Homework Assignment 3 Solution Set

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1 Homework Assignment 3 Solution Set PHYCS 44 6 Ferury, 4 Prolem 1 (Griffiths.5(c The potentil due to ny continuous chrge distriution is the sum of the contriutions from ech infinitesiml chrge in the distriution. For twodimensionl distriution we write V ( r = 1 σ( r 4πɛ r r da where the integrl is over the entire chrge distriution (in this cse flt, round disk. At ny point on the z-xis the distnce to n ritrry element of chrge σ(rrdr on the surfce is z + r, where r is the distnce from the center of the disk to the element of chrge. Therefore, V (z = 1 4πɛ = σ ɛ = σ ɛ S dr π σ(r r z + r dr(r + z 1 r [ ] (r + z 1 = σ( + z z ɛ Now, we know tht E( r = V ( r. However, in this cse we hve limited ourselves to the z-xis. Therefore, we cn only find V ( r E(z = ẑ z = σ ( z 1 ẑ ɛ + z Note: A clever wy to check your work on something tht is little messy like this is to verify the field in some limiting cses. For exmple, for the field 1

2 ove we see tht for z we get E(z σ ( 1 1 ẑ ɛ σ ẑ ɛ which is the field due to n infinite plne of chrge (s it should e. Also, for z we cn expnd in inomil series to otin E(z = σ ( 1 z( + z 1 ẑ ɛ = σ ɛ = σ ɛ σ 4ɛ 1 ( 1 ( ( ( ẑ z 1 4πɛ Q z ẑ ( z 1 ẑ ( + ẑ z which is the field of point chrge with the sme mgnitude s the totl chrge on the plte (s it should e. Prolem (Griffiths.1 We hve derived previously the electric field inside nd outside uniformly chrged sphere (if you don t rememer the result you cn use Guss s Lw to figure it out gin. The potentil is relted to the field y V ( r = E( r d r which is definitive up to n ritrry constnt tht depends on where we choose V =. Outside sphere of uniformly distriuted chrge we hve Thus, E( r = 1 4πɛ r ˆr. V ( r = = E( r d r 1 4πɛ r dr = 1 4πɛ r + C

3 Setting V ( = gives C =. Therefore, outside the sphere the potentil is the sme s for point chrge (just like the electric field is the sme s for point chrge: V (r = 1 4πɛ r. Thus, Inside the sphere the electric field is E( r = 1 4πɛ r 3 ˆr. V ( r = = E( r d r 1 r 4πɛ 3 dr = 1 1 r 4πɛ 3 + C Deciding on the vlue of C is slightly trickier here. The sphere is not infinite in size, so we cnnot tke this form of V (r nd set V ( =. However, we lredy fixed the zero of the potentil everywhere outside the sphere. Therefore, we should just mke the potentil continuous t the surfce of the sphere. We reuire V ( = 1 is 4πɛ, so C = 3 1 4πɛ V (r = 1 1 r 4πɛ πɛ.. Thus, the potentil inside the sphere Oviously the grdient of V gives us ck the electric field since we found V y integrting the field! Inside the sphere we hve nd outside we hve E(r = V = 1 4πɛ r ˆr E(r = V (r = 1 4πɛ r 3 ˆr. Below is plot of the potentil due to the uniformly chrged sphere with = 1 nd V ( = 1 in some ritrry unit system. Prolem 3 (Griffiths.3 3

4 Clculte U = 1 ρv dτ 1 r 4πɛ πɛ with V = 1 s found in the previous prolem. We get U = 1 ρ( 1 1 r 4πɛ πɛ dτ = 1 dr π = 1 3 4πɛ 16π 4 = 1 3 4πɛ 16π = 1 3 4πɛ 8 = 1 3 4πɛ 8 = πɛ π π π dr dφ 4 ( 1 1 r 3 π3 4πɛ πɛ r sin θ π π π dφ( sin θ 5 ( sin θ + sin θ 5 [ cos θ ] cos θ π 5 dφ( r4 sin θ + 3r sin θ + sin θ which certinly hs the right units. Let s check it ginst... U = ɛ E(r dτ ll spce We ll need two integrls: one for outside the sphere nd one for inside. Outside the sphere our integrl is U outside = ɛ = ɛ dr π π ( 1 dφ 4πɛ ( 1 π dr 4πɛ dr 1 r ( 1 = πɛ 4πɛ = 1 1 4πɛ. r π r sin θ dφ sin θ r To this we should dd the integrl over the interior of the sphere. Inside we hve U inside = ɛ π π ( 1 r dr dφ 4πɛ 3 r sin θ 4

5 = ɛ ( 1 = ɛ 1 4πɛ 3 ( 1 4πɛ = πɛ. Adding the two prts gives dr π π π π dφ sin θ U inside + U outside = πɛ which grees with our nswer to prt (. dφr 4 sin θ Prolem 4 (Griffiths.33 - optionl This time our integrl is over concentric sphericl shells, ech of which crries chrge d = ρ4πr dr. To ring such n infinitesiml chrge to the surfce of sphere of rdius r tkes n mount of work dw = 1 (rd 4πɛ r where (r = ρ 4 3 πr3 is the mount of chrge lredy ssemled. If the totl chrge on the ssemled sphere is nd the finl rdius is then we hve ρ = 4. Therefore we my 3 π3 write 1 (r U = 4πɛ r d V = dr( 16π ρ r 5 3 = 1 16π ( 4πɛ 3 4 r 4 dr 3 π3 = πɛ. Thus, we get the sme result s ove, ut with much less work! Prolem 5 (Griffiths.39 As efore, let s use the result of Guss s Lw tht we hve previously found for the electric field etween two such coxil cylinders. The field is E(s = 1 4πɛ λ s ŝ so the potentil difference etween the inner cylinder nd outer shell is V = 1 λ 4πɛ s ds = 1 4πɛ λ ln. 5

6 The cpcitnce is C = V, nd the chrge in length l of the inner cylinder is = λl. Therefore, the cpcitnce per unit length of the coxil configurtion is Prolem 6 (Griffiths.4 C l = 1 l V = πɛ ln. dw = F dx = P Adx = ɛ E Aɛ. Let u e the energy per unit volume nd dτ the infinitesiml chnge in volume due to the shift of the pltes y n mount ɛ. Then Prolem 7 (Griffiths.35 de = udτ = ɛ E Aɛ. From the sic priniciples of conductors discussed in clss nd in the text we get: σ( = 4π σ( = 4π σ( = 4π. There re few wys to look t this, ut to e complete let s look t the integrl of E(r from infinity to the center of the concentric configurtion. Since V ( is lredy zero in this pproch we won t need to correct the guge. The totl potentil is V = = = 4πɛ E(r outside dr 1 4πɛ r dr ( E(r shell dr (dr 1 4πɛ r dr 6 E(r etween dr (dr E(r inside dr

7 Note tht this potentil is smller thn the potentil of metl sphere of rdius nd chrge y n mount 1 1. The conductive shell pprently screens some of the potentil. c Now we ground the outer shell. Clerly σ( doesn t chnge ecuse the grounding of the outer shell doesn t ffect the inner sphere. Also, in order for the field inside the shell to e zero we still need σ( to lnce σ(, so it doesn t chnge either. Finlly, we wnt the potentil of the shell itself to e zero. This mens tht the field outside the shell should e zero. Thus, σ( = 4π σ( = 4π σ( =. Our potentil then is just due to the field etween the two conductors since V = = = 4πɛ E(r outside dr (dr ( 1 1. (dr E(r shell dr 1 4πɛ r dr E(r etween dr (dr E(r inside dr Note tht now the potentil is even smller, since we nulled the effect of one region y grounding the outer shell. Prolem 8 (Griffiths.36 - optionl We cn repetedly pply Guss s Lw nd the principles of conductors to otin σ = σ = σ = + c Ner point it s E(r outside = 1 + 4πɛ r E(r = 1 ˆr 4πɛ r ˆr center 7

8 nd ner point it s E(r = 1 4πɛ r ˆr where r & r re mesured from points & respectively. d F = for oth nd, since they re screened from ech other y the conductive mteril etween them, nd the induced chrges on the surfce of ech cvity is uniformly distriuted. e Bringing third chrge ner the conductor (ut exterior to it mkes some of the free chrges in the conductor move round. However, the surfce chrges σ nd σ re such tht they cncel the fields due to nd only. Also, the net chrge on the sphere remins zero. So, even though the chrges rerrnge on the surfce to cncel the field of the third chrge, the totl chrge σ on the surfce remins constnt (though the surfce chrge density is certinly different from point to point thn it ws efore. The field outside the conductor is definitely different. It is the sum of the field from the conductor nd the field from the third point chrge fter the chrges on the surfce hve redistriuted. Now, there is still zero field inside the conductor, nd the discontinuity in the field t the oundries of the inner cvities depends only on σ there, which we sid doesn t chnge. Therefore, the field inside ech cvity is the sme, nd the force on nd is still zero. So, the only things tht chnge re the field outside the sphere nd the distriution of chrge on the surfce (ut not the totl chrge on the surfce. 8

Homework Assignment 6 Solution Set

Homework Assignment 6 Solution Set PHYCS 440 Mrch, 004 Prolem (Griffiths 4.6 One wy to find the energy is to find the E nd D fields everywhere nd then integrte the energy density for those fields. We know