2iR 3/2 (Tr —T1/2)
2.252. q — o ,„_ =40 W/m2, where i = 3, d is the
9n"'"lclaNA JI M
effective diameter of helium atom.
2.253. X = 23 mm > 1, consequently, the gas is ultra-thin;
q = p (v) (t 2 — t1)I6T (7 —1) = 22 W/m2, where (v)
= 118RTInM, T = 1/2 (T1 ± T2).
2.254. T = 7'1 + 1nT(2 .11 - 2/T R10 In R ri
2.255. T =T1+ 1IR T21—T 11 1R2 ( R1 r)
2.256. T = T o
(R2 — r2) iv/4x.
2.257. T = T o + (R2 — r2) w/6x.
3.1. The ratio Fei/Fgr is equal to 4.1042 and 1.1036 respectively;
q/m = 0.86.10-10 C/kg.
3.2. About 2.1016 N.
3.3. dq/dt = 3/2a 2neomg//.
T
3.4.
-1/
qlq2
(F
17
V
(1TH-12)2 '
r3—
/714F2 •
3.5. AT— lq°
8naeor2 •
3.6. E = 2.7i — 3.6j, E = 4.5 kV/m.
ql
'1(Izeo (12±,2)3/2
3.8. E= 2312 60 q R2 — 0.10 kV/m.
l
q
3.9. E—
43180
For
r the strength E 4neos
(0+12
q
)3/2
in the case of a point charge. Eniax =
q
for 1=r1-11I.
6 j/ 180r2
R2
0
X
3.12. (a)
48X:R ; (b) E—
For xR the
460 (x2 + R93/2 •
strength E— 4n eo
P x3'
where p= nR2X0.
3.13. (a) E=
n
V
; (b) E=
In both cases
4Eor
a2 d-r2
go (r2 — a2) •
E— 4 ne
q
or2
for r >> a.
3.14.
•
ns
E= X 1r5 The vector E is directed at the angle 45° to
4 oy
the thread.
3.15. (a)
4
n17
eoR
72- • (b) E O.
3.7. Eas
3 q R2
3.10. E- 47Esox4 •
3.11. F —
43-EsoR •
2.252. q — o ,„_ =40 W/m2, where i = 3, d is the
9n"'"lclaNA JI M
effective diameter of helium atom.
2.253. X = 23 mm > 1, consequently, the gas is ultra-thin;
q = p (v) (t 2 — t1)I6T (7 —1) = 22 W/m2, where (v)
= 118RTInM, T = 1/2 (T1 ± T2).
2.254. T = 7'1 + 1nT(2 .11 - 2/T R10 In R ri
2.255. T =T1+ 1IR T21—T 11 1R2 ( R1 r)
2.256. T = T o
(R2 — r2) iv/4x.
2.257. T = T o + (R2 — r2) w/6x.
3.1. The ratio Fei/Fgr is equal to 4.1042 and 1.1036 respectively;
q/m = 0.86.10-10 C/kg.
3.2. About 2.1016 N.
3.3. dq/dt = 3/2a 2neomg//.
T
3.4.
-1/
qlq2
(F
17
V
(1TH-12)2 '
r3—
/714F2 •
3.5. AT— lq°
8naeor2 •
3.6. E = 2.7i — 3.6j, E = 4.5 kV/m.
ql
'1(Izeo (12±,2)3/2
3.8. E= 2312 60 q R2 — 0.10 kV/m.
l
q
3.9. E—
43180
For
r the strength E 4neos
(0+12
q
)3/2
in the case of a point charge. Eniax =
q
for 1=r1-11I.
6 j/ 180r2
R2
0
X
3.12. (a)
48X:R ; (b) E—
For xR the
460 (x2 + R93/2 •
strength E— 4n eo
P x3'
where p= nR2X0.
3.13. (a) E=
n
V
; (b) E=
In both cases
4Eor
a2 d-r2
go (r2 — a2) •
E— 4 ne
q
or2
for r >> a.
3.14.
•
ns
E= X 1r5 The vector E is directed at the angle 45° to
4 oy
the thread.
3.15. (a)
4
n17
eoR
72- • (b) E O.
3.7. Eas
3 q R2
3.10. E- 47Esox4 •
3.11. F —
43-EsoR •
