Solutions to Problems in Jackson, Classical Electrodynamics, Third Edition Homer Reid December 8, 999 Chapter : Problems - Problem A line charge with linear charge density τ is placed parallel to, and a distance R away from, the axis of a conducting cylinder of radius b held at fixed voltage such that the potential vanishes at infinity Find (a the magnitude and position of the image charge(s; (b the potential at any point (expressed in polar coordinates with the origin at the axis of the cylinder and the direction from the origin to the line charge as the x axis, including the asymptotic form far from the cylinder; (c the induced surface-charge density, and plot it as a function of angle for R/b=,4 in units of τ/b; (d the force on the charge (a Drawing an analogy to the similar problem of the point charge outside the conducting sphere, we might expect that the potential on the cylinder can be made constant by placing an image charge within the cylinder on the line conducting the line charge with the center of the cylinder, ie on the x axis Suppose we put the image charge a distance R < b from the center of the cylinder and give it a charge density τ Using the expression quoted in Problem 3 for the potential of a line charge, the potential at a point x due to the line charge and its image is Φ(x = τ R ln 4ɛ x τ R ln Rî 4ɛ x R î
Homer Reid s Solutions to Jackson Problems: Chapter = τ ln x R î 4ɛ x Rî We want to choose R such that the potential is constant when x is on the cylinder surface This requires that the argument of the logarithm be equal to some constant γ at those points: or x R î x Rî = γ b + R R b cos φ = γb + γr γrb cos φ For this to be true everywhere on the cylinder, the φ term must drop out, which requires R = γr We can then rearrange the remaining terms to find R = b R This is also analogous to the point-charge-and-sphere problem, but there are differences: in this case the image charge has the same magnitude as the original line charge, and the potential on the cylinder is constant but not zero (b At a point (ρ, φ, we have Φ = For large ρ, this becomes τ 4ɛ ln ρ + R ρr cos φ ρ + R ρr cos φ Φ τ ln R ρ cos φ 4ɛ R ρ cos φ Using ln( x = (x + x / +, we have Φ τ (R R cos φ 4ɛ ρ = τ R( b /R cos φ ɛ ρ (c Φ σ = ɛ ρ r=b = τ [ b R cos φ 4 b + R br cos φ b R cos φ b + R br cos φ [ = τ b b R cos φ b + b4 R b3 R cos φ b R cos φ b + R br cos φ
Homer Reid s Solutions to Jackson Problems: Chapter 3 Multiplying the first term by R /b on top and bottom yields [ σ = τ R b b R + b br cos φ = τ [ R b b R + b br cos φ (d To find the force on the charge, we note that the potential of the image charge is Φ(x = τ C ln 4ɛ x R î with C some constant We can differentiate this to find the electric field due to the image charge: E(x = Φ(x = τ 4ɛ ln x R î = τ 4ɛ (x R î x R î The original line charge is at x = R, y =, and the field there is E = τ ɛ R R î = τ R ɛ R b î The force per unit width on the line charge is F = τe = τ ɛ R R b tending to pull the original charge in toward the cylinder Problem Starting with the series solution (7 for the two-dimensional potential problem with the potential specified on the surface of a cylinder of radius b, evaluate the coefficients formally, substitute them into the series, and sum it to obtain the potential inside the cylinder in the form of Poisson s integral: Φ(ρ, φ = Φ(b, φ b ρ b + ρ bρ cos(φ φ dφ What modification is necessary if the potential is desired in the region of space bounded by the cylinder and infinity?
Homer Reid s Solutions to Jackson Problems: Chapter 4 Referring to equation (7, we know the b n are all zero, because the ln term and the negative powers of ρ are singular at the origin We are left with Φ(ρ, φ = a + ρ n {a n sin(nφ + b n cos(nφ} ( n= Multiplying both sides successively by, sin n φ, and cos n φ and integrating at ρ = b gives a = a n = b n = b n b n Φ(b, φdφ ( Φ(b, φ sin(nφdφ (3 Φ(b, φ cos(nφdφ (4 Plugging back into (, we find Φ(ρ, φ = { } Φ(b, φ + ( ρ n [sin(nφ sin(nφ + cos(nφ cos(nφ dφ b n= = { } Φ(b, φ + ( ρ n cos n(φ φ (5 b n= The bracketed term can be expressed in closed form x = (ρ/b and α = (φ φ Then For simplicity define + x n cos(nα = + [ x n e inα + x n e inα n= n= = + [ xe iα + xe iα = + [ xe iα xe iα + xe iα xe iα + x = [ + x cos α + x x cos α = x cos α x + + x x cos α = [ x + x x cos α Plugging this back into (5 gives the advertised result
Homer Reid s Solutions to Jackson Problems: Chapter 5 Problem 3 (a Two halves of a long hollow conducting cylinder of inner radius b are separated by small lengthwise gaps on each side, and are kept at different potentials V and V Show that the potential inside is given by Φ(ρ, φ = V + V + V V ( bρ tan b ρ cos φ where φ is measured from a plane perpendicular to the plane through the gap (b Calculate the surface-charge density on each half of the cylinder This problem is just like the previous one Since we are looking for an expression for the potential within the cylinder, the correct expansion is ( with expansion coefficients given by (, (3 and (4: a = = Φ(b, φdφ [V dφ + V dφ = V + V a n = [V b n sin(nφdφ + V sin(nφdφ = [ nb n V cos nφ + V cos nφ = nb n [V (cos n + V ( cos n { }, n even = b n = b n (V V /(nb n, n odd [V cos(nφdφ + V [ = nb n V sin nφ + V sin nφ = With these coefficients, the potential expansion becomes Φ(ρ, φ = V + V + (V V n odd cos(nφdφ ( ρ n sin nφ (6 n b
Homer Reid s Solutions to Jackson Problems: Chapter 6 Here we need an auxiliary result: n odd n xn sin nφ = i = = n odd n= n (iyn [e in e inφ ( n n + (x = iy [ (ye iφ n+ (ye iφ n+ [ tan (ye iφ tan (ye iφ (7 where in the last line we just identified the Taylor series for the inverse tangent function Next we need an identity: ( tan γ tan γ = tan γ γ + γ γ (I derived this one by drawing some triangles and doing some algebra With this, (7 becomes n xn sin nφ = ( iy sin φ tan + y n odd = ( x sin φ tan x Using this in (6 with x = ρ/b gives Φ(ρ, b = V + V + V V ( ρb sin φ tan b ρ (Evidently, Jackson and I defined the angle φ differently
Homer Reid s Solutions to Jackson Problems: Chapter 7 Problem 5 (a Show that the Green function G(x, y; x, y appropriate for Dirichlet boundary conditions for a square two-dimensional region, x, y, has an expansion G(x, y; x, y = g n (y, y sin(nx sin(nx n= where g n (y, y satisfies ( y n g n (y, y = δ(y y and g n (y, = g n (y, = (b Taking for g n (y, y appropriate linear combinations of sinh(ny and cosh(ny in the two regions y < y and y > y, in accord with the boundary conditions and the discontinuity in slope required by the source delta function, show that the explicit form of G is G(x, y; x, y = n sinh(n sin(nx sin(nx sinh(ny < sinh[n( y > n= where y < (y > is the smaller (larger of y and y (I have taken out a factor 4 from the expressions for g n and G, in accordance with my convention for Green s functions; see the Green s functions review above (a To use as a Green s function in a Dirichlet boundary value problem G must satisfy two conditions The first is that G vanish on the boundary of the region of interest The suggested expansion of G clearly satisfies this First, sin(nx is when x is or Second, g(y, y vanishes when y is or So G(x, y; x, y vanishes for points (x, y on the boundary The second condition on G is ( G = + x y With the suggested expansion, we have x G = n= G = δ(x x δ(y y (8 g n (y, y sin(nx [ n sin(nx y G = y g n(y, y sin(nx sin(nx n=
Homer Reid s Solutions to Jackson Problems: Chapter 8 We can add these together and use the differential equation satisfied by g n to find G = δ(y y sin(nx sin(nx n= = δ(y y δ(x x since the infinite sum is just a well-known representation of the δ function (b The suggestion is to take g n (y, y = { An sinh(ny + B n cosh(ny, y < y; A n sinh(ny + B n cosh(ny, y > y (9 The idea to use hyperbolic sines and cosines comes from the fact that sinh(ny and cosh(ny satisfy a homogeneous version of the differential equation for g n (ie satisfy that differential equation with the δ function replaced by zero Thus g n as defined in (9 satisfies its differential equation (at all points except y = y for any choice of the As and Bs This leaves us free to choose these coefficients as required to satisfy the boundary conditions and the differential equation at y = y First let s consider the boundary conditions Since y is somewhere between and, the condition that g n vanish for y = is only relevant to the top line of (9, where it requires taking B n = but leaves A n undetermined for now The condition that g n vanish for y = only affects the lower line of (9, where it requires that = A n sinh(n + B n cosh(n One way to make this work is to take Then = (A n + B n e n + ( A n + B n e n ( A n + B n = e n and A n + B n = e n B n = e n + A n A n = e n e n so A n = cosh(n and B n = sinh(n With this choice of coefficients, the lower line in (9 becomes g n (y, y = cosh(n sinh(ny +sinh(n cosh(ny = sinh[n( y ( for (y > y Actually, we haven t completely determined A n and B n ; we could multiply ( by an arbitrary constant γ n and ( would still be satisfied Next we need to make sure that the two halves of (9 match up at y = y: A n sinh(ny = γ n sinh[n( y (
Homer Reid s Solutions to Jackson Problems: Chapter 9 7 6 5 g(yprime 4 3 4 6 8 yprime Figure : g n (y, y from Problem 5 with n=5, y=4 This obviously happens when A n = β n sinh[n( y and γ n = β n sinh(ny where β n is any constant In other words, we have { g n (y, y βn sinh[n( y sinh(ny =, y < y; β n sinh[n( y sinh(ny, y > y = β n sinh[n( y > sinh(ny < (3 with y < and y > defined as in the problem Figure shows a graph of this function n = 5, y = 4 The final step is to choose the normalization constant β n such that g n satisfies its differential equation: ( y n g n (y, y = δ(y y (4 To say that the left-hand side equals the delta function requires two things: that the left-hand side vanish at all points y y, and that its integral over any interval (y, y equal if the interval contains the point y = y, and vanish otherwise The first condition is clearly satisfied regardless of the choice of β n The second condition may be satisfied by making g n continuous, which we have already done, but giving its first derivative a finite jump of unit magnitude at y = y:
Homer Reid s Solutions to Jackson Problems: Chapter y g y =y + n(y, y = y =y Differentiating (3, we find this condition to require nβ n [ cosh[n( y sinh(ny sinh[n( y cosh(ny = nβ n sinh(n = so (4 is satisfied if Then (3 is β n = n sinh(n g n (y, y = sinh[n( y > sinh(ny < n sinh(n and the composite Green s function is G(x, y; x, y = = g n (y, y sin(nx sin(nx n= sinh[n( y > sinh(ny < sin(nx sin(nx (5 n sinh(n n= Problem 6 A two-dimensional potential exists on a unit square area ( x, y bounded by surfaces held at zero potential Over the entire square there is a uniform charge density of unit strength (per unit length in z Using the Green function of Problem 5, show that the solution can be written as Φ(x, y = 4 3 ɛ m= sin[(m + x (m + 3 { } cosh[(m + (y (/ cosh[(m + / Referring to my Green s functions review above, the potential at a point x within the square is given by Φ(x = [ G(x ; x ρ(x dv + Φ(x G ɛ V S n G(x ; x Φ x n da x (6 In this case the surface integral vanishes, because we re given that Φ vanishes on the boundary, and G vanishes there by construction We re also given that
Homer Reid s Solutions to Jackson Problems: Chapter ρ(x dv = dx dy throughout the entire volume Then we can plug in (5 to find Φ(x = ɛ n= n sinh(n The integrals can be done separately The x integral is sinh[n( y > sinh(ny < sin(nx sin(nx dx dy (7 sin(nx sin(nx dx = sin(nx [cos(n n { ( sin(nx /n, n odd =, n even (8 The y integral is y sinh[n( y sinh(ny dy + sinh(ny sinh[n( y dy y = { sinh[n( y cosh(ny y n sinh[ny cosh[n( y } y = n {sinh[n( y cosh(ny + sinh(ny cosh[n( y sinh(ny sinh[n( y } = n {sinh[n sinh[n( y sinh(ny } (9 Inserting (8 and (9 in (7, we have Φ(x = 4 3 ɛ n odd sin(nx n 3 { sinh[n( y } + sinh(ny sinh(n The thing in brackets is equal to what Jackson has, but this is tedious to show so I ll skip the proof
Homer Reid s Solutions to Jackson Problems: Chapter Problem 7 (a Construct the free-space Green function G(x, y; x, y for twodimensional electrostatics by integrating /R with respect to z z between the limits ±Z, where Z is taken to be very large Show that apart from an inessential constant, the Green function can be written alternately as G(x, y; x, y = ln[(x x + (y y = ln[ρ + ρ ρρ cos(φ φ (b Show explicitly by separation of variables in polar coordinates that the Green function can be expressed as a Fourier series in the azimuthal coordinate, G = e im(φ φ g m (ρ, ρ where the radial Green functions satisfy ( ρ ρ ρ g m ρ m ρ g m = δ(ρ ρ ρ Note that g m (ρ, ρ for fixed ρ is a different linear combination of the solutions of the homogeneous radial equation (68 for ρ < ρ and for ρ > ρ, with a discontinuity of slope at ρ = ρ determined by the source delta function (c Complete the solution and show that the free-space Green function has the expansion G(ρ, φ; ρ, φ = 4 ln(ρ > m= m ( ρ< ρ > where ρ < (ρ > is the smaller (larger of ρ and ρ m cos[m(φ φ (As in Problem 5, I modified the text of the problem to match with my convention for Green s functions (a R = [(x x + (y y + (z z / Integrating, [a + u /, a = [(x x + (y y /, u = (z z Z Z du [a + u / = ln [ (a + u / + u +Z Z
Homer Reid s Solutions to Jackson Problems: Chapter 3 = ln (Z + a / + Z (Z + a / Z = ln ( + (a /Z / + ( + (a /Z / ln + a Z a Z = ln 4Z + a a = ln[4z + a ln a Since Z is much bigger than a, the first term is essentially independent of a and is the nonessential constant Jackson is talking about The remaining term is the D Green s function: G = ln a = ln[(x x + (y y in rectangular coordinates = ln[ρ + ρ ρρ cos(φ φ in cylindrical coordinates (b The d Green s function is defined by G(ρ, φ; ρ, φ ρ dρ dφ = but G = at points other than (ρ, φ These conditions are met if G(ρ, φ; ρ, φ = ρ δ(ρ ρ δ(φ φ ( You need the ρ on the bottom there to cancel out the ρ in the area element in the integral The Laplacian in two-dimensional cylindrical coordinates is = ( ρ ρ ρ ρ ρ φ Applying this to the suggested expansion for G gives G(ρ, φ; ρ, φ = { ( ρ ρ ρ g } m ρ m ρ g m e im(φ φ If g m satisfies its differential equation as specified in the problem, the term in brackets equals δ(ρ ρ /ρ for all m and may be removed from the sum, leaving G(ρ, φ; ρ, φ = = ( δ(ρ ρ ρ ( δ(ρ ρ ρ δ(φ φ e im(φ φ
Homer Reid s Solutions to Jackson Problems: Chapter 4 (c As in Problem 5, we ll construct the functions g m by finding solutions of the homogenous radial differential equation in the two regions and piecing them together at ρ = ρ such that the function is continuous but its derivative has a finite jump of magnitude /ρ For m, the solution to the homogenous equation is Thus we take g m = { ( ρ ρ ρ } ρ m ρ f(ρ = f(ρ = A m ρ m + B m ρ m { Am ρ m + B m ρ m, ρ < ρ A m ρ m + B m ρ m, ρ > ρ In order that the first solution be finite at the origin, and the second solution be finite at infinity, we have to take B m = A m = Then the condition that the two solutions match at ρ = ρ is which requires A m ρ m = B m ρ m A m = γ m ρ m B m = ρ m γ m for some constant γ m Now we have ( γ ρ m m ρ, ρ < ρ g m = ( m γ ρ m ρ, ρ > ρ The finite-derivative step condition is dg m dρ dg m dρ = ρ =ρ ρ ρ =ρ+ or mγ m ( ρ + ρ = ρ so Then g m = = m γ m = m ( ρ m m ρ, ρ < ρ ( ρ m ( ρ< ρ > m ρ, ρ > ρ m
Homer Reid s Solutions to Jackson Problems: Chapter 5 Plugging this back into the expansion gives G = ( m ρ< e im(φ φ 4 m = m ρ > ( ρ< ρ > m cos[m(φ φ Jackson seems to be adding a ln term to this, which comes from the m = solution of the radial equation, but I have left it out because it doesn t vanish as ρ Problem 8 (a By finding appropriate solutions of the radial equation in part b of Problem 7, find the Green function for the interior Dirichlet problem of a cylinder of radius b [g m (ρ, ρ = b = See (4 First find the series expansion akin to the free-space Green function of Problem 7 Then show that it can be written in closed form as [ ρ ρ + b 4 ρρ b cos(φ φ G = ln b (ρ + ρ ρρ cos(φ φ or [ (b ρ (b ρ + b ρ ρ G = ln b ρ ρ (b Show that the solution of the Laplace equation with the potential given as Φ(b, φ on the cylinder can be expressed as Poisson s integral of Problem (c What changes are necessary for the Green function for the exterior problem (b < ρ <, for both the Fourier expansion and the closed form? [Note that the exterior Green function is not rigorously correct because it does not vanish for ρ or ρ For situations in which the potential falls of fast enough as ρ, no mistake is made in its use (a As before, we write the general solution of the radial equation for g m in the two distinct regions: { g m (ρ, ρ Am ρ = m + B m ρ m, ρ < ρ A m ρ m + B m ρ m, ρ ( > ρ The first boundary conditions are that g m remain finite at the origin and vanish on the cylinder boundary This requires that B m =
Homer Reid s Solutions to Jackson Problems: Chapter 6 and so A m b m + B m b m = A m = γ m b m for some constant γ m Next, g m must be continuous at ρ = ρ : B m = γ m b m m ( b ρ m [ (ρ A m ρ m = γ m b A m = γ [ (ρ ( m m m b ρ m b ρ With this we have [ (ρ ( m m ( b ρ g m (ρ, ρ m = γ m, ρ b ρ < ρ ρ [( ρ m ( m b = γ m b ρ, ρ > ρ so and or Finally, dg m /dρ must have a finite jump of magnitude /ρ at ρ = ρ = dg m ρ dρ dg m ρ =ρ + dρ ρ =ρ [ [ ρ m (ρ ( = mγ m b m + bm m m b ρ m+ mγ m b ρ ρ ( m b = mγ m ρ ρ g m (ρ, ρ = m = m g m (ρ, ρ = m γ m = ( ρ m m b [( ρρ m ( ρ m b ρ [( ρρ m ( m ρ b ρ [( ρρ m Plugging into the expansion for G gives G(ρ, φ, ρ, φ = n= m b [( ρρ m b ( ρ< ρ > ( ρ< ρ >, ρ < ρ, ρ > ρ m m cos m(φ φ (
Homer Reid s Solutions to Jackson Problems: Chapter 7 Here we need to work out an auxiliary result: [ x n xn cos n(φ φ = u n du cos m(φ φ n= n= { } x = u n cos n(φ φ du u n= x { cos(φ φ } u = + u u cos(φ φ du = ln( u cos(φ φ + u x = ln[ x cos(φ φ + x (I summed the infinite series here back in Problem The integral in the second-to-last step can be done by partial fraction decomposition, although I cheated and looked it up on wwwintegralscom We can apply this result individually to the two terms in (: G(ρ, φ; ρ, φ = [ + (ρρ 4 ln /b (ρρ /b cos(φ φ + (ρ < /ρ > (ρ < /ρ > cos(φ φ = [( ρ 4 ln > b 4 + ρ ρ ρρ b cos(φ φ b 4 ρ > + ρ < ρ < ρ > cos(φ φ = [( ρ 4 ln > b 4 + ρ ρ ρρ b cos(φ φ b 4 ρ > + ρ < ρ < ρ > cos(φ φ = ( ρ 4 ln > b 4 ln [ b 4 + ρ ρ ρρ b cos(φ φ b (ρ + ρ ρρ cos(φ φ (3 This is Jackson s result, with an additional ln term thrown in for good measure I m not sure why Jackson didn t quote this term as part of his answer; he did include it in his answer to problem 7 (c Did I do something wrong? (b Now we want to plug the expression for G above into (6 to compute the potential within the cylinder If there is no charge inside the cylinder, the volume integral vanishes, and we are left with the surface integral: Φ(ρ, φ = Φ(b, φ G ρ da (4 ρ =b where the integral is over the surface of the cylinder For this we need the normal derivative of (3 on the cylinder: G ρ = { 4 ρ ρ ρb cos(φ φ b 4 + ρ ρ ρρ b cos(φ φ ρ ρ cos(φ φ ρ + ρ ρρ cos(φ φ }
Homer Reid s Solutions to Jackson Problems: Chapter 8 Evaluated at ρ = b this is G ρ = { ρ =b ρ b b(ρ + b ρb cos(φ φ In the surface integral, the extra factor of b on the bottom is cancelled by the factor of b in the area element da, and (4 becomes just the result of Problem (c For the exterior problem we again start with the solution ( Now the boundary conditions are different; the condition at gives A m =, while the condition at b gives A m = γ m b m B m = γ m b m From the continuity condition at ρ = ρ we find A m = γ m ρ m [ (ρ b m ( b ρ m The finite derivative jump condition gives [ (ρ ( m m [ b (ρ ( m m b mγ m b ρ ρ mγ m + b ρ ρ = ρ or γ m = m ( m b ρ Putting it all together we have for the exterior problem g m = [( b m m ρρ ( ρ< ρ > m This is the same g m we came up with before, but with b and ρρ terms flipped in first term But the closed-form expression was symmetrical in those two expressions (except for the mysterious ln term so the closed-form expression for the exterior Green s function should be the same as the interior Green s function }