Irodov – Problems in General Physics

pot

38 o

and E

3.16. E = — 1 / 3 ar/e 0.

3.17. E = — 1 / 3 ka 0 /8 0 , where k is the unit vector of the z axis

with respect to which the angle 0 is read off. Clearly, the field inside

the given sphere is uniform.

3.18. E = — 1 / 6 aR (^2) /e 0.
3.19. I (1) I = 1 / 2 kR/e (^0). The sign of (I) depends on how the direc-
tion of the normal to the circle is chosen.
3.20. I (DI eo
171+ (R//) 2 /^1 • The sign of 413 depends on
how the direction of the normal to the circle is chosen.
3.21. I (I)I = 1 / 3 3tpro (R 2 — 4)/8 0.
3.22. Emax = Xhted.
3.23. E = 1 / (^2) a 0 /8 (^0) , with the direction of the vector E corres-
ponding to the angle cp =
3.24. (1) = 421.//a.
(b) Emax 1 / 9 poR/e (^0) for r,„= 213 R.
3.26. q= 22tR 2 a, E = 1 / 2 a/8 0.
3.27. E=-- 132
--1 °r2 r2 (1—e-ar 3 ). Accordingly, E
Po
Uoar 2 •
3.28. E = 1 / (^3) ap/so-
3.29. E = (^1) /2ap/g (^0) ,
axis of the cavity.
(^1) )
ateon
,
1+ (a /R)2
3.30. Acp r, 1
3.31. cpl — cp 2 = — In = 5 kV. 288o
3.32. (p. = 1 / 2 aR/ec,, E= 1 14c (^1) /Eo.
3.33. co
e°

al Cjil + (R/02-1),a /,, ± R.

l—,- 0, then (p= , E= 2 a 80 ; when / ),R, then cp

E z 4aq 8012 ' where q=-- aaR 2.

3.34. IT = GR/neo.

3.35. E = —a, i.e. the field is uniform.

3.36. (a) E (^) —2a (xi — yj); (b) E = —a (yi — xj). Here i, j
are the unit vectors of the x and y axes. See Fig. 16 illustrating the
case a >0.
3.37. E = —2 (axi ayj bzk), E = 217a2 (x^2 + y2) + b2z^2.^
(a) An ellipsoid of revolution with semiaxes licp/a and 1/4/b. (b)
In the case of (ID >0, a single-cavity hyperboloid of revolution;
when = 0, a right round cone; when cp < 0, a two-cavity hyper-
boloid of revolution.
3.38. (a) (p 3q
o meoR ; (b) cpc, — 312 ) , r<R.
3.25. (a) E----- 123 - 3 :(1- 8 1T3r 1 - for r<R, E--
8, P°R3 71 for 12 r>.- R;
where the vector a is directed toward the
When
q
43180 /
20*