Kirk ph501set2

Princeton University Ph501 Electrodynamics Problem Set 2 Kirk T. McDonald (1998) [email protected] http://puhep1.princeton.edu/~mcdonald/examples/ Princeton University 1998 Ph501 Set 2, Problem 1 1 1. Show that the electromagnetic energy of a dielectric subject to fields EandD=/epsilon1Eis U=1 8π/integraldisplay E·DdVol, (1) by considering the model of atoms as springs (Problem 8b, set 1). The energy Uthen has two parts: U1=1 8π/integraldisplay E2dVol, (2) stored in the electric field, and U2=/integraldisplay nkx2 2dVol, (3) stored in the spring-like atoms ( nis the number of atoms per unit volume). Assume n is small so that the dielectric constant /epsilon1is nearly 1. Princeton University 1998 Ph501 Set 2, Problem 2 2 2. (a) Show that the energy of a quadrupole in an external electric field E, Uquad=−1 6Qij∂Ej ∂xi, (4) in terms of its quadrupole tensor Qij, can be rewritten as Uquad=−Qxx 4∂Ex ∂x, (5) if the quadrupole is rotationally symmetric about the xaxis. Give an expression for the force Fon the quadrupole. (b) A rotationally symmetric quadrupole of strength Qxx(zero net charge, zero dipole moment) is located at distance rfrom a point charge q. What is the force on the quadrupole if: i. The xaxis is along the line joining Qxxandq? ii. The xaxis is perpendicular to the line joining Qxxandq? For your own edification, confirm your answer by considering the simple quadrupole: t −q1y 2q1t −q1 Princeton University 1998 Ph501 Set 2, Problem 3 3 3. The principle of an electrostatic accelerator is that when a charge eescapes from a conducting plane that supports a uniform electric field of strength E0, then the charge gains energy eE0das it moves distance dfrom the plane. Where does this energy come from? Show that the mechanical energy gain of the electron is balanced by the decrease in the electrostatic field energy of the system. Princeton University 1998 Ph501 Set 2, Problem 4 4 4. (a) Two point dipoles of strength pare aligned along their line of centers, and distance 2dapart. Calculate the force between the dipoles via F=(p·∇)E,a n db ym e a n s of the Maxwell stress tensor. (b) A spherical conducting shell of radius acarries charge q. It is in a region of zero external field. Calculate the force between two hemispheres in two different ways. Princeton University 1998 Ph501 Set 2, Problem 5 5 5. (a) Two coaxial pipes of radii aandb(a<b) are lowered vertically into an oil bath: If a voltage Vis applied between the pipes, show that the oil rises to height h=(/epsilon1−1)V2 4πρgln/parenleftBig b a/parenrightBig (b2−a2), (6) where gis the acceleration due to gravity. (b) Recalling prob. 1(c) of set 1, discuss qualitatively how the force arises the pulls the liquid up into the capacitor. Princeton University 1998 Ph501 Set 2, Problem 6 6 6. According to a theorem of Green, the potential φ(x) in the interior of a volume V can be deduced from a knowledge of the charge density ρ(x) inside that volume plus knowledge of the potential and the normal derivative ∂φ/∂n of the potential on the surface Sthat bounds the volume, φ(x)=/integraldisplay Vρ(x/prime) RdVol/prime+1 4π/integraldisplay S/bracketleftBigg φ(x/prime)∂ ∂n/prime/parenleftbigg1 R/parenrightbigg −1 R∂φ(x/prime) ∂n/prime/bracketrightBigg dS/prime, (7) where R=|x−x/prime|is the distance between the point of observation and the element of the integrand. However, further insights of Green indicate that it suffices to specifyonly one of φor∂φ/∂n on the bounding surface to determine the potential within. As a particular example, show that the potential within a charge-free sphere of radius a, centered on the origin, can be determined from knowledge of only the potential φon its surface according to (Poisson, 1820) φ(x)=a 2−x2 4πa/integraldisplay Sφ(x/prime) R3dS/prime. (8) Green (1828) gave a derivation of Poisson’s integral (8) that can be generalized to many other problems in electrostatics. Recall that a key step towards eq. (7) is the identity /integraldisplay V∇·(ψ∇φ−φ∇ψ)dVol =/integraldisplay V(ψ∇2φ−φ∇2ψ)dVol =/integraldisplay S(ψ∇φ−φ∇ψ)·dS=/integraldisplay S/parenleftBigg ψ∂φ ∂n−φ∂ψ ∂n/parenrightBigg dS. (9) For problems in which the interior of volume Vis charge free the potential obeys ∇2φ= 0 there. To have a nonzero potential φinside Vthere must, of course, be charges on the surface of or exterior to volume V. If function ψalso obeys ∇2ψ=0 inside V(and so might be the potential for some other distribution of charges exterior toV), then the identity (9) reduces to 0=/integraldisplay S/parenleftBigg ψ∂φ ∂n−φ∂ψ ∂n/parenrightBigg dS. (10) Hence, we could combine eqs. (7) and (10) to yield the relation φ(x)=1 4π/integraldisplay S/bracketleftBigg φ(x/prime)∂ ∂n/prime/parenleftbigg1 R+ψ/parenrightbigg −/parenleftbigg1 R+ψ/parenrightbigg∂φ(x/prime) ∂n/prime/bracketrightBigg dS/prime =1 4π/integraldisplay S/bracketleftBigg φ(x/prime)∂G(x,x/prime) ∂n/prime−G(x,x/prime)∂φ(x/prime) ∂n/prime/bracketrightBigg dS/prime, (11) where G(x,x/prime)=1 R+ψ. (12) Princeton University 1998 Ph501 Set 2, Problem 6 7 IF the Green’s function G(x,x/prime) vanishes on the surface S, then we have the desirable relation between the potential φin the interior of Vand its value on the bounding surface S, φ(x)=1 4π/integraldisplay Sφ(x/prime)∂G(x,x/prime) ∂n/primedS/prime. (13) Green noted that the auxiliary potential ψcan be thought of as due to exterior charges that bring the surface Sto zero potential when there is unit charge at position xinside volume V,a n dGas the total potential of that charge configuration. Further, we may think of the bounding surface Sas being a grounded conductor for the purposes of determining the potentials ψandG, in which case the “exterior” charges reside on the surface S. Hence, it is plausible that these exist for interesting physical surfaces S (although it turns out that mathematicians have constructed examples of surfaces for which a Green’s function does not exist). Since the function Gis the potential for a specifiable charge configuration, the normal derivative −∂G/∂n corresponds to the electric field (whose only nonzero component isEn)a tt h es u r f a c e Sproduced by those charges. If we consider surface Sto be a grounded conductor when determining function G, then the charge density σGat position x/primeon that surface, caused by the hypothetical unit charge at x,w o u l db e σG(x,x/prime)=En/4π=−(1/4π)∂G/∂n . Green emphasized this phyisical interpretation in his original work, and wrote eq. (13) as φ(x)=−/integraldisplay SσG(x,x/prime)φ(x/prime)dS/prime. (14) Turning at last to Poisson’s integral (8), we see that the needed Green’s function for a sphere corresponds to the potential at x/primedue to unit charge at xin the presence of a grounded conducting sphere of radius a. Use the method of images to construct the Green’s function and its normal derivative, and thereby verify Poisson’s result. Princeton University 1998 Ph501 Set 2, Problem 7 8 7. A parallel-plate capacitor is connected to a battery which maintains the plates at constant potential difference V0. A slab of dielectric constant /epsilon1is inserted between the plates, completely filling the space between them. (a) Show that the battery does work Q0V0(/epsilon1−1) during the insertion process, if Q0 is the charge on the plates before the slab is inserted. (b) What is the change in the electrostatic energy of the capacitor? (c) How much work is done by the mechanical forces on the slab when it is inserted? Is this work done by, or on, the agent inserting the slab? Suppose the battery was disconnected before the dielectric was inserted. (d) Repeat (b). (e) Repeat (c). Princeton University 1998 Ph501 Set 2, Problem 8 9 8. (a) Find the “escape velocity” of an electron initially 1 A above a grounded conducting plate. (b) Point electric dipoles p1andp2lie in the same plane at a fixed distance apart. If p1makes angle θ1to their line of centers, show that the equilibrium angle θ2of p2is related to θ1by tanθ1=−2tanθ2. (15) Princeton University 1998 Ph501 Set 2, Problem 9 10 9. We may define the capacity of a single conductor with respect to infinity as C=Q/V, where Vis the potential (with respect to potential φ=0a t ∞) when charge Qis present on the conductor. Calculate the capacity of a conductor composed of two tangent spheres of radius a. Princeton University 1998 Ph501 Set 2, Problem 10 11 10. A grounded conducting sphere of radius ais placed in a uniform external field E=E0ˆz. (This field changes after the sphere is added.) This problem may be solved by the method of images if we suppose the field E0is due to two charges ±Qat positions z=±R,w i t h QandRappropriately large. (a) Show that the image of the source of E0is then a dipole p=a3E0located at the center of the sphere. (b) Give an expression for the potential φ(r, θ) in spherical coordinates ( r, θ, ϕ) cen- tered on the sphere. Sketch the electric field lines. (c) Show that the induced charge distribution on the sphere is σ=3 4πE0cosθ. (d) Show that the force between the two hemispheres with equator perpendicular to E0isF=9 16a2E0. Princeton University 1998 Ph501 Set 2, Problem 11 12 11. A hollow infinite rectangular conducting tube of sides aandbhas two faces grounded and two faces at potentials V1andV2as shown: abφ=0 φ=0φ=φ=VV2 1 xy Find the potential φ(x,y) inside the tube. Remember to use a sum of products of all solutions to the separated equations which do not violate the boundary conditions. Princeton University 1998 Ph501 Set 2, Problem 12 13 12. A hollow rectangular conducting box has walls at x=0a n d a,a ty=0a n d b,a n da t z=0a n d c. All faces are grounded except that at z=c,f o rw h i c h φ=V: φ=0 xyz φ=0 φ=0 φ=0φ=0φ=V c b a Find the potential φ(x,y,z ) inside the box. (Choose the signs of the separation constants carefully!) Princeton University 1998 Ph501 Set 2, Solution 1 14 Solutions 1. The energy stored in a dielectric composed of spring like atoms can be written in two parts, U=U1+U2=1 8π/integraldisplay E2dVol +/integraldisplay nkx2 2dVol, (16) where Eis the applied electric field, and where nis the number of molecules per unit volume. The displacement xin the spring-like atom is related by kx=eEon atom ,w h e r e eis the charge of an electron. Then, U2=/integraldisplay ne2E2 on atom 2kdVol =1 2/integraldisplay ne2E2 on atom mω2dVol =1 2/integraldisplay nαE2 on atom dVol,(17) where ω=/radicalBig k/mis the frequency of oscillation of the electron of mass m,a n d α= e2/mω2is the atomic polarizability introduced in eq. (69) of set 1. On p. 20 of the Notes, we argued that Eon atom =E+4πP/3, in terms of the applied fieldEand the induced polarization P.B u t , P=nαEon atom ,s o P=nα 1−4πnα/3E,and Eon atom =E/parenleftBigg 1+nα 1−4πnα/3/parenrightBigg ≈E, (18) where the approximation holds for small n.I nt h i sc a s e , U2≈1 8π/integraldisplay 4πnαE2dVol, (19) and U≈1 8π/integraldisplay (1 + 4 πnα)E2dVol≈1 8π/integraldisplay /epsilon1E2dVol =1 8π/integraldisplay E·DdVol, (20) using the Lorenz-Lorentz approximation for the dielectric constant /epsilon1in terms of the polarizability α, and supposing that D=/epsilon1E. Princeton University 1998 Ph501 Set 2, Solution 2 15 2. (a) As argued on p. 13 of the Notes, rotational symmetry of a charge distribution about the xaxis implies that its quadrupole tenson Qijcan be written Qij=⎛ ⎜⎜⎜⎜⎜⎝Qxx 00 0−Qxx/20 00 −Qxx/2⎞ ⎟⎟⎟⎟⎟⎠, (21) and hence, from (4), U=−Qxx 6/parenleftBigg∂Ex ∂x−1 2∂Ey ∂y−1 2∂Ez ∂z/parenrightBigg =−Qxx 6/parenleftBigg3 2∂Ex ∂x−1 2∇·E/parenrightBigg =−Qxx 4∂Ex ∂x, (22) using∇·E= 0, assuming that the external field is produced by charges not at the location of the quadrupole. The force on the quadrupole is: F=−∇U=Qxx 4/parenleftBigg∂2Ex ∂x2,∂2Ex ∂x∂y,∂2Ex ∂x∂z/parenrightBigg . (23) ( b ) i .C o n s i d e rap o i n tc h a r g e qat at the origin and the quadrupole at ( x,y,z )= (R,0,0). The x-component of the electric field from qobserved at ( x,y,z )i s Ex=qx r3,where r2=x2+y2+z2. (24) Then, ∂Ex ∂x=qr2−3x2 r5, (25) and the force is evaluated from (23) at ( R,0,0) as F=/parenleftbigg3 2qQxx R4,0,0/parenrightbigg . (26) Let us check this for the simple quadrupole shown in the picture. t −q1y 2q1t −q1 Suppose the distance between −q1and 2q1isa. The force on the quadrupole due to charge qat distance Rfrom the center of the quadrupole, and along the latter’s axis, is Fx=−q1q (R−a)2+2q1q R2−q1q (R+a)2=−6a2q1q R4(1 +O(a/R)). (27) This agrees with (26), since Qij=/integraldisplay ρ/prime(3r/prime ir/prime j−r/prime2δij)dVol/prime⇒Qxx=/summationdisplay 2q/primer/prime2=−4q1a2.(28) Princeton University 1998 Ph501 Set 2, Solution 2 16 ii. If, instead, the quadrupole is at (0 ,R ,0) (but still oriented parallel to the x axis), eqs. (23) and (25) combine to reveal that only the derivative ∂2Ex/∂x∂y is nonvanishing, and F=/parenleftbigg 0,−3qQxx 4R4,0/parenrightbigg . (29) Again, we can directly compute the force on the simple quadrupole: Fy=2/parenleftBigg q1q R2−q1qR (R2+a2)3/2/parenrightBigg ≈3qq1a2 R4=−3qQxx 4R4, (30) using (28). Princeton University 1998 Ph501 Set 2, Solution 3 17 3. Once the charge has reached distance dfrom the plane, the static electric field Eeat an arbitrary point rdue to the charge can be calculated by summing the field of the charge plus its image charge, Ee(r,d)=er1 r3 1−er2 r3 2, (31) where r1(r2) points from the charge (image) to the observation point r, as illustrated below. The total electric field is then E0ˆz+Ee. The charge eand its image charge −eat positions ( r, θ, z)=( 0 ,0,±d)w i t h respect to a conducting plane at z= 0. Vectors r1andr2are directed from the charges to the observation point ( r,0,z). It turns out to be convenient to use a cylindrical coordinate system, where the obser- vation point is r=(r, θ, z)=(r,0,z), and the charge is at (0 ,0,d). Then, r2 1,2=r2+(z∓d)2. (32) The part of the electrostatic field energy that varies with the position of the charge is the interaction term, Uint=/integraldisplayE0ˆz·Ee 4πdVol =eE0 4π/integraldisplay∞ 0dz/integraldisplay∞ 0πdr2/parenleftBiggz−d [r2+(z−d)2]3/2−z+d [r2+(z+d)2]3/2/parenrightBigg =eE0 4/integraldisplay∞ 0dz⎛ ⎜⎝⎧ ⎪⎨ ⎪⎩2i f z>d −2i fz<d⎫ ⎪⎬ ⎪⎭−2⎞ ⎟⎠ =−eE0/integraldisplayd 0dz=−eE0d. (33) When the particle has traversed a potential difference V=E0d, it has gained energy eVand the electromagnetic field has lost the same energy. In a practical “electrostatic” accelerator, the particle is freed from an electrode at potential −Vand emerges with energy eVin a region of zero potential. However, the particle could not be moved to the negative electrode from a region of zero potential by purely electrostatic forces unless the particle lost energy eVin the process, leading to zero overall energy change. An “electrostatic” accelerator must have an essentialcomponent (such as a battery) that provides a nonelectrostatic force that can absorbthe energy extracted from the electrostatic field while moving the charge from potentialzero, so as to put the charge at rest at potential −Vprior to acceleration. Princeton University 1998 Ph501 Set 2, Solution 4 18 4. (a) First, we calculate the force directly. The electric field from one of the dipoles, taken to be at the origin and with moment p=pˆx,i s E=3(p·ˆr)ˆr−p r3=3pxˆr r4−pˆx r3. (34) For a second dipole at ( x,y,z )=( 2 d,0,0), also with moment p=pˆx,w eh a v e F=(p·∇)E=p∂E ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle (2d,0,0)=−3p2 8d4ˆx. (35) The minus sign indicates that the dipole’s attract. As an aside, we can also calculate F=−∇U,w h e r e Uis the energy of interaction of the two dipoles. First, the energy of a charge q2at position r2in the field of a dipole p1at position r1is U=q2p1·r r3, (36) where r=|r|=|r2−r1|, as on p. 12 of the Notes. A point dipole p2is the limit of a pair of charges ±q2at positions r2andr2−swhere s=sˆp2, and the product q2sis held constant at value p2. Thus, the interaction energy of two point dipoles is obtained from (36) as U= lim s→0,q2s=pq2/parenleftBigg p1·r r3−p1·r/prime r/prime3/parenrightBigg =(p2·∇2)p1·r r3, (37) where r/prime=|r2−s−r1|.F o r p1=p2=pˆxseparated by distance 2 dalong x, (37) reduces to U=p2∂ ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=2d1 x2. (38) Then, the force F=−∇Uis along xwith magnitude F=−p2∂2 ∂x21 x2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=2d=−3p2 8d4, (39) as found in (35). Now, let us calculate the force via the Maxwell stress tensor. The force on the charges within a (closed) surface Sis given by Fi=/contintegraldisplay STijdSj, (40) as on p. 33 of the Notes, where the Maxwell tensor in empty space is given by Tij=1 4π/parenleftbigg EiEj−1 2δijE2/parenrightbigg . (41) In the problem with two dipoles, it is convenient to choose the surface as the midplane perpendicular to the line connecting two dipoles (the xaxis), closing Princeton University 1998 Ph501 Set 2, Solution 4 19 the surface at infinity around one of the dipoles. On this plane ( x=d)t h eo n l y nonzero component of EisEx, and this is twice Exfrom the dipole at x=0 . A t radius rfrom ( d,0,0) in the symmetry plane, the total field is then Ex(r)=2p/parenleftBigg 3d2 [r2+d2]5/2−1 [r2+d2]3/2/parenrightBigg =2p2d2−r2 [r2+d2]5/2. (42) The Maxwell stress tensor is thus, Tij=1 8π⎛ ⎜⎜⎜⎜⎜⎝E2 x00 0−E2 x0 00 −E2 x⎞ ⎟⎟⎟⎟⎟⎠. (43) We take our surface element to be dS=( 2πrdr,0,0) in cylindrical coordinates, the sign of which implies that the surface Sencloses the dipole at x= 0. Then, (40) and (43) indicate that only Fxis nonzero, and it is given by Fx=1 4/integraldisplay∞ 0rd r E2 x=p2/integraldisplay∞ 0rd r(r2−2d2)2 (r2+d2)5 =p2 2/integraldisplay∞ 0dt(t−2d2)2 (t+d2)5=p2 2/integraldisplay∞ 0dt[(t+d2)−3d2]2 (t+d2)5(44) =p2 2/integraldisplay∞ 0dt/bracketleftBigg1 (t+d2)3−6d2 (t+d2)4+9d4 (t+d2)5/bracketrightBigg =p2 2/bracketleftbigg1 2d4−2 d4+9 4d4/bracketrightbigg =3p2 8d4. This agrees with (35), noting that since the dipoles attract, the force on the dipole atx= 0 is in the + xdirection. (b) The electric field outside the conducting sphere of radius aisE=qˆr/r2.T h e pressure (= force per unit area) on the surface charges is P=σE/2, where σis the surface charge density; hence, P=q2ˆr/8πa4.( T h e c o e ffi c i e n t 1 /2 is needed because Eis the field outside the surface, while the field inside the sphere is zero, thus the average field inside the charge layer is E/2.) To find the force between two hemispheres, we integrate the component of pressure normal to the equatorialplane ( Pcosθ) over one hemisphere: F=/integraldisplay1 02πa2dcosθq2cosθ 8πa4=q2 8a2. (45) Now, let us calculate the force using the Maxwell stress tensor. We integrate Fzover the x-yplane separating our sphere into two hemispheres. Since dS= (0,0,2πr dr ) there, and the only nonzero components of Eon that surface are ExandEy,o n l y Tzz=−E2/8π=−q2/8πr4contributes to the force. Integrating fromr=ato∞, we find Fz=/integraldisplay∞ a2πr drT zz=q2 4/integraldisplay∞ adr r3=q2 8a2, (46) in agreement with (45). Princeton University 1998 Ph501 Set 2, Solution 5 20 5. (a) The electrical force Frequired to pull oil of density ρinto a cylindrical capacitor of inner and outer radii aandb, respectively, to height habove the bath is equal to the force of gravity: F=ρgh(b2−a2). (47) A second relation for Fcan be computed from the balance of electrical energy, noting that the capacitor is held at constant voltage by a battery. Suppose weincrease the height of the oil by δh. Then, work Fδhis done on the oil, the energy U=CV 2/2 stored in the capacitor changes by δU, and the battery loses energy VδQ. Conservation of energy implies 0=Fδh+δU−Vδ Q . (48) Since V=Q/C, we find for constant voltage, VδQ=V2δC=2δ/parenleftBiggCV2 2/parenrightBigg =2δU. (49) Together, (48) and (49) imply that F=+∂U ∂h/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle V. (50) As the liquid is drawn intothe capacitor, the energy for this must come from elsewhere; yet, the energy of the capacitor increases because the battery loses energy in twice the amount of work done on the liquid. We now calculate the stored energy Uby integrating the electric field energy density. By cylindrical symmetry and Gauss’s law, the electric field between thepipes has form E r(r)=α/r,w h e r e αis fixed by V=/integraldisplayb aErdr=αlnb a,or α=V lnb a. (51) Suppose the total height of the capacitor (above the bath) is H. Then, the energy of the electric field in the capacitor is: U=1 8π/epsilon1h/integraldisplayb aE22πr dr +1 8π(H−h)/integraldisplayb aE22πr dr, (52) where the first term on the right is the contribution from the space filled with the oil whose dielectric constant is /epsilon1, while the second term is from the empty space above. Evaluating the integrals: 2π/integraldisplayb aE2rd r=2πV2 ln2b a/integraldisplayb adr r=2πV2 lnb a, (53) we then find: U=1 4V2 lnb a[(/epsilon1−1)h+H]. (54) Princeton University 1998 Ph501 Set 2, Solution 5 21 The force is obtained from (50) and (54): Fel=∂U ∂h/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle V=V2(/epsilon1−1) 4lnb a. (55) Equating this to the force of gravity, (47) we obtain the height hof the oil column: h=(/epsilon1−1)V2 4πρgln/parenleftBig b a/parenrightBig (b2−a2). (56) (b) The force on the liquid arises from the effect of gradients of the electric field on the molecular dipoles in the liquid. The spatially varying electric field Eresults in a bulk dielectric polarization given by P=χE=/epsilon1−1 4πE, (57) where χis the dielectric susceptibility and /epsilon1is the dielectric constant. The energy density associated with the induced polarization is u=−P·E=−/epsilon1−1 4πE2, (58) and so the force density on the liquid is given by f=−∇u=/epsilon1−1 4π∇E2. (59) The gradient ∇E2in the fringe field of the capacitor points from the outside to the interior of the capacitor, with a generally vertical component for the liquid below the capacitor in the present problem. It is interesting to consider a variant on this problem: a capacitor with horizontal plates completely immersed in a dielectric liquid. Here, the fringe fields of the capacitor pull the liquid in from all sides, “trapping” it inside the capacitor. That is, work would be required to pull the liquid out of the capacitor in any direction. Is this an example of electrostatic trapping – which is claimed not to exist? No! The “trapping” in the direction perpendicular to the capacitor plates is not pro- vided by purely electrostratic fields, but by the material of the capacitor plates(whose stability is not a result of purely electrostatic effects). See prob. 7 of set4 for further discussion. We have concluded that the liquid is drawn into the interior of the capacitor and that the liquid near the middle of the capacitor is forced up against the capacitorplates by electrostatic forces on the induced dipoles. If we drill a hole in thecenter of one capacitor plate, would liquid squirt out? (If yes, we would have aperpetual motion machine.) No, the fringe fields around the hole will pull liquid into the interior of the capacitor creating a static equilibrium much as before. Princeton University 1998 Ph501 Set 2, Solution 6 22 6. We work from eq. (13), for which we first need the potential φat a point rinside a grounded conducting sphere of radius awhen unit charge is located at x, also inside the sphere. Then we need the normal derivative of this potential on the inner surfaceof the sphere, i.e.when|r|=r=a. The image method for a grounded conducting sphere tells us that the potential inside the sphere can be calculated as that due to unit charge at xtogether with charge −a/x at position x /prime=a2x/x2. We denote the angle between vectors randxasθ,s ot h a t R=|r−x|=√ r2+2rxcosθ+x2, (60) and R/prime=|r−x/prime|=/radicalBigg r2+2ra2 xcosθ+a4 x2. (61) We see that when r=a,t h e n R/prime=a xR. (62) The potential inside the sphere can now be written φ(r)=1 R−a R/primex, (63) The normal derivative of the potential on the inner surface of the sphere is the negative of its radial derivative when r=a, ∂φ ∂n=−∂φ(r=a) ∂r=a+xcosθ R3−a[a+(a2/x)co sθ] R/prime3x=a2−x2 aR3, (64) using eq. (62). Inserting this in eq. (13), we obtain Poisson’s integral, φ(x)=a2−x2 4πa/integraldisplay Sφ(x/prime) R3dS/prime. (65) Princeton University 1998 Ph501 Set 2, Solution 7 23 7. (a) The capacitance Cof a parallel-plate capacitor of area A, gap thickness dand dielectric constant /epsilon1is C=/epsilon1A d≡Q V. (66) Adding the dielectric increased the capacitance to Cf=/epsilon1C0, (67) and hence the charge also increase, if the voltage is kept fixed. Thus, the work done by the battery as the dielectric is inserted, ΔWbatt=V0ΔQ=V2 0ΔC=V2 0(/epsilon1−1)C0=Q0V0(/epsilon1−1), (68) is positive. (b) As the dielectric is inserted, the field energy U=CV2/2 stored in the capacitor changes by ΔU=1 2ΔCV2 0=1 2C0V2 0(/epsilon1−1) =1 2Q0V0(/epsilon1−1). (69) (c) The work done by the battery, (68), is only partly accounted for in increase in the field energy, (69). The rest of the work done by the battery is done onthe external agent that held the dielectric during insertion (the external agent gainedenergy): ΔW on agent =1 2Q0V0(/epsilon1−1). (70) (d) If the battery had been disconnected before the dielectric was inserted, then the charge Q0would be constant. From (66) we see that the final voltage would be onlyV0//epsilon1. Recalling (67), the change in the electrostatic field energy would then be ΔU=1 2/epsilon1C0/parenleftbiggV0 /epsilon1/parenrightbigg2 −1 2C0V2 0=1 2Q0V0/parenleftbigg1 /epsilon1−1/parenrightbigg <0. (71) (e) By conservation of energy, the work done on the external agent that held the dielectric during insertion is equal and opposite to the change in stored energy.Hence the work done on the agent is again positive, but now with the value ΔW on agent =1 2Q0V0/epsilon1−1 /epsilon1. (72) That is, the dielectric is pulled into the capacitor whether or not the battery is still connected. Princeton University 1998 Ph501 Set 2, Solution 8 24 8. (a) The field energy associated with an electron at distance rfrom a grounded con- ducting plane is 1/2 that associated with the corresponding image charge, i.e., with that electron plus a positron at distance −r, in the absence of the conducting plane. Hence, U=−1 2e2 2r=−e2 4r. (73) The fields in the image solution have reality only outside the conducting plane; there is no energy associated with the “fictitious” image fields inside the conduc-tor. Equation (73) indicates that an electron is “bound” to the conducting plane, and so to escape, must have a minimum velocity related by v min=/radicalBigg 2|U| m=/radicalBigg e2 2mr=/radicalBigg e2c2 2mc2r=c/radicalbigg re 2r, (74) where re=e2/mc2=2.8×10−13cm is the classical electron radius. Thus, for r=1 A, vmin c=/radicalBigg 2.8×10−13 2×10−8=0.0037. (75) (Notice that the nonrelativistic approximation suffices.) The “binding energy” can be estimated from (73) as U=−e2 4mc2rmc2=−re rmc2 4=−2.8×10−13 10−85.11×105eV 4=−3.6e V.(76) (b) In equilibrium, the torque on dipole p2must vanish, and so p2will be directed along the electric field created by dipole p1. The electric field of the latter is given by E=3(p1·ˆr)ˆr−p1 r3. (77) The projection of Eonto the line connecting two dipoles is E/bardbl=E·ˆr=2p1 r3cosθ1. (78) The orthogonal projection is E⊥=E−E/bardbl, (79) leading to E⊥=−p1 r3sinθ1, (80) where the minus sign indicates that E⊥is directed opposite to p1,⊥. The angle of the field line, and hence of p2is tanθ2=E⊥ E/bardbl=−1 2tanθ1. (81) Princeton University 1998 Ph501 Set 2, Solution 9 25 9. We solve the problem of the capacity of two tangent, conducting spheres of radii aby the method of images. We first find the image-charge distribution needed to bring one sphere to potential V, but leaving the other at zero potential. Then, we complete the solution by superposingthe mirror distribution, obtained by reflection symmetry about the plane through the point of tangency of the two spheres. (a) Place charge q = aV at the center of sphere 1, bringing its surface to otential V. (b) To bring sphere 2 to zero potential, place charge −q(a/2a)=−q/2a td i s t a n c e a 2/2a=a/2 from the center of sphere 2, following the prescription on p. 41 of the Notes. (c) The image charge (b) takes sphere 1 away from potential V. To bring it back, add an image charge (c) inside sphere 1 so that this sphere is at zero potential underthe effect of charges (b) and (c). That is, add charge −(−q/2)(a/(3a/2)) = + q/3 at distance a 2/(3a/2) =a/3 from the center of sphere 1. (d) Add charge −q/4a ta/4 from the center of sphere 2 to bring it back to zero potential. (e) .... q q-q/2 -q/2 ... ... The total charge needed to bring both spheres to potential Vis double that described in the sequence above. Hence, Q=2q/parenleftbigg 1−1 2+1 3−1 4+.../parenrightbigg =2aVln2, (82) and the capacitance is C=Q/V=2aln2 = 1 .386a. (83) Note that since the dimensions of potential are [charge]/[length], capacitance has the dimension of [length] in Gaussian units. Thus, we expect that C≈afor this problem, sinceais the only relevant length. Princeton University 1998 Ph501 Set 2, Solution 10 26 10. (a) The uniform field E0=E0ˆzis approximated as being due to charges ±Qat z=∓R,w h e r e Q→∞ andR→∞ in such a way as to keep Q/R2constant. In the limit, the field in the region of the sphere is homogeneous and equal toE=( 2Q/R 2)ˆz. According to the image method, we can make the potential on the sphere vanish by adding charge q/prime=−Qa/R atz=−a2/Rand−q/primeat z=a2/R. Thus, the perturbation to the field due to the sphere is effectively that due to a dipole with the moment p=2a2 RQa Rˆz=a3E0. (84) (b) The potential outside the sphere is thus, φ=φ0+φdipole=−E0rcosθ+E0a3cosθ r2. (85) The field lines bend in to be normal to the sphere at r=a: p (c) We find the surface charge density σfrom the normal component of the electric field at the surface of the sphere: Er(a,θ)=−∂φ ∂r/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle r=a=3E0cosθ, (86) and so σ(θ)=Er(a,θ) 4π=3E0cosθ 4π. (87) (d) The force acting on the surface charge density σisF=σEr(a)ˆr/2=E2 r(a)ˆr/8π (where the latter form follows immediately from the Maxwell stress tensor). Theforce on the right hemisphere is directed along zand is obtained by integrating thezcomponent of F: F z=1 8π/integraldisplay1 02πa2dcosθ(3E0cosθ)2(cosθ)=9 16a2E2 0. (88) Since the force on the hemisphere at z>0 is positive, the hemispheres repel each other. Princeton University 1998 Ph501 Set 2, Solution 11 27 11. We seek solutions to Laplace’s equation in 2 dimensions, ∇2φ(x,y) = 0, of the form φ=X(x)Y(y). This leads to solutions of the form e±kxe±ikyore±ikxe±ky. Since the boundary conditions include φ=0a t y=0a n d b, it is advantageous to consider functions Yof the type e±iky, which can be immediately restricted to the form: Y(y)=s i n ky, where k=nπ/b, n =1,2,.... (89) This also fixes the separation constants k. The general expression for the potential is now: φ(x,y)=/summationdisplay nXn(x)Yn(y)=/summationdisplay n/parenleftBig Anenπx/b+Bne−nπx/b/parenrightBig sinnπy b. (90) The boundary conditions at x=0a n d x=aare φ(0,y)=V1=/summationdisplay n(An+Bn)s i nnπy b, (91) φ(a,y)=V2=/summationdisplay n/parenleftBig Anenπa/b+Bne−nπa/b/parenrightBig sinnπy b. (92) A straigthforward approach to find AnandBnis to multiply(91) and (92) by sin( nπy/b ) and integrate from y=0t o b: /integraldisplayb 0φ(0,y)s innπy bdy=−bV1 nπcosnπy b/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleb 0=bV1 nπ⎧ ⎪⎨ ⎪⎩2,nodd 0,neven=b 2(An+Bn),(93) and similarly, bV2 nπ⎧ ⎪⎨ ⎪⎩2,nodd 0,neven=b 2/parenleftBig Anenπa/b+Bne−nπa/b/parenrightBig . (94) Thus, for neven,An=Bn= 0, while for nodd, An=2 nπsinhnπa b/parenleftBig V2−V1e−nπa/b/parenrightBig ,B n=2 nπsinhnπa b/parenleftBig V1enπa/b−V2/parenrightBig .(95) Finally, we get for the potential: φ(x,y)=4 π/summationdisplay noddsinnπy b nπsinhnπa b/bracketleftBigg V2sinhnπx b+V1sinhnπ(a−x) b/bracketrightBigg . (96) To verify that this solution satisfies the boundary conditions, note that (91) and (93) combine to yield the expansion: 1=4 π/summationdisplay nodd1 nsinnπy b. (97) Princeton University 1998 Ph501 Set 2, Solution 11 28 We also note that the potential is symmetric about the midplanes, φ(x,y)= φ(a−x,y)=φ(x,b−y), which could have been invoked as far back as (90) to show that only odd ncontributes. Remark: This problem could also usefully be solved as the superposition of two cases, each with three walls at potential zero and the fourth at a nonzero value. The form of the solution (96) displays this superposition. Princeton University 1998 Ph501 Set 2, Solution 12 29 12. Since φ=0a t x=0,aandy=0,b, solutions φ=X(x)Y(y)Z(z) must have the form Xm(x)=s i nmπx a,and Yn(y)=s i nnπy b, (98) where nandmare positive integers (and odd, recalling the remark at the end of problem 9). The functions Z(z) then have the form e±kz.S i n c e φ=0a t z=0 ,w e can make the further restriction: Zmn(z)=s i n h kmnz, (99) where kmnis determined by inserting the trial solutions into Laplace’s equation, yield- ing k2 mn=/parenleftbiggmπ a/parenrightbigg2 +/parenleftbiggnπ b/parenrightbigg2 . (100) The general solution satisfying all the boundary conditions except for the one at the facez=cis: φ(x,y,z )=/summationdisplay m,nAmnsinmπx asinnπy bsinh/radicalBigg /parenleftbiggmπ a/parenrightbigg2 +/parenleftbiggnπ b/parenrightbigg2 z. (101) The remaining boundary condition tells us that V=/summationdisplay m,nAmnsinmπx asinnπy bsinh/radicalBigg /parenleftbiggmπ a/parenrightbigg2 +/parenleftbiggnπ b/parenrightbigg2 c. (102) To find Amn, multiply (102) by sinmπx asinnπy band integrate from 0 to ainxand from 0t obiny. Similarly to (93), we find Amn=16V mnπ2sinh/radicalbigg /parenleftBig mπ a/parenrightBig2+/parenleftBig nπ b/parenrightBig2c, (103) for odd mandn, and 0 otherwise. Hence, φ(x,y,z )=16V π2/summationdisplay m,n odd1 msinmπx a1 nsinnπy bsinh/radicalbigg /parenleftBig mπ a/parenrightBig2+/parenleftBig nπ b/parenrightBig2z sinh/radicalbigg /parenleftBig mπ a/parenrightBig2+/parenleftBig nπ b/parenrightBig2c. (104) Note that we have demonstrated the expansion 1=/summationdisplay m4 mπsinmπx a/summationdisplay n4 nπsinnπy bfor 0 <x<a , 0<y<b . (105) Since this follows from (97), we could have used it to go from (102) to (103) without performing the integrations.