Name SQL A) S - (1) [25 pointsj A spherical shell, radius R, has a surface charge arranged so that there is +0 on a on the larger remainder of the spherical cap covering angles from 8= 0 to r14, and spherical surface. Assume o is positive. — (a) Find the dipole moment of this charge distribution, measured from the center of the sphere. of (b) Find the electric field, with direction, measured a long distance z directly above the center the sphere, in the direction 0 = 0 (e.g. find the leading term at this point). ‘7i ( p J rccs9 G(r--R.)) rAr ;i9 o(&4 2i’ ; (yi-R) R3Tcoc - —‘ U - w L. a — 2R ‘ () fL (1/ — 1/ o 3 irR ;-, / = -4 — = 2R zR Yo -4Le I/) - 1eL- 4 J 11rf :: -1Z 2 — ‘4, SL ) TR(O’7A) Name___________ 2 where C is a (2) In some region of space, a steady magnetic field is given by, B = Cx constant. (a) Find the current density in this region. ulation, (b) Show that this current density satisfies the continuity equation with no charge accum as expected for a steady current distribution. (c) Find the vector potential associated with this current. [1 suggest using the gauge choice discussed in class, though as always for such problems there are other solutions that may work as weLl.I , --r - C) V‘3 ‘\Thr ‘ %, D Z _)_ i\ — () / $ j V = (C) A VA — —) io / 3 A =- Cx L ___ ___ Name dielectric sphere, of (3) A grounded metal sphere, radius R, is placed in the center of a linear By an arrangement of external fields, a radius 2R. The relative permittivity of the dielectric is (at 2R), where a is a potential V = acosO is generated at the outer surface of the dielectric constant. dielectric, R < r < 2R. (a) Write the general form for the electric potential in the region of the in terms of a and (b) Using appropriate surface conditions, solve for the potential in this region R. (c) Find the electric field and the displacement field in this region. and find the (d) Find the free charge distribution on the grounded sphere at the center, s. distribution of bound charges in the dielectric and on the inner and outer surface . Sie (q) oCC ? / (C V, 2 r. - ) R J r=2R Vcs (2/ /y- V .= / Cc-) - ( = 1 2R A -- E’ 4_1( (ck) —I AA $ ,• 1 JR i f 0 0 I-J--; => y r 7 R, ,c “ = = 4 —( 3 -;I r-- L - p — s (2R) -tA, / = P p— U - 3) ( -() Name (4) A hollow sphere has a single layer of windings around the z axis ( direction), with the turn 0 is a constant. density (turns/length) equal to no sin8 where n 0 is large, the resulting current density can be approximated a (a) If the wires are very thin and n surface current. Find K as a function of 0 if the current in each wire is 1. rsinO) inside the 1 (b) In this situation, we can show that the vector potential is given by, (C 2 are constants. For these vector 1 and C 2 sinO / r2) outside the sphere, where C sphere, and (c , , potentials, solve for the corresponding magnetic fields. . 2 1 and C (c) Using appropriate boundary conditions, solve for the constants C &(cf) EE K I-1Y10Sl C C) ih ;& A r. 5(3 v>i6 / \ - C CI 1- 2C 1 VLt,1t r )-u5 ( 3 V 1 t, / _2Cco6L ‘ (.7 — > 3 R A. ifJ 1c Cl 5 )5 Ca5((i) yL 134 ( 3 = ,A(QT4AR Name ,$cL AJ S ion. The shaded region (radius R) carries uniform (5) Consider a long wire, shown in cross-sect circle is a hollow region (radius R!2) running current density J into the page, while the smaller the length of the wire. inside and outside of a solid wire carrying (a) First show how to calculate the magnetic field were no hollow region. uniform current, such as the shaded wire if there ibution shown, at points along the horizontal (b) Now find the magnetic field of the current distr the hollow region. [Hint: think diameter, both in the solid region of the wire and in superposition.I C) s AJe tS.e 4L± fk1j L2 o4it4 6- 1 ? cLoj€. k\4-e 3- fe-- ru Z rrf .:/4(J Jir $ 1;? / c 0 C L v Lj 1— ( ft (L -ej 4 4 L / 14 (Qh ) c c f3’-S (JIL±( rD Z (LLo S 4 t-9 J ) ::. i:I t (V & — - Z (siR/L) — (y÷) Vj ÷(-s) C&pLl C 1 (7 1. +i 1/ L 5 R) /