∇2Φ =-ρ0.The potential has no angular dependence,therefore,ρ=-01rd2dr2(rΦ) =-q4πrd2dr2he-αr1 +αr2i=-qα38πe-αrThis is valid forr >0. Forr→0, the potential Φ diverges and alsoρdiverges. Note that inthis limitlimr→0Φ(r) =q4π0rBecause∇2 1r=-4πδ(r) we have to add this piece to the solution. Finally, we getρ(r) =qδ(r)-qα38πe-αr4. a) The potential satisfies the Poisson equation Φ00(x) =-ρ/0.The boundary conditions areΦ(0) = 0 and Φ0(d) =σ0/0. The solution isΦ(x) =-ρ20x2+ρd+σ00x.b) The threshold shift is Φ(
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