Question

1 Potential of concentric spheres A spherical shell with internal radius Rį and external radius R2 has a potential in its surfaces given by 0(R1,0,0) = Vi sin (20) sin(0) and (R2,0,0) = V2 sin (20) sin(0) (V1 and V2 are constants). If there are not electric charges any where inside or outside the shell R R2 (a) Write the general solution for the electric potential o in each of the three regions of interest: r < R1, R; <r < R2 and r > R2 in terms of sums of the radial functions and spherical harmonics. (b) Use the boundary conditions and continuity for finding the relations between the coefficients that appear en the solutions for the three regions. (c) Find the particular solution in each of the three regions using the boundary condi- tions of the shell. Some useful spherical harmonics Y;' = JAH Y;" = cos e y+ = +V sin beti Y; 152 (3 cos? 0 – 1) yl = = sin 6 cos 6etiYz? = 15 sinº 6e+2i® 871 32

Answer 1

1 ) RI R2 (R10,0) = vi singo) sing = 2V, sino coro sing = 9Vsino 2080 (eit - e-io iv 8T ET iv, sino coso (e-ig - eim) 15 sinocoso e 10 - sino eso eit ivije (-Y5(0,0) - Y(0.5)) 50, (Ruj 0,0) = -ivi vet ( 43(0,4) + Yz'(0,0) And I (R2,0,0) = (1) V2 singo sing - Liv> VE (YA (0,4) + X 10.60) -(3)

The potential is continions at so= R1 The general solution for the eleettoie polential kal in region no < R is, 1 $, (0,0%) = Ž Ž Alm rol yum (0,0) 1=0 m = -1 The potential in region P, (0,0,0) = Ž Ž (Bm roft Cum general solution for the electric Ri< ro< Rq is, l 1=0 m =-l rarti ya 0,0) And the general solution for the electric potential in roegion roy Rz is, 93 (m,0,0)= You (0,0) l Dem 1=0 m =-d molti b) Boundary conditions %, (R1,0%) = R.,80 > Z Ž Amru yn (0,4) 1=0 m =-d N d i Å (BimR" + Cam R;!H) you (050) 1=0 m =-d

How equating the coefficients of B the methogonal fimetions, yom (6,4), we yet, l 1 Aim Ri Bim R, " + Cam atl Rieti continious at so = R2 Now, the potential is P2 (R2,0,0) = 3 (R2,0,0) → E (BimRy + YAM (8,8) RATI l cem 1=0 m = -1 e Dim ك ك YM (050) l=0 m =-1 Reti Equating the coefficients of Yom (6,4) we get, Bim Rot Com R₂1 lti Dim R 4+1 (4) There is no free som face change selation between and Eq (4) are the the coefficients.

Rest of all Alm. C) How from (R1,0,0)= - IVV eq (a) we get, Y(056) + Y3% (0.05) RT 15 -ive 87 15 X30 l l 1=0 m =-d => B11 R1,0,0) = % (R1,0,0) (Ya ty Z E Am R, y m (of) --ivi vet (9. (09) • +42' (0-9) both sides composing get, -ivi > Asi > As, i = - 11/27 - R? ivija we Ri Asi - - ; 8T VI And RiTAs, 15 A2,-1 As,1 = -V 85 vi R. 15 moe Zero.

We also have, (R2,0,0) = (R2,0,0) l → E E Dim Yom (6,4) ati 1=0 ml Live 1910.8) 4 X 110,00 comparing both sides we gets Dal 15 R} 7 Dal - 87 V 15 V₂ R And -iet va R2 Ds,- Dall ( 37 3 VR Rest of all Dim me zero: How froom eq(3) we get, 2 Azi Ri By Ri (2) + R3 - Bal+ Cai R5 A21 (5)

similarly, - B2,-) + C2,- R5 A2,-1 Asl (6) And from eq (4) we get B2, R 2 + C21 R3 Da! - R 3 > BRI + 21 Dal R25 (7) similarly, B2,-1+ (2,-1 Dzi1 Dal R25 RE R5 (8) From eq (5) and eq (7) we eget, Cal (21 R5 Dzi Rz AR1 R シ Cal Dal -A21 RS RE R,5 11 - i 87 V2 15 R² Vi R, RS Ris

SO) Car = 8T 15 ( V2 R2 Vi R2 1 ㅗ Rob R,5 Similarly froom eq (6) and eq(8) we V2 (2,-1 (9) 1 VE OT 15 le VI R2 1 R5 volt How, from en (5) we get, Bal AR) - (2) 1,5 Live 8T VI 15 ti 85 l Corona Vi 15 R5 Roz R2 RS RAS NI VE 8t 15 V2 R2 R? V R, 2 R,5 R5 -1 8T he VI R5 R2 R₂5 V2 R² R5 R35 - 1 Vi -i 8T 15 RRR V2 R2R5 R2 R5

Similarroly from eq (7) we We get, B2,-1 = Vz -1 87 15 ilis RRR RZR,5 I RE Rio All others $0 Bem, clam ane zero. the particulars solutions in the three region in is, Foro no <RI l $1(1,0,0) = ² ² Amrot you (0.8). d=0 m =-d Asi Ya + A2, m² 2 + A2, -10² Y 1 - 3) (4:10,6) + Y3 (0:0) As, Toro so y R2, Dim $3 (0,01$) = 3 = Y" ( 0 ) molti l=0 m =-d Dal you + Day-1 n3 13 Y

8T 15 vzleta) (42+y5' → 03180,0,0) 80 15 ()'( %d(0.8) +y=40.6) And foro Rino LR2 $, (-0,0,0) = Ž Bimrol + Camym 1=0 ma-l put Ball ² y + B2 - 1 Y 2 + C2,1 401 - B21 M² Y + B2, 102 Y = 1 + Cal 23 Y₂ + C2,- -Y m3 - 80 81 Vi RRE 15 V2 R²R 5 102 CY_' + Y.') - RO R,5 8T Live V2 Ro 1. R2 -) R5 (13+ *>') 1 3 RE