Theoretische Physik 2: Elektrodynamik (Prof. A.-S. Smith) Home assignment 4

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1 WiSe Prof. Dr. A.-S. Smith Dipl.-Phys. Ellen Fischermeier Dipl.-Phys. Mtthis Sb m Lehrstuhl für Theoretische Physik I Deprtment für Physik Friedrich-Alexnder-Universität Erlngen-Nürnberg Theoretische Physik : Elektrodynmik (Prof. A.-S. Smith Home ssignment 4 Problem 4.1 Green s reciproction theorem Green s reciproction theorem sttes tht if Φ is the potentil due to volume-chrge density ρ within volume nd surfce-chrge density σ on the conducting surfce S bounding the volume, nd Φ is the potentil due to nother chrge distribution ρ nd σ, then ρφ d + σφ ds = ρ Φ d + σ Φ ds. S Two infinite, grounded, prllel conducting plnes re seprted by distnce d. A point chrge q is plced between the plnes. Use the reciproction theorem of Green to prove tht the totl induced chrge on one of the plnes is equl to ( q times the frctionl perpendiculr distnce of the point chrge from the other plne. Hint: As your comprison electrosttic problem with the sme surfces choose one whose chrge densities nd potentil re known nd simple. b Consider potentil problem in the hlf-spce defined by, with Dirichlet boundry conditions on the plne = (nd t infinity. (i Write down the pproprite Green s function G( x, x. (ii Suppose tht the potentil on the plne = is specified to be Φ = inside circle of rdius centered t the origin, nd Φ = outside tht circle. (1 Find n integrl expression for the potentil t the point P specified in terms of the cylindricl coordintes (ρ, φ,. ( Show tht, long the xis of the circle (ρ =, the potentil is given by ( Φ = 1 + (3 Show tht t lrge distnces (ρ + the potentil cn be expnded in power series in (ρ + 1, nd tht the leding terms re Φ = 3 (ρ + 3/ 1 4(ρ + + 5(3ρ + 4 ] 8(ρ erify tht the results of prts ( nd (4 re consistent with ech other in their common rnge of vlidity. S

2 Problem 4. Method of imges Using the method of imges, discuss the problem of point chrge q inside hollow, grounded, conducting sphere of inner rdius. Find (i the potentil inside the sphere; (ii the induced surfce-chrge density; (iii the mgnitude nd direction of the force cting on q. Is there ny chnge in the solution if the sphere is kept t fixed potentil? If the sphere hs totl chrge Q on its inner nd outer surfces? b A conducting sphere is plced in uniform externl field. Clculte the potentil Φ nd the surfce chrge density σ. Problem 4.3 Hollow cube A hollow cube hs conducting wlls defined by six plnes x =, y =, =, nd x =, y =, =. The wlls = nd = re held t constnt potentil. The other four sides re t ero potentil. Find the potentil Φ(x, y, t ny point inside the cube. b Evlute the potentil t the center of the cube numericlly, ccurte to three significnt figures. How mny terms in the series is it necessry to keep in order to ttin this ccurcy? c Find the surfce-chrge density on the surfce =. Due dte: Tuesdy,

WiSe 1 8.1.1 Prof. Dr. A.-S. Smith Dipl.-Phys.

Smith Solutions to Home ssignment 4 Solution of Problem 4.

3 WiSe Prof. Dr. A.-S. Smith Dipl.-Phys. Ellen Fischermeier Dipl.-Phys. Mtthis Sb m Lehrstuhl für Theoretische Physik I Deprtment für Physik Friedrich-Alexnder-Universität Erlngen-Nürnberg Theoretische Physik : Elektrodynmik (Prof. A.-S. Smith Solutions to Home ssignment 4 Solution of Problem 4.1 Green s reciproction theorem Let us plce the two conducting plnes perpendiculrly to the -xis, one through the origin, nd the other through the point = d, while the chrge q is t the distnce from the left plne (see Figure 1. The plnes re grounded Φ = = Φ =, =d while the point chrge density is ρ( x = qδ(xδ(yδ(. On the plnes, the surfce chrge densities σ 1, σ re being induced. Fig. 1: A chrge q between conducting plnes. Consider the following problem: the chrges q nd q re uniformly distributed on the two infinitely long, conducting plnes. This configurtion describes prllel plte cpcitor, for which we know the potentil t generl point to be (Figure 1 Φ = ( d (where is the potentil difference between the pltes d Φ = = Φ =d =, We now use Green s reciproction theorem. Here, the unprimed vribles represent the quntities in the cse of there being chrge between the two plnes, nd the primed vribles represent those in the cse of the prllel plte cpcitor configurtion, without chrge present in between. We get ρφ d + σφ ds = ρ Φd + σ ΦdS S S

ρ Φd = (s ρ = S σ ΦdS = (s the plnes re grounded. Putting the result for these four terms bck into the Green s reciprocity reltion, we get where Q is the chrge on the plne =.

4 Now we clculte the different terms in the bove eqution. ρφ d = qδ(xδ(yδ( q ( d d = ( d d d σφ ds = σ 1 Φ ds + σ Φ ds = σ 1 ds = Q S = =d }{{ } = = where Q is the totl chrge induced on the = plne, which we wnt to find. ρ Φd = (s ρ = S σ ΦdS = (s the plnes re grounded. Putting the result for these four terms bck into the Green s reciprocity reltion, we get where Q is the chrge on the plne =. Q = q ( d, d b (i The Green s function for the specified region is equl to the solution for the potentil in the problem of the unit chrge ner n infinite grounded conducting plne. If the unit chrge is t the point with coordintes (x, y,, nd its imge is t the point (x, y,, then the Green s function reds (ii (1 G( x, x = 1 (x x + (y y + ( 1 (x x + (y y + ( +. (1 Rewrite the Green s function in cylindricl coordintes x = ρ cos φ y = ρ sin φ = Fig. : Unit chrge ner conducting, grounded plne. (x x + (y y + ( = (ρ cos φ ρ cos φ + (ρ sin φ ρ sin φ + ( G( x, x = = ρ + ρ ρρ cos(φ φ + ( 1 ρ + ρ ρρ cos(φ φ + ( 1 ρ + ρ ρρ cos(φ φ + ( +. The generl solution of the Dirichlet problem without the spce distribution of chrge ρ(x reds Φ( x = 1 Φ( x G( x, x 4π S n ds.

5 The norml is directed outside of the volume nd therefore n = e (see fig. 3. The derivtion over the norml t the plne = is G( x, x n = = n G( x, x = G( x, x = = { ( 1 = ( ( 1 ρ + ρ ρρ cos(φ φ + ( ] 3/ } ( 1 ( + ρ + ρ ρρ cos(φ φ + ( + ] 3/ = ρ + ρ ρρ cos(φ φ + ] 3/ At the surfce t infinity the potentil is ero. The finl solution is then Φ(ρ, φ, = π π dφ = ρ dρ ρ + ρ ρρ cos(φ φ +. ( ] 3/ Note tht the potentil distribution is imuthlly symmetric on the edge nd hence the potentil is independent of the coordinte φ. So, in the bove solution we cn put φ =. ( For ρ =, the solution in ( becomes Φ( = π dφ ρ dρ ( π ρ + ] 3/ = 1. + (3 We will expnd the solution ( into its Tylor series in powers of ρ +. For ρ + the expnsion will converge fst. ρ ρ + ρ ρρ cos φ + ] 3/ ρ 1 = ρ + ] 3/ 1 + ρ ρρ cos φ = ρ ρ + ] 3/ ρ + ] 3/ { 1 3 ρ ρρ cos φ ρ + ( ρ ρρ cos φ ρ + Integrtion of the first term gives π Integrtion of the second term gives 3 (ρ + ρ + ] 3/ Integrtion of the third term gives π +...} dφ. Fig. 3: Are of the disc shpe t the potentil on the conducting, grounded plne. ρ dρ ρ + ] 3/ = π ρ + ] 3/ (3 dφ ρ (ρ ρρ cos φ dρ 3π 4 = 4(ρ + ρ + ] 3/ (4 π 15 8(ρ + ρ + ] 3/ dφ ρ (ρ ρρ cos φ dρ 15 ( π = 8(ρ + ρ + ] 3/ π 4 ρ (5 3

6 utting in the results from (3, (4 nd (5 into (, we get 1 Φ(ρ, = (ρ + 3/ 3 4(ρ + + 5(3ρ + 4 8(ρ ]. (6 erifiction. Does the solution (6 for ρ = mtch with the one we got in prt ( for? From (6, Φ(ρ =, = ] The solution from prt ( for becomes + = Φ( = = = 1 1 ( ] ] Hence we hve shown tht the solutions mtch in the specified rnge. Solution of Problem 4. Method of imges (i The solution for the potentil inside sphericl cvity is of the form (Fig. 4. Φ( x = q r + d rd cos θ + q r + d rd cos θ where q is the vlue nd d is the position of the imge chrge. Both q nd d hve to be evluted from the boundry condition Φ( x r= =. Using the ssumed form of the potentil with the boundry condition, Φ( x r= = q + d d cos θ + q + d d cos θ =, = q + d d cos θ = q + d d cos θ = q ( + d q ( + d ] + cos θ dq d q ] = Functions 1 nd cos θ re linerly independent, thus the coefficients multiplying them hve to be both in order for the upper equlity to be stisfied. From the two obtined equtions we now find q nd d : q = q d n d q q d d = d. The solution for the potentil inside the sphere is then q Φ( x = r + d rd cos θ d Fig. 4: Point chrge inside of the sphericl cvity in the conductor. q r + 4 r d d cos θ. 4

7 (ii We re going to evlute the induced surfce chrge density using the boundry condition for the norml component of the electric field. Here we lso hve to tke into ccount the direction of the norml (Fig. 4. σ = 1 E 4π n ( n = e r r= σ = 1 4π Φ r = q 4π d = q r= 4π 1 r d cos θ (r + d rd cos θ 3/ + rd d cos θ d 3 (r + 4 r d d cos θ 1 + d ( ] 1 d d cos θ ] 3/, d <. As check, we cn clculte the induced chrge on the inner surfce: π π q ind = σ(θds = dφ dθ sin θ σ(θ Sphere = q ( ] π sin θdθ 1 d d ( ] 3/ 1 + d d cos θ = q ] ( d ( 1 1 d d 1 + d d cos θ = q ( ] ( d 1 d + d d d = qd ( ] d 1 d d = q. We could hve come to the sme conclusion using Guss lw nd by inspecting surfce tht lies completely inside the conductor nd surrounds the sphericl cvity. The electric field inside the conductor must equl, which mens tht the totl chrge inside tht surfce is. The induced chrge is then equl to q ind = q. (iii The force on the chrge q is clculted using the imge chrge method. qq F = (d d = q 3 (1 d d. The direction of the force is prllel to the direction of the unit vector e. The existence of n induced chrge with sign opposite to tht of the chrge q on the inner surfce implies tht the force is ttrctive. (iv If the sphere is kept t fixed potentil, the solution for the potentil chnges to the following form: Φ = Φ = Φ +. π 3/ r= 5

8 This solution stisfies the boundry condition Φ r= =. Thus, the surfce chrge density nd the force on the chrge q remin the sme. If there is chrge q inside the sphericl cvity nd the sphere is insulted, then one cn obtin the solution in the following wy. First we ground the sphere. The solution of tht problem is lredy known: chrge q is induced on the inner surfce, while on the outer surfce the chrge equls. Now we insulte the sphere nd we dd chrge Q + q to it. The totl chrge on the inner nd outer surfces of the conductor is now Q. The dded chrge Q + q is distributed uniformly on the outer surfce of the sphericl conductor. The solution inside the sphere is now chnged to Φ = Φ = Φ + Q + q b where b is the outer rdius of the sphericl conductor. This chnge in the potentil does not hve ny influence on the force exerted on chrge q nd on the surfce chrge density of the inner surfce. Note tht the chrge Q + q cnnot distribute on the inner surfce, otherwise the Guss lw for the surfce tht lies inside the conductor would not hold. b Suppose tht two opposite chrges ±Q re locted on the -xis t = ±R. Within sphere of rdius much smller thn R, the field is pproximtely constnt. If we now tke the limits R nd Q such tht kq R is held constnt t E, we obtin uniform field E t the origin. Now we let the conducting sphere be plced t the origin, nd llow R to be finite. In this cse, both Q nd Q hve imge chrges inside the sphere: imge of Q imge of Q = q 1 = Q R = q = Q R t = R t = R The potentil is the superposition of the potentils due to ll the chrges Fig. 5: Configurtion of the system. ] Φ( r = k Q q 1 r r(q + r r(q1 + q 1 r r(q Q r r( Q Thus, it follows tht +Q = R, r E θ Q =R Φ( r k = Q (r + R + rr cos θ 1/ Q (r + R rr cos θ 1/ Q + R 1/ + (r + 4 R + r R cos θ Q R (r + 4 R r R cos θ 1/ Since r R: ( r + R + rr cos θ 1/ = R 1 (1 + r R + r 1/ R cos θ = R (1 1 r R r R 1/ (r + 4 R + r R cos θ = r (1 1 4 r R cos θ +... rr cos θ

potentil relted to the chrge induced on the surfce. r σ = 1 φ 4πk r = 1 φ 4πk r r= = 1 ] E (1 + 3 4πk 3 cos θ. Thus, σ = 3 4πk E cos θ.

9 The potentil now becomes Φ( r = k Q 1 r R R r R k Q rr cos θ 1 + r R r R cos θ 1 4 r R rr cos θ 1 + = k Qr Q cos θ + k R rr 4 r R rr cos θ +... = k Q R ] +... ] rr cos θ +... (r 3 r cos θ ( ll Q terms, where n >, Rn re dropped In the limit R, Q ( Φ( r = E r 3 r cos θ, where E r cos θ = E, which is equl to the potentil of the uniform field, nd where 3 E cos θ is the potentil relted to the chrge induced on the surfce. r σ = 1 φ 4πk r = 1 φ 4πk r r= = 1 ] E ( πk 3 cos θ. Thus, σ = 3 4πk E cos θ. The surfce potentil of this chrge density vnishes, so there is no difference between grounded nd n insulted sphere. Solution of Problem 4.3 Hollow cube We hve to solve Lplce s eqution in rectngulr coordintes, Φ x + Φ y + Φ =, (7 inside cube. A solution of this prtil differentil eqution cn be found in terms of three ordinry differentil equtions, ll of the sme form, by the ssumption tht the potentil cn be represented by product of three functions, one for ech coordinte Φ(x, y, = X(xY (yz(. (8 Fig. 6: Substitution of (8 into (7 nd division by X(xY (yz( yields 1 d X X(x dx + 1 d Y Y (y dy + 1 d Z Z( d =. 7

10 If the bove eqution is to hold for rbitrry vlues of the independent coordintes, then ech of the three terms must be seprtely constnt: 1 d X X(x = α dx 1 d Y Y (y = β dy (9 1 Z( d Z d = γ where α, β nd γ re complex numbers, such tht α + β = γ. The solutions of the three differentil equtions (9 re exp(±iαx, exp(±iβy nd exp(± α + β. The potentil (8 cn thus be built up from the product solutions Φ = exp(±iαx exp(±iβy exp(± α + β. The coefficients α nd β cn be determined from the boundry conditions on the potentil. In the problem we re given tht the wlls = nd = re held t constnt potentil while the other four fces of the cube re t ero potentil. We will split the problem into two superposble prts nd first consider the cse in which the fce = is t the potentil Φ(x, y, = =, nd ll the other fces re t the potentil ero. Strting from the requirement tht Φ = for x =, y =, = it is esy to see tht the required forms of X, Y, Z re X = sin(αx Y = sin(βy Z = sinh( α + β In order tht Φ = t x =, y = it is necessry tht α = mπ nd β = nπ. With the definitions α m = mπ ; β n = nπ ; γ mn = π m + n. we cn write the prtil potentil Φ mn Φ mn = sin(α m x sin(β n y sinh(γ mn. The potentil cn be expnded in terms of these Φ mn with initilly rbitrry coefficients (to be chosen to stisfy ll boundry conditions Φ(x, y, = n,m=1 A mn sin(α m x sin(β n y sinh(γ mn. Similrly we find the expression for the cse in which the side = is t the potentil. In conclusion, the generl solution for the interior of the cube in which the fce = or is t the potentil Φ(x, y, =, nd ll the other fces re t the potentil ero (see figure 6 reds Φ (1 (x, y, = Φ ( (x, y, = = n,m=1 n,m=1 = n,m=1 A mn sin(α m x sin(β n y sinh(γ mn for = A mn sin(α m x sin(β n y sinh(γ mn (1 B mn sin(α m x sin(β n y sinh(γ mn ( for = n,m=1 B mn sin(α m x sin(β n y sinh(γ mn (11 8

11 The equtions (1 nd (11 comprise of double Fourier series. From (1, the coefficient A mn sinh(γ mn of the Fourier series is given by 4 ( mπ ( nπ A mn sinh(γ mn = sinh(π dx dy sin m + n x sin y 4 ( mπ ( nπ A mn = sinh(π dx dy sin m + n x sin y 4 = sinh(π ( 1( 1 m 1] m + n mπ ( 1( 1 n 1]. nπ Similrly, A mn = B mn = { ; if m or n even 16/π mn sinh(π m + n ] ; if m nd n both odd { ; if m or n even 16/π mn sinh(π m + n ] ; if m nd n both odd The totl solution of the problem is obtined by the superposition of Φ (1 (x, y, nd Φ ( (x, y, Φ(x, y, = Φ (1 (x, y, + Φ ( (x, y, = 16 1 π mn sin(α mx sin(β n y sinh(γ mn + sinh(γ mn ( ] sinh(π. (1 m + n m,n odd Using the trigonometric identities sinh(x + sinh(y = sinh( x + y cosh( x y sinh(x = sinh(x cosh(x we cn write π ] sinh(γ mn + sinh(γ mn ( = sinh m + n sinh(π m + n = sinh π m + n ] cosh π cosh m + n ( ] π ] m + n This simplifies the frction in (1. Also, since the sum in (1 is over only the odd positive integrl vlues of m nd n, we cn mke it run over ll the positive integers by replcing m nd n by m 1 nd n 1, respectively. Therefore, Φ(x, y, = 16 π m,n=1 π cosh (m 1π 1 sin (m 1(n 1 cosh π (m 1 + (n 1 ( ]}. ] ] x sin (n 1π y ] (m 1 + (n 1 b The coordintes of the cube s center re (/, /, /, so the potentil is ] ] Φ,, = 16 1 sin (m 1π sin (n 1π π (m 1(n 1 m,n=1 cosh π ]. (13 (m 1 + (n 1 The numericl results of the summtion of the first m n terms of the expnsion (13 re given in Tble 1 in units of. 9

12 Tble 1: The potentil t the center of the box in units of. m, n We see tht it is enough to consider the first 9 terms (m, n = 3 to chieve n ccurcy up to the third deciml plce. c The surfce chrge density on the side = is σ = = 1 Φ 4π = = 4 π (m 1 + (n 1 sin π 3 m,n=1 (m 1(n 1 cosh π π ]} sinh (m 1 + (n 1 { = 4 1 π (m π (n 1 tnh m,n=1 ] ]} (m 1π (n 1π sin x sin y. ] (m 1π x sin (n 1π ] (m 1 + (n 1 ] y (m 1 + (n 1 ] 1

Problem Set 3 Solutions

Msschusetts Institute of Technology Deprtment of Physics Physics 8.07 Fll 2005 Problem Set 3 Solutions Problem 1: Cylindricl Cpcitor Griffiths Problems 2.39: Let the totl chrge per unit length on the inner