Irodov – Problems in General Physics

that charge and parallel to the plates, the charges ql and 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 /8 0.
3.69. F = q 32xce 0.5 kN.

2

oR2 —^

0

3.70. F = 1 / 4 nR 2 aV8^0.

3.71. N — (^) (e-nel) 1) 4E — 3.10^3 ,^ Fig. 19.
where no is the concentration of molecules.
3.72. F= 3131)2 4:1 260 / 7
1.1R (attraction)
3.73. (a) x = R/17- ; (b) x = { 0.29R (repulsion). See Fig. 19.
3.74. P — eel 1 r, q
,
i =^ eel^ 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 cos^2 ao + 8^2 sin^2 ao = 5.2 V/m;
tan a = 8 tan ao, hence, a = 740;
al = go (81)
(^) e E 0 cos ao =
64 pC/m 2.
3.79. (a) () E dS =-- 8— e 1 nR 2 E 0 cos 0; (b) § D dr=--so (6-1) X
X 1E 0 sin 0.
1 pd p/1 lc ee 00 for 1<d, f — p/ 2 /288 0 for 1<d,
3.80. (a) E= for / > d, (1)=- 1 — (d/2e + 1— d) pd/e 0
for 1>-d.
The plots E x (x) and p (x) are shown in Fig. 21. (b) a' = pd (e — 1)/8,
p' = —p (a — 1)/8.
1 pr/38 0 8 for r < R,
3.81. (a) E =
pR 3 /38 0 r 2 for r > R;
(b) p' = —p (8 — 1)/8, a' = pR (8 — 1)/3e. See Fig. 22.
3.82. E = —dP/48^0 R.
3.83. E = —Po (1 — x^2 1d^2 )18^0 , U = 4dP0/3co.