Jackson 2.26 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell


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1 Jckson 2.26 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: The twodimensionl region, ρ, φ β, is bounded by conducting surfces t φ =, ρ =, nd φ = β held t zero potentil, s indicted in the sketch. At lrge ρ the potentil is determined by some configurtion of chrges nd/or conductors t fixed potentils. β () Write down solution for the potentil, tht stisfies the boundry conditions for finite ρ. (b) Keeping only the lowest nonvnishing terms, clculte the electric field components E ρ nd E φ nd lso the surfcechrge densities σ(ρ, ), σ(ρ, β), nd σ(, φ) on the three boundry surfces. (c) Consider β = π ( plne conductor with hlfcylinder of rdius on it). Show tht fr from the hlfcylinder, the lowest order terms of prt b give uniform electric field norml to the plne. Sketch the chrge density on nd in the neighborhood of the hlfcylinder. For fixed electric field strength fr from the plne, show tht the totl chrge on the hlfcylinder (ctully chrge per unit length in the z direction) is twice s lrge s would reside on strip of width 2 in its bsence. Show tht the extr portion is drwn from regions of the plne nerby, so tht the totl chrge on strip of width lrge compred to is the sme whether the hlfcylinder is there or not. SOLUTION: We cn think of the chrges wy from this rounded corner s externl to the problem, so tht they simply crete some boundry condition on the potentil t lrge ρ. The region ner the corner hs no chrges nd is described by the Lplce eqution. 2 = In polr coordintes, the Lplce eqution becomes: 2 2 = 2
2 Using the method of seprtion of vribles, the generl solution is found to be,= b ln A B b A e i B e i, Apply the boundry condition,== = b ln A b A B, To hold true for ll vlues of ρ we must hve A = nd B = A. The solution now becomes:,= b ln B b A sin, Apply the boundry condition,== = b ln B b A sin, To hold true for ll vlues of ρ we must hve B = nd = n where n =, 2, 3... which gives,= n= n n / b n n / A n sin n Apply the boundry condition =,= = n= n n / b n n / A n sin n To hold true for ll vlues of ngles we must hve = n n / b n n / which leds to: b n = n 2n/ The solution t this point tkes the form (where severl constnt fctors hve been combined with the lst remining undetermined constnt):,= n= An / n n / sin n (b) Keeping only the lowest nonvnishing terms, clculte the electric field components E ρ nd E φ nd lso the surfcechrge densities σ(ρ, ), σ(ρ, β), nd σ(, φ) on the three boundry surfces. The electric field defined in terms of the electric potentil is: E=
3 In polr coordintes this becomes: E= or presented differently: E = nd E = E = A n= n n / n / sin n n E = A n n= n / n / sin n The lowest nonvnishing term is: E = A / / sin The other component cn lso be esily done: E = n= An n / n/ sin n n E = An n= n / n / cos n The lowest nonvnishing term is: E = A / / cos The totl electric field with ll the terms is then: E= E E n E= [ An n= n / n/ sin n n/ n / cos n ] The surfce chrge density σ(ρ, ) is found using the pillbox Gussin surfce, which yields:,=[ E n] n=n
4 ,=[ E ] =,=[ E ] =,=[ A / / cos ]=,= A / / The surfce chrge density σ(ρ, β) is found in similr mnner:,=[ E ] =,=[ E ] =,=[ A / / cos ]=,= A / / The chrge densities on the two flt surfces re equl, which is wht we would expect becuse of the symmetry of the problem. The surfce chrge density σ(, φ) obeys:,=[ E ] =,=[ E ] =,=[ A / / sin ]=,= A 2 sin
5 (c) Consider β = π ( plne conductor with hlfcylinder of rdius on it). Show tht fr from the hlfcylinder, the lowest order terms of prt b give uniform electric field norml to the plne. Sketch the chrge density on nd in the neighborhood of the hlfcylinder. For fixed electric field strength fr from the plne, show tht the totl chrge on the hlfcylinder (ctully chrge per unit length in the z direction) is twice s lrge s would reside on strip of width 2 in its bsence. Show tht the extr portion is drwn from regions of the plne nerby, so tht the totl chrge on strip of width lrge compred to is the sme whether the hlfcylinder is there or not. If β = π, the solution for the totl electric field reduces to: n E= [ A n n= n n sin n n n ] cos n the lowest order term is: [ E= A 2 sin 2 ] cos Fr wy from the hlfcylinder, ρ >>, nd thus >> (/ρ) nd >> (/ρ) 2 so tht E= A [ sin cos ] E= A j This is just uniform electric field in the y direction, norml to the conducting plne. The chrge density on the hlf cylinder hs the form,= sin where = A 2 nd on the sides:,=,= 2 2
6 The totl chrge on the hlfcylinder (per unit length in the z direction) is 2 Q hlfcyl = A sin d Q hlfcyl = A 4 If the sme electric field were used ( E= A / j s derived bove) nd the cylinder were replced with strip 2 wide, the chrge density on the strip would be y==[ E j ] y= y== A / The totl chrge (per unit length in the z direction) on the strip is: Q strip = 2 Q strip = A 2 So tht Q strip = 2 Q hlfcyl Consider lrger strip with width l tht includes the centrl region lredy focused on. The totl chrge with the hlfcylinder included is:
7 l Q =2 A 2 d Q hlfcyl Q = 2A l 2 A /l /Q hlfcyl Q =2 A [ l l ] For l >> : Q = 2l A The totl chrge without the cylinder is = 2l =[ E y ] y= 2l = A / 2l = 2 l A It now becomes pprent tht Q =. The totl chrge over lrge region is the sme with nd without the hlfcylinder. This mens tht the extr chrge on the hlfcylinder is drwn from the regions nerby on the plnes.
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