3.135. (a) W = 3q 2 /20ne^0 R; (b) W 1 /W^2 = 1/5,
3.136. W = (q^2 /8ne^0 e) (1/a — 1/b) = 27 mJ.
3.137. A = (018218 0 ) (11R 1 — 11R 2 ).
3.138. A—q(4o+q12)^1
47E8 0 \1
R1 R2 )
3.139. F 1 = a 2 /2e 0.
3.140. A = (q 2 /8ne 0 ) (1/a — 1/b).
3.141. (a) A = q 2 (x 2 — x 1 )/2e 0 S;
(b) A -----8^0 SV^2 (x^2 — xl)/2x^1 x^2.
3.142. (a) A = '1 2 CV 2 11(1 — 1)2 = 1.5 mJ;
(b) A = 112 CV 2 ria (a — WEE — 1 (a — 1)1^2 = 0.8 mJ.
3.143. Ap = 8 0 8 (a — 1) V^2 /2d^2 = 7 kPa = 0.07 atm.
3.144. h = (a — 1)6 2 /2a^0 epg
3.145. F = nRe 0 (a — 1) V 2 /d.
3.146. N = (8 — 1) e^0 R^2 V^2 /4d.
3.147. I = 2na 0 aEv = 0.5 RA.
3.148. I 27(6 0 (a — 1) rvV Id = 0.11 p,A.
3.149. (a) a = (al la 2 )/(1 i); (b) a (a 2 rlai)/(1 -1- 1).
3.150. (a) 516 R; (b) 7 / 12 R; (c) 3 / 4 R.
3.151. Ric = (ij — - 1).
3.152. R = (1 -1-1/ 1 + 4R21R1) R 1 12=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 0 R 0 , where I is the current carried by the
lead wires, 10 is the current carried by the conductor AB.
The current / 0 can 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 = R^0 /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 = 2a1V1pr 3 ; (b) R = p/4na.
3.159. (a) j = Z VI2pr 2 In (11a); (b) RI = (p/n) In (11a).
3.160. I = VC/paa 0 = 1.5 [LA.
3.161. RC = pea,.
3.162. a = D, = D cos a; j = D sin cr:/ (^88) 01)•
3.163. I = VS (a 2 — al)/d In (6 2 /a 1 ) = 5 nA.
3.165. q= so (p2 — pi)
3.166. a = c (^0) V ( (^8) 21)2 — eiPi)/(Pidi p 2 d 2 ), a = 0 if gip,. =
E2P2•
3.167. q = 801 ( 8 21)2 —
3.168. p = 2e 0 V (rt — 1)/d^2 (7) + 1).