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3.45. cp ;--.:.: -I- --:.--r/ - (1 x^ ) , E,•,=:-.
aiR ,, If xR, 2e0 1 1 s 2-^1 -^112 2e0(x2 + R2)3," ' then cp ,-...z.-, ± 40:x2 and E ,--:-.', 2318Poxs , where p= nR2al.^ In the
formulas for the potential cp the plus sign corresponds to the space adjoining the positively charged plate and the minus sign to the space adjoining the negatively charged plate.
3.46. (a) F = 0; (b) F = n _XP 2 ; (c) F -= .., 4 3p2^4 ngor^ Itsr 3.47. F = (^) 2neot 4 = 2.1.
-16 N.
3.48. cp = —axy const. 3.49. cp = ay (3 — x2 ) + const. 3.50. cp = —y (ax^ bz)^ const. 3.51. p = 6e0ax. 3.52. p = 2e0AT/c/2; E = pdieo. 3.53. p = —6e0a. 3.54. q= 41 17 neokx. 2 3.55. A= q 16aso/
3.56. (a) F = (2 i8!: 311 2 ) q2 ; (b) E= 2 (
3.57. (^) F = (2 (^) 323t8012q
3.58. F = (^3132) 327[8,0 '
3.59. a -=
ql 2 7t (12 +r2)-^ 3/2^ t^ qind =^ q• 3.60. (a)^ F1= 47t)'820/ ; (b) a =^ (4x2).
3.61. (a) a =^ (b)a^ (r)^ 2n 1712+r
3.62. (a) (/2 + R2)3/
— (b) E= , 0
lq 422 L8 41 +1/4 (R/1)93/ 1 `= 471E0 (^) _R_ (^1) 1/1+4 (i/R)
3.63. (^) (p = (^) 4neol •
3.64. cp=
1 1 _L 1
4aeo Ri - 1- Rg
11r-114 if a < r < b,
3 '65' q2 a q1; (1—bla) r if r > b.
3.66. (a) E 23 = Acp/d, E12 = E34 = 112 E23; (b) al = = 1/2804M, az = I as I = 3/2804/d. 3.67. q1 = —q (1 — x)I1, q2= —qx/l. Instruction.^ If the charge q is imagined to be uniformly spread over the plane passing through
q
5 lig 1 neo
that charge and parallel to the plates, the charges qland^ q 2 remain, obviously, unchanged. What changes is only their distribution, and the electric field becomes easy to calculate. fx 3.68. dFldS = 1 /262/80. 3.69. F = (^) 32xce q 0.5^ kN.
2 oR2 —^
3.70. F = 1 /4nR2aV80. 3.71. N — (^) (e- nel) 1) 4E —^ 3.103,^ Fig. 19. where nois the concentration of molecules. 3.72. F=3131)2 4:1260/ 1.1R (attraction) 3.73. (a) x = R/17- ; (b) x = { 0.29R (repulsion). See Fig. 19.
3.74. P — ee l 1 r, q
, i =^ e^ e^ l^ q. 3.76. qinn = — q (8-1)1e, q;,,,t =q^ (e —1)/8.
(a)
r
Fig. 20.
3.77. See Fig. 20.
3.78. E cos2 ao + 82sin2 ao= 5.2 V/m;
tan a = 8 tan ao, hence, a = 740;^
al = go (8_1) e E0 cos ao =
64 pC/m2.
3.79. (a) () E dS =-- 8— e 1 nR2E0 cos 0; (b) § D dr=--so^ (6-1) X X 1E0 sin 0.^ 1 pdp/1lcee 00 for (^) 1<d, f — p/2/2880 for^ 1<d, 3.80. (a) E=^ for / > d, (1)=-1 — (d/2e + 1— d) pd/e for 1>-d._ The plots Ex(x) and p (x) are shown in Fig. 21. (b) a' =^ pd (e —^ 1)/8, p' = —p (a — 1)/8. 1 pr/3808 for r < R, 3.81. (a) E = pR3/380r2 for r >^ R; (b) p' = —p (8 — 1)/8, a' = pR (8 — 1)/3e. See Fig. 22. 3.82. E = —dP/480R. 3.83. E = —Po(1 —^ x21d2)180, U =^ 4dP0/3co.
3.105. C = 4asoa/In (R2/R1). 3.106. When siRiElm = 82R2E2m.• 3.107. V = (^) [ln (R2/R1) (^) (81/ 82) In (R3/R2) 3.108. C 2T80 In (b/ a). 3.109. C 2Itso/ln (2b1a).
3.110. C z 2neosa. Instruction. When^ b >> a, the charges can
be assumed to be distributed practically uniformly over the sur- faces of the balls. 3.111. C 4asoa. 3.112. (a) (^) Ctotal Cl C2 + (^) C3; (b) (^) Ctotai = C. 3.113. (a) C = 2s0S/3d; (b) C = 380S/2d. 3.114. V V1(1 Ci/C 2)^ = 9 kV. 3.115. U = 61(1 + 3rd + 12) = 10 V. 3.116. C x = C (11/5 — 1)/2 = 0.62C. Since the chain is infinite, all the links beginning with the second can be replaced by the ca- pacitance Cxequal to the sought one. 3.117. V1 = q/C1 = 10 V, (^) V2 = (^) q/C2 = 5 cpB (^) + 6) C1C2I(C1 + C2).
V 1= (62— 60/(1 + C11C2), V2 =(6., — 62)41 + c2/c1).
q = 61 — 621 c1c2/(c1 + C2).
c2cs—cic
TA — TB= (^) (C 1 C2) (C3+c4) In the ease when^ C1/C2=
V 3.121. q-— (^) 1/C11/C2 1/C3_ 0.06 mC. 3.122. q1 = (^) gC2, q2= — gUtC2/(Ci -FC2)• 3.123. q1 - = 6C1 (C1 — C2)/(C1^ =^ —^ 24 .tC, 42= gC2 (C1 —C2)/(C1 + C2) — 36 I.LC, q3 =^^6 (C2— C1) = +^ 6011C. 3.124. (^) WA— TB (C2g2—Cigi)/(C1 1- C2+ CO-
3.125. (pi =W2C2 + W1C3+ W1 (C2+ CS) ci d-c,-Fc
W3c3— W2 (C1+ C3) W1 C1W2C2 -43 (C1+ C2) (4)2 = (^) C1+C2±C3 (^) , W3 (^) Ci+C2±C
3.126. Ctotal = 2C1C2 +C3^ (C1 +C2) ci+C2+2c
q2143180a. 21n 2 q 2 4:180 a • —q2/85t80 /. 41q2/4aso /.
= — 1/2172C1C2/(C1 + C2) =
e2cco/(2c +
1/2ce22. It is remarkable
W2 +
V, where q = = (WA — 3.118. 3.119. 3.120. C3/C4.
3.127. (a) W = (4/- -P 4) q2/4neoa; (b) W=(1/2-4)= 2 q2/43teoa; (c) W= — 172 3.128. W —
3.129. W = 3.130. W = 3.131. AW 3.132. Q = 3.133. Q = independent of 3.134. W =
1 q? 4n60 \ 2R
4_ q2 q1q 2R2 (^) R2 •
that the result obtained is
—0.03 mJ.
312
3.135. (a) W = 3q2/20ne0R; (b) W1/W2= 1/5, 3.136. W = (q2/8ne0 e) (1/a — 1/b) = 27 mJ. 3.137. A = (0182180) (11R1— 11R2). 3.138. A—q(4o+q12)^
_47E80 _
1 R1 R2 ) 3.139. F1= a2/2e0. 3.140. A = (q2/8ne0) (1/a — 1/b). 3.141. (a) A = q2(x2— x1)/2e0S; (b) A -----80SV 2(x2— xl)/2x1x2. 3.142. (a) A = '12CV211(1 — 1)2 = 1.5 mJ; (b) A = 1 12CV2ria (a — WEE — 1 (a — 1)12 = 0.8 mJ. 3.143. Ap = 808 (a — 1) V2/2d2= 7 kPa = 0.07 atm. 3.144. h = (a — 1)62/2a0epg 3.145. F = nRe0(a — 1) V2/d. 3.146. N = (8 —^ 1) e0R2V 2/4d.
3.147. I = 2na0aEv = 0.5 RA.
3.148. I 27(60(a — 1) rvV Id =^ 0.11 p,A. 3.149. (a) a = (al la2)/(1 i); (b) a (a2 rlai)/(1 -1- 1). 3.150. (a) 516R; (b) 7/12R; (c) 3/4R. 3.151. Ric = (ij-— 1). 3.152. R = (1 -1-1/ 1 + 4R21R1) R112=6^ Q. Instruction.^ Since the chain is infinite, all the links beginning with the second can be replaced by the resistance equal to the sought resistance R. 3.153. Imagine the voltage V to be applied across the points A and B. Then V = (^) IR = I 0R0, where I is the current carried by the lead wires, 10 is the current carried by the conductor AB. The current / 0can be represented as a superposition of two cur- rents. If the current I flowed into point A and spread over the
infinite wire grid, the conductor AB would carry (because of symmet-
ry) the current 1/4. Similarly, if the current I flowed into the grid from infinity and left the grid through point B, the conductor AB would also carry the current I/4. Superposing both of these solutions, we obtain /0= 1/2. Therefore, R = R0/2. RAc= (^) iRo 3.154. R = (p/2a/) In (b/a). 3.155. R = p (b — a)/4aab. In the case of b^ oo R =^ p/4.1-Ea. 3.156. p = 4nAtabl(b — a) C In 3.157. R = p/2na. 3.158. (a) j = 2a1V1pr3; (b) R = p/4na. 3.159. (a) j = Z VI2pr2 In^ (11a);^ (b) RI= (p/n) In^ (11a). 3.160. I = VC/paa0 = 1.5 [LA. 3.161. RC = pea,.
3.162. a = D, = D cos a; j = D sin cr:/ 8801)•
3.163. I = VS (a 2— al)/d In (62/a1) = 5 nA. 3.165. (^) q= so(p2 — (^) pi) 3.166. a = c0V (821)2 — eiPi)/(Pidi p 2d2), a = 0 if gip,. = E2P2• 3.167. (^) q = 801 (8 21)2 — 3.168. p = 2e0V (rt — 1)/d2(7) + 1).
