Fig. 21. Fig. 22.13.84. (a) E 1 = 2sE 0 /(e + 1), E^2 = 2E^0 /(e + 1), D 1 = D2 =2860 E 0 /(e + 1); (b) E (^1) = Eo, E2 = EVE, D (^1) = D2 = a0E0.
3.85. (a) (^) = E 2 = Eo, D1= 8 0E0, D2 = 8 D1; (b) E2
= 2E 0 /(e + 1), D^1 = 2e^0 E^0 /(e + 1), D2 =^8131.
3.86. E = q/2ne, (e + 1) r 2.
3.87. p = poe/(e — 1) = 1.6 g/cm^3 , where a and po are the per-
mittivity and density of kerosene.
3.88. cr;flax = (e — 1) NE = 3.5 nC/m^2 ,^ nR^2 (e — 1)^80 E=
= 10 pC.
3.89. (a) Since the normal component of the vector D is contin-
uous at the dielectric interface, we obtain
= —ql (e — 1)/2nr^3 (e + 1), for^1 0 and a' -4- 0;
(b) q' = —q (8 — 1)/(e + 1).
3.90. F = q 2 (a — 1)/161te^012 (e + 1).
q/2n (1 + e) r 2 in vacuum,
3.91. D =
eq/2n (1 + e) r 2 in dielectric;
E = q/2neo (1 + e) r 2
cp = q/2neo (1 + e) r
both in vacuum and in dielectric.
3.92. a' = ql (8 — 1)/2nr 3 e (e -I- 1); for^1 0 and a'^ 0.
3.93. a' = ql (a — 1)/2nr^3 e.
3.94. E^1 = Ph/eod (between the plates), E2. = —(1 — h/d)P/co,
= D2 = Phld.
3.95. p' = —2a, i.e. is independent of r.
3.96. (a) E =
3.97. E 0 = E — P/36 0.
3.98. E = 3E 0 /(e + 2), P = 3e^0 E^0 (e — 1)/(a + 2).
3.99. E —P/28 0.
3.100. E = 2E 0 /(e + 1); P = 28^0 E^0 (e^ 1)/(e + 1).
3.101. C — 4neoaR1
1-1-(8-1) R^1 /R^2 '
3.102. The strength decreased^1 / 2 (e + 1)^ times; q =
=^112 C6 (8 — 1)/(e^ 1).
S (^) 81- 82
3.103. (a) C = d1/e1
(^80) ±d2/e2 ; (b) a' = eoT 7 8,42+ 82di.
3.104. (a) C= co (e^2 — el) Sid In (e^2 /e^1 ); (b) p' = — q(e^2 — ei)/dSe2.