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Homework Solutions: 1 Note: In these solutions, generally bold fonts denote vectors, and the corresponding symbol in plain font denotes its magnitude. E.g. The electric eld E and its magnitude E. Ex. 1.1: Capacitor (4) (a) (This should be a familiar exercise, so I will not include diagrams.) First consider an in nite plane, with uniform surface charge density s . Applying Gauss s law, with a Gaussian surface of uniform cross-section, having area A, we have 2EA = E dA = 4 s dA = 4 s A , (1) E = 2 s . Ignoring edge e ects, our capacitor is equivalent to two in nite parallel planes, with surface charges of opposite sign. Then from (1), the superposition principle immediately implies that the electric eld between the plates has magnitude E = 4 s . (2) (b) Consider one in nite sheet in our capacitor. The view looking edge-on to this sheet is shown below. The force on the sheet is due to the Lorentz + s Eother ? interaction with the electric eld of the other plate, F = qEother , where q is the charge on the sheet and Eother = 2 s from above. By de nition the pressure is p = dF/dA, and note that also by de nition s = dq/dA. Since the electric eld is constant, then p= dq Eother = s = 2 2 . s dA (3) Note this pressure is attractive. (c) Including dielectrics polarizations, Gauss s law in integral form becomes D dA = 4 qf , (4) 1 where qf are the free charges. The D eld depends only on free charges, which in this system lie on the two sheets forming the capacitor with density s : there are no free charges inside or on the dielectric. Hence, similarly to part (a) we nd D = 4 s between the plates, both inside and outside the dielectric. Now, outside the dielectric in the vacuum, D = E, so Eout = 4 s . (5) The dielectric material has dielectric constant , and by de nition D = E. So inside the dielectric 1 Ein = 4 s . (6) (d) Remember that to nd the potential, you should carefully de ne a coordinate system, shown below. Then potential at b with respect to a is + s 6 x E ? b c s a b Vba = a c E dx b =+ a Ein dx + c Eout dx = 4 s d/2 + (1/ )4 s d/2 = 2 s d(1/ + 1) , which is positive, as expected. (7) Ex. 1.2: Spherical Cavity (8) (a) Consider a sphere S of radius R with uniform polarization P = P0 ez , shown below. First, the polarization is uniform so that clearly P = 0, and hence the volumetric bound charge density b = 0. The surface bound charge density sb = n P is nonzero, however. Now, in the spherical coordinates (r, , ), where and are the azimuthal and inclination angles respectively, the unit normal vector is n = (cos sin , sin sin , cos ) , (8) and so sb = P0 cos . 2 (9) P 6 0 n Coulomb s law then tells us that the eld at the origin due to the charge on an di erential area dA is simply sb (10) dE(0) = n 2 dA = nP0 cos d , R since dA = R2 d on the surface of a sphere and we have used (9). Note the eld points inwards for positive s whence the minus sign. Applying the principle of superposition, we integrate over the spherical surface S, so that 2 1 E(0) = P0 d 0 d(cos ) cos (cos sin , sin sin , cos ) 1 1 = 2 P0 ez 1 d(cos ) cos2 (11) = 4 4 P0 ez = P . 3 3 (b) Consider two spheres labelled 1 and 2, with respective centers located at /2 and /2 and respective uniform charge densities + 0 and 0 . Outside each sphere, the electric eld due the sphere in question looks like that of a point charge, with charge q = 0 V . The dipole moment is then clearly just q , so that the dipole moment per unit volume - the polarization - is P = 0 . (12) Hence we need P0 = 0 for these two systems to have the same polarization. The electric eld inside sphere 1 is E1 (r) = and for sphere 2 E2 (r) = 4 0 r + /2 , 3 (14) 4 0 r /2 , 3 (13) 3 We assume that R, so that the overlap of the spheres is approximately complete, so the intersection of the spheres is approximately our original uniformly polarized sphere. The eld inside this sphere is then, by superposition 4 0 , (15) E(r) = E1 + E2 = 3 which is a constant. (c) Finally, consider a spherical cavity in a polarized medium, with uniform polarization P . The spherical cavity is equivalent to superimposing a sphere of polarization P on the medium, as shown below. From (11), 6 P 6 P P ? the eld in the cavity due to the P polarization is Epol = (4 /3)P . Hence, if the eld in the polarized material E, is then by superposition the eld in the cavity is Ecav = E + 4 P. 3 (16) Ex 1.3: Ring of Current (6) Consider a ring of current centered at the origin, having radius R and carrying current I. We label a position on the symmetry axis z and a position on the ring r. Using the usual cylindrical coordinates (r, , z), we then have in Cartesian coordinates z = (0, 0, z) r = (R cos , R sin , 0) . (17) Similarly, an elemental piece of the ring is dl = dle , where e denotes the unit tangent vector to the ring at position . That is dl = Rd ( sin , cos , 0) . By the Biot-Savart law, the magnetic eld B(z) = = I c 2 (18) dl (z r) z r 3 I R (z cos , z sin , R) d 2 c0 (R + z 2 )3/2 IR2 2 = e. 2 + z 2 )3/2 z c (R 4 (19) Now, B(z)dz = 2 I c R2 dz . (R2 + z 2 )3/2 (20) Let z = R tan , and then we have B(z)dz = 2 I c /2 /2 R3 sec2 d (R2 sec2 )3/2 = /2 2 I cos d c /2 4 I. = c (21) This is precisely the magnetic line integral around any Amp`ran loop through e which the current loop passes, except that the z-axis, over which we have integrated, does not seem to be a closed loop . The resolution of this problem can be understood in various equivalent ways. One is to consider the z axis to be the limit of a circular Amp`ran loop enclosing with radius , or a e square loop with side length etc, noting the eld is zero as r . A more mathematical understanding is to note that the Amp`ran loop really just e needs to be the boundary of a well-de ned surface, and here our surface is a half-in nite plane whose boundary is the z-axis. Ex 1.4: Coaxial Solenoids (4) (Again, this exercise should be familiar, so I ll spare some details.) Consider two in nitely long coaxial solenoids, shown below in sectional view. We assume the solenoids are wound with the same orientation. For each solenoid, applying n2 n1 6 a2 a1 6 -x Amp`re s law using a loop of the form indicated by the dashed line, having nite e side length L (the other sides extend o to in nity), one nds B2 L = B dx = (4 /c)Ienc = (4 /c)n2 LI . (22) Hence the eld due to the outer solenoid is B2 = (4 /c)n2 Iex inside the solenoid, and one can deduce the eld outside is zero by using a similar rectangular loop with only nite sides. 5 Applying the superposition principle, and noting that the currents ow in di erent directions in each solenoid (a) B(r < a1 ) = (4 /c)I(n1 n2 )ex . (b) B(a1 < r < a2 ) = (4 /c)In2 ex . (c) B(r > a2 ) = 0 . (25) (24) (23) Ex 1.5: Newton s 3rd Law (8) Consider two current elements I1 dl1 and I2 dl2 , which are located at l1 and l2 by de nition and which belong to circuits 1 and 2 respectively. Applying the Biot-Savart Law, the magnetic eld due to 1 at 2 and due to 2 at 1 is respectively 1 I1 dl1 c 1 dB2 (l1 ) = I2 dl2 c dB1 (l2 ) = The Lorentz force of 1 on 2 is then dF12 = (l2 l1 ) I1 I2 dl2 dl1 c l2 l1 3 , (27) (l2 l1 ) l2 l1 3 (l1 l2 ) . l2 l1 3 (26) where the square brackets are needed as the triple cross-product is not associative, and dF21 is found by exchanging the subscripts everywhere. Applying the Lagrange expansion of the triple cross-product, we may write the di erential force as dF12 = I1 I2 (l2 l1 ) dl1 dl2 c l2 l1 3 (l2 l1 ) dl2 dl1 l2 l1 3 . (28) In this form it is clear that dF12 = dF21 since although the second term clearly just changes sign under exchange of subscripts, the rst term is a di erent di erential form altogether. In integral form, we have F12 = I1 I2 c dl1 1 dl2 2 e(l2 l1 ) l2 l1 2 dl2 2 dl1 1 e(l2 l1 ) l2 l1 2 . (29) Note l1 and l2 are now dummy variables: to nd F21 we instead exchange the integral domains 1 and 2 . For convenience, the vector terms have also been rewritten in terms of the unit vector e(l2 l1 ) = e(l1 l2 ) . 6 Now, in the rst term we have a closed path integral over the conservative eld er /r2 . That is e(l2 l1 ) dl2 =0, (30) l2 l1 2 2 for any l1 . Hence we have F12 = I1 I2 c I1 I2 = c I1 I2 =+ c = F21 , as expected. dl2 2 dl1 1 dl1 2 dl2 1 dl2 1 dl1 2 e(l2 l1 ) l2 l1 2 e(l1 l2 ) l1 l2 2 e(l2 l1 ) l2 l1 2 (swap variables) (swap integrals) (31) 7

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