, the action per unit length. We use g = 1 and will use the function. gψd 2 x = A 36. Ψ 2 d 2 x = A2 45
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1 Gbriel Brello - Clssicl Electrodynmics.. For this problem, we compute A L z, the ction per unit length. We use g = nd will use the function Ψx, y = Ax x y y s the form of our pproximte solution. First we compute Ψ = A x y y x y y, y x x y x x, Ψ = A y y x + x x y And note tht gψ = Ψ. The integrls compute to be v v gψd x = A 36 Ψ d x = A 45 Then the minimum of the difference of these is given when A = 5 4. Indeed, the second derivtive is positive t this vlue of A, so it is indeed minimum if tht wsn t obvious from the form. b. Figure : This is comprison of the true potentil in blue nd the pproximte potentil dshed The left plot shows y =.5 nd the right shows y =.5. The verticl xis is in units of ev 4πɛ
2 .4 Since this setup is π/ rdins rottionlly symmetric ech of the corner points nd ech of the edge points re equivlent, so we relly only need to compute one corner, one edge nd the middle. Let us lbel these points {, b, e} respectively. Recll the best-verge formul for the poisson eqution Φ p = Φ + h 5 g + h g c where Φ = 4 5 Φ c + 4 Φ s And the c nd s stnd for cross nd squre verge. the cross verge is the one tht mkes plus sign on the grid. Note, too tht h = /6. Now we re redy to get things lined up. Note the following tble: ij i+,j i,j+ i-,j i,j- i+,j+ i-,j- i+,j- i-,j+ b b e b e b b e b b b b After some fussing, the itertive equtions we get re: i+ = e i 6 + 4b i b i+ = i 5 + e i 5 + b i e i+ = 4b i Using the jcobin method with ɛ Φ = on ll sites initilly gives Note tht my nswers differ by fctor of 4π from the book s exct solution, becuse I computed Φ ɛ wheres Jckson gve 4πɛ Φ.
3 b. After ten itertions of the guss-seidel method using the sme initil vlues we get c. Itertion b e We fctored out /ɛ t the begining, so we re ctully reporting the vlue ɛ Φ. The following plot compres our pproximte solution to the exct one Figure : This is log plot of the proximte solution using the jcobin method blue nd the guss seidel method red compred to the exct solution lines. Unfortuntely the lowest exct vlue is obscured by the log plot, but it is there, I promise. The verticl xis is in units of Φ ɛ. As you cn see the guss-seidel method converges fster nd is more ccurte fter itertions. 3
4 .. The proprite imge chrge is q t position d,. Then the electric field t point, y is E, y = q dˆx + yŷ dˆx yŷ 4πɛ d + y 3 = q d ˆx πɛ d + y 3 This is relted to the surfce chrge by the eqution σ = ɛ E. So, Here is plot σ = qd d + y 3 π Figure 3: This shows the nondimentionlized chrge density with d = b. The force on the chrge is the sme s the force tht would result from the presence of the imge chrge. c. F = q 4πɛ d = q 6πɛ d The totl force cting on the plne cn be computed by computing the integrl F p = qd 8π ɛ = qd 8πɛ π d = 6πɛ q d u 3 du d + r 3 rdrdθ u d + r 4
5 d. The work is given by the integrl e. The potentil energy is given by the integrl W = F dl = q 6πɛ = q 6πɛ d d x dx d PE = F dl = q 4πɛ = q 8πɛ d d x dx This is fctor of two lrger, of course this is becuse s the chrge moves wy from the plne the chrge on the plne disperses, s if the imge chrge were moving wy t the sme rte. We usully define potentil energy by holding one chrge fixed while moving the other chrge in from infinity. In this cse, it is s if we re moving both chrges in from infinity t the sme time, resulting in fctor of two difference. One could lso think of it in terms of the fields. Potentil energy is stored in the field, but in the current sitution, only hlf the field ctully exists the other hlf is cncelled out by the conductor so there is only hlf s much energy stored in the field s in the corresponding sitution with rel chrge insted of conducting plne. f. q = e =.6 9 C d = A = m ɛ = F/m W = ev..9 The rection of the sphere rdius to uniform electric field E = Eẑ cn be modeled using imge chrges of chrge ± Q R t positions ± R with Q πɛ R = E nd tking the limit R. In the limit, this becomes n electric dipole, with dipole moment p = Q3 R ẑ = 4πɛ 3 Eẑ. Which gives surfce chrge density σ = 3ɛ E cosθ Unchrged σ q = σ + q 4π Chrged with chrge q Where θ is mesured from the positive z xis. Finlly, recll tht the force per unit re on differentil element is: F /A = σ ɛ 3 To find the force on ech hemisphere we must only integrte the bove formul over ech hemisphere. Note tht the force on the hemisperes will be equl nd opposite, since the chrge densities re equl nd opposite. Furthermore, note tht the component of force on surfce element not in the ẑ direction will cncel due to the zimuthl symmetry so we cn integrte insted only the z components of the force. Applying these observtions, the force required to keep the hemispheres from seprting is given by F = π π σ cosθ sinθdθ 4 ɛ 5
6 With 4 nd, we cn clculte the force for the two scenrios given:. F = 9πE ɛ π = 9 4 πe ɛ cosθ 3 sinθdθ b. For chrged sphere we get F q = π π = F + π π = F + q 4ɛ = F + q 3ɛ E cosθ + ɛ π q 4π 6ɛ E cosθ ɛ 6ɛ E cosθ + q 4ɛ 8π ɛ E cosθ sinθdθ q 4π + q 6π 4 cosθ sinθdθ q cosθ sinθdθ 4π And the totl force is given by F q = F + q q 4ɛ 8π ɛ E 5.7. The free spce greens function for three dimentions prt from n rbitrry multiplictive constnt is Gx, y, z; x, y, z = x x = x x + y y + z z By defining = x x + y x we cn write the -dimentionl free-spce greens function s Gx, y; x, y = + x dx x = z z Which hs the well-known solution prt from n inessentil, though infinite constnt. This constnt is hnd-wved wy by insted integrting to lrge vlue Z insted of to infinity, but I think this will be good enough. b. Where the lst line follows from the lw of cosines. Gx, y; x, y = logx x + y y = log + cosφ φ The generl solution to lplce s eqution is given by free-spce solutions for > nd < which re then ptched together t =. To this end, we define two new functions, R nd Φ, by the ssumption G = R, Φφ, φ. Let us compute the fourier coefficients of the Φ function for the lplce eqution sourced by delt functions. R, π e imφ Φφ, φ dφ R, π m e imφ Φφ, φ dφ = 4π π δ e imφ 6
7 The surfce terms in the second term of the l.h.s. cnceled becuse of the cyclic boundry conditions on Φ. We cn divide by e imφ nd pull out n overll fourier integrl to rrive t π R, m R, e imφ φ Φφ, φ dφ = 4πδ The fourier-trnsform fctor is constnt w.r.t. nd likewise the rest is constnt w.r.t. φ so w.l.o.g. we cn set the fourier trnsform fctor equl to constnt, we choose, so tht π Thus we rrive t the generl solution in the form e imφ φ Φφ, φ dφ = π Φφ, φ = R, m R, = 4π δ π eimφ φ where R m stisfies the bove eqution. G = π e imφ φ R m, c. To complete the solution we solve the bove eqution in the homogenious cse nd then ptch together the solution t =. The generl solution of the homogenious eqution is given in Jckson s R, = m m m Z\{} R, = + b log m = We need two solutions one for > nd one for <, we will distinguish them with subscript > nd < respectively on the coefficients. Then, the two full solutions re give by R <, = < + b < log + R >, = > + b > log + m Z\{} m Z\{} m< m m> m There re few conditions we cn impose right off the bt. Nmely, the solution needs to be well-behved t zero, nd be logrithmic s. With this we cn determine tht b < =, nd tht R < hs no negtive powers of nd R > hs no positive powers of. This leves us with R <, = < + m< m m> R >, = > + b > log + m> m m< Next, the solution must be symmetric. Tht is to sy tht if < then R <, = R >, nd likewise, for >, R >, = R <,. These conditions stte tht < + m< m = > + b > log + m> m m> m< > + b > log + m> m = < + m< m m< m> 7
8 Note tht the solutions for vrious m must independently obey this symmetry, since they re ll linerly independent. Then this tells us tht < = A log, > = nd b > = A, with A yet to be determined constnt. In ddition it must be tht m> = A m m nd m< = A m m with A m constnts tht will be determined by considering the derivtive discontinuity. So tht we finlly hve the solutions, us to n overll constnt in ech term R <, = A log + m> R >, = A log + m> A m m A m m The lst considertion to mke is tht the difference in the derivtives t = should be 4π/. evluted with respect to ner = re The derivtive R <,, = m> A m m R >,, = A m> A m m Ech term stisfies the discontinuity condition individully so we get tht A = 4π nd tht A m = π m. So tht the two solutions cn be written in the form Which cn be conveniently expressed s R <, = π log + 4π m m> R >, = π log + 4π m m> R <, = π log > + 4π m> m m m m < > With < > the lesser greter of the two rdii. Lstly we cn plug this bck into our originl solution. Note tht the full solution for ech m of the rdil eqution hs positive nd negtive power of m. We threw out the poorly-behved powers of m, but for the well behved ones we get contributions, one from the positive m in the series, nd one from the negtive m in the bove series. Thus this mmounts to dding the positive-m term to the negtive m term, cusing the imginry prts of the exponentil to cncel s needs to be the cse to itnerprit this s potentil, leving fctor of cos, so in the end we get G = log > + m < > m cosmφ φ.. Using the greens function of problem.7 the potentil is given s σ, φ Φ, φ = G, φ;, φ ddφ 4πɛ = λ 3 n log 4πɛ > + n= m= m < > m cosmφ nπ/ 8
9 Where > < is the greter lesser of nd, but since the log term hs no φ dependence it cncels when the sum over n is performed, thus we cn write the solution s Φ, φ = λ πɛ m= m < > m 3 n cosmφ nπ/ n= for moment lets focus on the sum of cosines. Writing this out in full nd mnipulting we find tht cosmφ cosmφ mπ/ + cosmφ mπ/ cosmφ 3mπ/ = cosmφ cosmφ mπ/ + m So tht we cn write the solution s = { cosmφ if m = 4k +, k Z else Φ, φ = λ πɛ k= k + < > 4k+ cos4k + φ b. By mnipulting our solution for Φ we quickly rrive t Φ, φ = λ 4k+ > Re e iφ πɛ k + < Note the identity pge 75 of jckson Where for our cse we hve Z = >e iφ < Φ = nodd so tht k= Z n n = + Z log Z λ + > Re log < e iφ πɛ > < e iφ = λ Re log πɛ i > < e iφ + i > < e iφ > < e iφ + > < e iφ In the interior we hve tht < = nd > = nd the solution tkes the form Φ = λ z iz + i Re log = Rewz πɛ z z + In the exterior < = nd > = nd we hve Φ = λ iz + iz Re log πɛ z + z But we cn multiply the interior by i = without chnging the rel prt since the rel prt of log depends only on the mgnitude of z. 9
10 Φ = λ i + zi z Re log πɛ z + z = λ z ii + z Re log πɛ z z + = Rewz Thus, we cn identify Φ = Rewz So tht the potentil is exctly the rel prt of the funtion wz defined in Jckson. Now, s is stted in problem.3 of Jckson, the potentil of n infinite line chrge is the logrithm of the distnce. This solution reflects tht beutifully by converting the crtesin components of position into complex vrible nd then tking the rel prt. The four fctors in the rgument of teh logrithm correspond exctly to two positive line chrges t ±i, or ±ŷ, nd two negtive line chrges t ±, or ±ˆx, where x nd y hve been identified with the rel nd imginry xes respectively. c. The crtesin components of the field re given by Note tht for < E x = cosθ Φ sin θ Φ φ E y = sinθ Φ + cos θ Φ φ So the full expressions re E x = λ πɛ k= = λ πɛ k= E y = λ πɛ k= = λ πɛ k= Φ = λ πɛ Φ = λ πɛ k= 4k+ cos4k + φ k= 4k+ sin4k + φ 4k+ cos4k + φ cosθ + sinθ sin4k + φ 4k+ cos4k + φ 4k+ cos4k + φ sinθ cosθ sin4k + φ 4k+ sin4k + 3φ Keeping only the first two terms, we get E x = λ πɛ E y = λ πɛ cosφ + sin3φ + 5 cos5φ 5 sin7φ
11 When y =, φ = or π nd the x component of the electric field is given by E x = ± λ πɛ ± 5 with the ± corresponding to the negtive or positive xis respectively. The reltive strength of the k = contribution nd k = contribution is k = k = = ± 4.6 Consider the -dimentionl region, φ bounded by conducting surfces held t zero potentil.the lrge vlue of the potentil Φ is determined by some unknown chrge distribution t >> nd thus this my be considered s lplce problem with dirichlet boundry conditions.. The generl solution to lplce s eqution in polr coordines is given, fter seprtion of vribles, by Rp = ν + b ν, Ψφ = A cosνφ + B sinνφ ν > Rp = + b log, Ψφ = A + B φ ν = Given the boundry conditions we cn conclude tht A = B = A = nd tht ν = n π. So, only the sin term of the ν > solution will pper. Lstly we hve the condition tht Φ, φ = which gives us the condition b. Which gives us the generl form of the solution Φ, φ = n Z Keeping only the lowest nonvnishing term we get ν + b ν = b = ν C n nπ Φ, φ = C π nπ π nπ sin nπ φ π sin π φ By which we cn clculte the electric fields t the surfces vi the eqution Eˆn = Φ ˆn to be E, φ = C π π π + E φ, φ = Φ ˆn = Φ φ = C π π π + sin π φ π π cos π φ
12 With E zero t φ =, nd E φ = t =. Then, the surfce chrge density is given by σ = ɛ E, or c. πɛ σ = = C π sin π φ πɛ σ φ= = C π π Consider = π. The liner chrge density looks like this π = σ φ= Figure 4: This plot shows the liner chrge density. Note tht the chrge density on the cylinder goes s sinφ but since it is hlf-circle, it works out tht the liner chrge density is constnt! Of course, this is plot of the non-dimentionlized function with = Now lets convert the polr vectors into crtesin vectors! I will use ˆx prllel to the plne nd ŷ perpendiculr to the plne. Then the electric field in x nd y coordintes, for lrge s is required to be fr wy from the cylinder, is given by Ex E y = E cosφ E φ sinφ E sinφ + E φ cosφ sinφ cosφ cosφ sinφ = C cosφ + sinφ = E So, indeed the field is constnt nd perpendiculr to the plne fr from the cylinder. Now, we hve the chrge density on the cylinder, we just need to integrte over φ. This works out to be φ σ = = E ɛ sinφ = 4E ɛ. Now, the surfce chrge density on n infinite conducting plne is σ = E ɛ so indeed the surfce chrge on the cylinder is twice tht contined on strip on width on n infinite conducting plne. Of course, ll the surfce chrge densities re ctully chrge density per unit length in the z direction. Next we need to show tht the extr chrge density is tken from the surrounding re. To show this we will tke the difference between the surfce chrge density of this system nd tht of n infinite conducting plne, nd integrte this from = to, nd multiply by, of course. This works out to be:
13 σ φ σ d = E ɛ = E ɛ d Exctly wht we needed! Now, just for completeness ske, note tht the difference between these two surfce chrges goes to zero for lrge so indeed the sttment tht the chrge is tken from nerby points on the plne is ccurte. This lso nswers the lst question regrding the surfce chrge on wide strip being the sme wether or not the bump is present. Since the difference in surfce chrge densities between the two cses goes to zero s becomes lrge, the totl surfce chrge within tht lrge strip will be the sme between the two cses, since the totl chrge density on the two surfces is the sme, nd the surfce chrge density t lrge is the sme. 3
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