Chng IX. in mi
CHNG IX. IN MIin mi l nhng cht bnh thng hu nh khng khng dn in.
Theo quan nim ca vt l c in, khc vi kim loi v cht in phn, trong in
mi khng c cc ht mang in tch t do, l nhng ht chuyn ng nh hng di tc
dng ca in trng to thnh dng in. in mi c th l tt c cc cht kh khng b
ion ha, mt s cht lng (nh benzen, nc ct, du m, du thc vt...) v cc
cht rn ( thy tinh, s, mica...). in tr sut ca cc cht in mi 10 6 1015
.m , 10 8 10 6 m . Tuy nhin, khi t in mi vo trong in trng ngoi, th
c in mi v in trng u c nhng bin i. in mi c nhiu ng dng quan trng v
rng ri trong k thut v i sng. Hu ht cc hp cht oxyt u l in mi v c s
dng ph bin trong k thut in - in t. Cc t in u phi dng cht in mi ghp
vo gia hai bn cc. Cc linh kin in t nh transistor trng (FET), vi mch
CMOS.v.v... trong cu to u phi s dng n in mi ixyt silic (SiO2),
nguyn l hot ng ca chng l da trn hin tng phn cc in mi ca lp SiO2
trong in trng phn cc bn ngoi. Do kho st v in mi v cc tnh cht ca in
mi v nhng bin i ca n trong in trng l ni dung ca chng ny. 1. S PHN
CC IN MI 1. Hin tng phn cc in mi Khi a thanh in mi ng cht v ng hng
vo trong in trng gy bi vt mang in A th ti cc mt gii hn ca thanh in
mi c xut hin nhng in tch tri du. Mt gii hn gn vi A c xut hin cc in
tch tri du vi A, cn mt gii hn ca thanh xa
A
+
-B
+ C+
Hnh 9-1 Hin tng phn cc in mi
A xut hin cc in tch cng du vi A. i vi thanh in mi ng cht v ng
hng th trong lng thanh khng xut hin in tch, cn i vi thanh in mi
khng ng cht v ng hng th trong lng thanh c xut hin in tch. Hin tng
trn gi l hin tng phn cc in mi. Hin tng ny b ngoi c v ging hin tng
in hng trong vt dn kim loi, song xt v bn cht hai hin tng hon ton
khc nhau. Trong in mi khng c in tch t do, cc in tch xut hin trn cc
mt gii hn ca thanh u l cc in tch lin kt. i lng c trng cho cht in mi
l hng s in mi . Cht c hng s in mi cng ln th hin tng phn cc cng mnh.
1. Phn loi in mi 176
Chng IX. in mi Trng tm in tch m: C th coi tc dng ca cc e trong
phn t tng ng nh mt in tch q t ti mt im gi l trng tm in tch m. Trng
tm in tch dng: c th coi tc dng ca ht nhn nh mt in tch +q t ti trng
tm in tch dng. Mi phn t ca cht in mi gm hai phn: ht nhn mang in dng
v cc electron mang in m. Bnh thng cc phn t trung ho v in. Cn c vo s
phn b ca cc electron quanh ht nhn, ngi ta phn in mi lm hai loi: -
Loi th nht: l cht in mi c phn t t phn cc. Trong loi ny, cc phn t c
phn b electron khng i xng quanh ht nhn nn tm in tch m cch tm in tch
dng mt khong l. Mi phn t t hnh thnh mt lng cc in c m men lng cc phn
t r r pe = q l . Bnh thng m men lng cc ca cc phn t sp xp hn lon i
vi nhau. l cc cht nh H2O, HCl, NH3, CH3Cl.v.v... - Loi th hai: l
cht in mi c phn t khng phn cc. Trong phn t, cc electron c phn b i
xng quanh ht nhn khin tm in tch m trng vi tm in tch dng. Phn t ca
in mi loi ny khng phi l lng cc in. l cc cht nh H2, N2, Cl2, kh
him,.v.v... Ring cc cht in mi tinh th (rn) c cc ion dng sp xp mt
cch trt t v lin kt cht ch vi nhau. Ta c th xem ton b tinh th in mi
rn nh mt phn t khng l m mng ion dng v mng ion m lng vo nhau. l cc
hp cht nh NaCl, CsCl.v.v... 3. Qu trnh phn cc in mi a. in mi c phn
t t phn cc - Khi cha t khi cht in mi trong in trng ngoi: cc phn t
sp xp hn lon, do chuyn ng nhit. Trong mt th tch bt k, tng mmen lng
cc ca cc phn t bng khng. Ton b khi in mi cha tch in (hnh 9+ 2a). +
+ - Khi t cht in mi vo in trng ngoi th cc a) mmen lng cc phn t s
quay theo chiu in + + r trng, hng ti v tr cn bng ( pe E 0 ). in
trng E 0 cng mnh v chuyn ng nhit ca cc phn t cng yu (nhit cht in mi
cng thp) th s nh hng ca cc mmen lng cc cng mnh m. Nu in trng ngoi
ln, cc lng cc phn t c th xem nh nm song song nhau theo phng+ ++
-
-
+ +
-
+
+
+
-
+
Eo Hnh 9-1. ie mophn n tpha ccc n mii pha t phn c n Hnh 9-2. in
t E0 . a) Khi cha atrong inntrng ngoi. i t n a) Khi cha t trong ie
trg ngoa t n n - Kt qu: trong lng cht in mi, cc tm in b) Khi atrong
ie trngg ngoaEo b) Khi t trong in tr ngoi. itch dng v m ca cc phn t
trung ho nhau nn 177
-
+
-
+
-
+
b)
Chng IX. in mi khng xut hin in tch. Cn trn cc mt gii hn c th xut
hin cc in tch tri du (hnh 9-2b): mt gii hn m cc ng sc in trng i vo
xut hin in tch m, mt m cc ng sc in trng i ra xut hin in tch dng . y
l cc in tch lin kt, chng khng t do dch chuyn c. Ta ni rng cht in mi
b phn cc. b. in mi c phn t khng phn cc - Khi cha t khi cht in mi
trong in trng ngoi: cc tm in tch dng v m ca phn t trng nhau. Trong
cht in mi khng c cc lng cc phn t, do trong ton khi in mi cng khng c
in tch no c. - Khi t cht in mi vo in trng ngoi E 0 , in trng s tc
dng ln cc tm in tch ca mi phn t: tm in tch m b y ngc chiu vi E 0 ,
cn tm in tch dng b ko cng chiu vi E 0 . r - Kt qu l phn t tr thnh
lng cc in c mmen lng cc pe cng hng vi
E 0 . Qu trnh xy ra ging vi trng hp trn: cht in mi b phn cc. c.
in mi tinh th rn Di tc dng ca in trng ngoi, cc mng ion dng b x dch
theo chiu in trng, cn cc mng ion m b x dch ngc chiu in trng, gy ra
hin tng phn cc in mi. Dng phn cc ny gi l phn cc ion. Tm li, d l in
mi loi no, khi c t trong in trng ngoi th ti hai mt gii hn i din ca
n u xut hin hai lp in tch tri du, gi l cc in tch phn cc hay in tch
lin kt. Mt in tch phn cc ln hay b (cht in mi b phn cc mnh hay yu)
ph thuc vo bn cht ca cht in mi v vo cng in trng ngoi. 4. Vct phn cc
in mi a. nh ngha: c trng cho mc phn cc cht in mi, ngi ta nh ngha
khi nim vct phn cc in mi, k hiu l Pe : Vct phn cc in mi Pe l tng
vct mmen r lng cc in pe ca cc phn t c trong mt n v th tch ca cht in
mi. Biu thc: Xt mt th tch V ca cht in mi b phn cc (bi trng ngoi),
th tch ny ln cha mt s ln n cc phn t. Khi theo nh ngha trn th: n r
pei Pe = i =1 (9-1) V Vct Pe hng dc theo chiu ca vct cng in trng E
0 . i vi loi in mi c phn t khng phn cc t trong in trng ngoi th mi
phn t in mi u c cng vc t mmen in, khi vc t phn c in mi c xc nh: r r
npe r r r r (9.2) Pe = = n0 pe = n0 0 E hay Pe = 0 e E V 178r r
Chng IX. in mi Trong e= n0 l h s phn cc ca mt n v th tch in mi
(hay gi l cm in mi) i vi loi in mi c phn t phn cc, ngi ta cng chng
minh c trong trng hp in trng ngoi yu cng thc trn vn ng, nhng trong
: e = n0 pe2 , trong k l 3 0 kT
hng s Bnzman. Trong trng hp in trng ngoi mnh ti mt gi tr ln th
tt c cc vc t mmen in u song song vi in trng ngoi, khi d tng in trng
ngoi th Pe khng tng, ta ni hin tng phn cc in mi t trng thi bo ha. i
vi in mi tinh th ngi ta chng minh cng thc (9.2) vn ng. i vi in mi
cu to bi cc phn t t phn cc (in mi phn cc cng): c xy ra hin tng phn
cc bo ha, qu trnh phn cc chu s nh hng ca nhit . i vi in mi cu to bi
cc phn t khng t phn cc (in mi phn cc mm): khng c hin tng phn cc bo
ha, qu trnh phn cc khng chu s nh hng ca nhit . 5. Lin h gia vct phn
cc in mi vi mt in tch lin kt Trong khi in mi ng nht ta tng tng tch
ra mt khi tr xin c ng sinh song song vi vct cng in trng tng hp E
trong in mi (tc l song song vi Pe ) c hai y song song vi nhau v c
din tch l S, ng sinh c chiu di L (hnh 9-2). Gi n l php tuyn ngoi ca
y mang in tch dng v l gc gia n v E , - v + l mt in tch trn hai y.
in tch tng cng xut hin hai y l +(S) v -(S). Ta xem ton b khi tr nh
mt lng cc in khng l m ln mmen in ca n bng S.L. Theo nh ngha ca vct
phn cc in mi, ta c: n r ' +' pei i =1 S Pe = E0 V Pe Trong
pi =1
n
ei
= S.L,
L Hnh 9-3.
P en
n
cn th tch ca khi tr xin V = S.L.cos. .S .L Do : = Pe = S .Lcos
cos Suy ra: = Pecos = Pen
Vi Pen l hnh chiu ca vc t ph n cc in mi trn phng php tuyn n Vy:
Mt ca cc in tch phn cc xut hin trn mt gii hn ca khi in mi c gi tr
bng hnh chiu ca vct phn cc in mi trn php tuyn ca mt gii hn .
179
Chng IX. in mi 2. IN TRNG TNG HP TRONG CHT IN MI 1. Vct cng in
trng E trong cht in mi Xt mt t in phng, mt in tch trn hai bn cc l +
v - gy ra in trng u E 0 trong lng t in. t mt khi in mi ng cht, ng
hng vo gia hai bn cc ca t in (hnh 9-3). Khi in mi b phn cc: xut hin
hai lp in tch phn cc c mt l + v -. Cc in tch phn cc ny sinh ra in
trng ph E ngc chiu vi in trng E 0 . V vy in trng tng hp trong lng
cht in mi s l: E = E 0 + E ' E = E0 - E Chiu biu thc trn ln phng ca
E 0 ta c: V hai lp in tch phn cc li c th xem nh hai bn cc ca mt t
in phng mi m ln ca cung in trng do n sinh ra l E =//0 = Pen = 0 e
En = 0 e E E = = e E
0
Do vy E = E0 E hay E0 E Hay: E = = 0 1 + e
/
E = E0 - eE
(9-3) (9-4)
Vy in trng tng hp trong cht in mi gim i ln so vi in trng trong
chn khng. Ta cng d dng tm c mi lin h gia vct in cm
D v vct phn cc in mi Pe . Mun vy, theo nh ngha tac D = 0 E , vi
= 1 +e
D = 0 (1 +e) E r r hay D = 0 E + o e E = 0 E + PeDo 4. IN MI C
BIT 1. in mi Scnht
(9-5)
Khong nhng nm 1930 1934 hai nh Vt l Nga l Cucsatp v Cbic tm thy
mt hp cht tinh th c cng thc NaK(C4H4O6).4H2O (bitctrat natri Kali
ngm nc), gi tt l mui Scnht, c nhiu tnh cht c bit. Sau ngi ta tm thy
mt nhm nhng in mi tinh th khc cng c cc tnh cht tng t v gi tn chung
cho chng l in mi Scnht. Sau y l nhng tnh cht c bit ca in mi Scnht:
a. Trong mt khong nhit no , hng s in mi ca in mi Scnht rt ln, c th
t ti 104. S ph thuc ca BaTiO3 vo nhit c biu din trn hnh v (Hnh
9-5a). Hng
180
Chng IX. in mi s in mi c gi tr gn 2000 ti nhit khong 120OC, nhng
li tng vt ti gn 6000 khi nhit gim ti 80OC. Khi nhit gim xung na th
li gim xung.
3600
Pe
Peb 1600o
70 Hnh 9-5a S ph thuc ca vo nhit
500
C
0
E
Hnh 9-5b S ph thuc ca vo E
E Eb Hnh 9-5c S ph thuc ca Pe vo E
0
b. Hng s ca sc nht khng nhng ph thuc vo nhit m cn ph thuc vo in
trng E trong in mi ( hnh 9-5b). Do , i vi in mi sc nht vc t phn cc
in mi Pe khng ph thuc bc nht vo in trng. Hnh 9-5c cho ta thy s ph
thuc ca Pe voP B
E . T hnh v, ta nhn thy khi E tng ti mt gitr Eb no , th Pe khng
tng na. Lc ny s phn cc in mi ca sc nht t ti trng thi bo ha; Pe =
const. T cng thc D = O E + Pe ta suy ra vi nhng ga tr E > Eb khi
Pe = const, th cm ng D ph thuc bc nht vo E . c. Vc t phn cc Pe ca
in mi sc nht khng nhng ph thuc vo in trng E m cn ph thuc vo trng
thi phn cc trc ca in mi ( hnh 9-5c). Khi t in mi cha b phn cc vo
trong mt in trng ngoi E thc nghim chng t, khi tng in trng ti gi tr
Eb th Pe t gi tr bo ha. Vn gi nguyn chiu in trng nhng gim Ek Eb
Pd
O
Ek
Eb
E
Pd
Hnh 9 - 6 Chu trnh in tr
E xung ti gi tr bng khng th Pe khng gim bng khng m vn cn bng mt
gi tr Pd no . Hin tng ny gi l hin tng phn cc cn d hay hin tng in
tr. Sau ta i 181
Chng IX. in mi
chiu ca in trng E v tng ti gi tr Ek th Pe mi t gi tr bng khng, s
phn cc ca in mi sc nht b kh i v Ek c gi l in trng kh. Nu tip tc tng
E ti gi tr - Eb, sau gim in trng t Eb v khng, ri li i chiu in trng
t khng ti Eb th ng biu din khng theo ng OB, m theo ng pha di OB to
thnh ng cong khp kn gi l chu trnh in tr ( hnh 9 -6). S bin i tun
hon ca qu trnh phn cc ca in mi sc nht c tiu tn nng lng. Phn nng lng
ny s ta thnh nhit lm nng in mi. Din tch ca chu trnh in tr t l vi
nhit lng ta ra trong mt n v th tch khi sc nht sau mi chu trnh phn
cc. d. Khi nhit in mi t ti nhit Tc no ( gi l nhit Curie) th in mi
sc nht mt ht cc tnh cht phn cc trn v tr thnh in mi phn cc bnh thng.
Nhit Tc ca mt s in mi sc nht: Titanat bari c Tc = 133OC, LiNbO3 c
Tc = 1210OC. Mui sc nht ch c tnh cht ca in mi sc nht trong khong
nhit Tc t -18OC n 24OC in mi sc nht c hng s in mi ln, do n c dng ch
to cc t in c kch thc nh nhng in dung ln, rt cn thit cho cc ngnh k
thut in v v tuyn in. 2. Hiu ng p in a. Hiu ng p in thun Nm 1880 hai
nh Vt l Php l Pie Curie v Gic Curie tm thy mt hin tng mi: khi nn
hoc ko dn mt s tinh th in mi theo nhng phng c bit trong tinh th th
trn cc mt gii hn ca tinh th xut hin nhng in tch tri du, tng t nh
nhng in tch xut hin trong hin tng phn cc in mi. Hin tng ny gi l hiu
ng p in thun, xy ra vi cc tinh th thch anh, tuamalin, mui scnht,
ng, titanat bari,.v.v... Nu i du ca bin dng (t nn sang dn hoc ngc
li) th in tch xut hin trn hai mt gii hn cng i du (hnh 9-7). Do c in
tch tri du xut hin nn gia hai mt gii hn ny c mt hiu in th. Hiu ng p
in thun c p dng trong k thut bin i nhng dao ng c (m) thnh nhng dao
ng in.
b. Hiu ng p in nghch Trong cc tinh th nu trn, ngi ta cn quan st
thy hin tng p in nghch: Nu ta t ln hai mt i din ca mt tinh th mt
hiu in th th n s b dn ra hoc b nn li. Nu y l mt hiu in th xoay chiu
th tinh th s b dn, nn lin tip v s dao ng 182
Chng IX. in mi theo tn s ca hiu in th xoay chiu. Tnh cht ny c ng
dng ch to cc ngun pht siu m. HNG DN HC CHNG IX I. MC CH - YU CU 1.
Phn bit c hin tng phn cc in mi v hin tng hng ng in ca vt dn. 2. Phn
bit c hai loi phn t in mi: phn t t phn cc v phn t khng t phn cc.
Phn bit c ba loi: Phn cc nh hng, phn cc eelectron v phn cc ion. 3.
Nm c mi lin h gia vc t phn cc in mi v mt in tch lin kt. 4. Tm c in
trng tng hp trong cht in mi ng cht v ng hng. 5. Nm c hiu ng p in v
ng dng ca n. II. TM TT NI DUNG in mi l nhng cht khng dn in trong iu
kin bnh thng. Trong in mi khng c cc in tch t do. in tr sut ca cht
in mi rt ln ( 10 6 10 5 m) trong khi in tr sut ca kim loi rt nh (
10 8 10 6 m) . Khi t trong khi in mi ng cht v ng hng vo trong in
trng ngoi, th ti cc mt gii hn ca cht in mi c xut hin nhng in tch
tri du. Nhng khc vi hin tng hng ng in ca vt dn, cc in tch xut hin y
l nhng in tch lin kt. gii thch hin tng phn cc in mi, ngi ta da theo
s phn b cc eelectron quanh ht nhn v chia ra hai loi phn t in mi:
loi phn t khng t phn cc, i vi loi ny, trng tm in ch m v trng tm in
tch dng trng nhau, v loi phn t t phn cc, i vi loi ny, trng tm in
tch m v trng tm in tch dng trng tm in tch dng trng nhau, v loi phn
t t phn cc, i vi loi ny, trng tm in tch m v trng tm in tch dng khng
trng nhau, m cch nhua mt on l. Do , ngay khi khng t trong in trng
ngoi Eo , phn t t phn cc c coi nh mt lng cc in. Ta gii hn ch xt cht
in mi ng cht ng hng. Khi t cht in mi vo trong in trng ngoi, c ba s
phn cc khc nhau: a. Phn cc nh hng xy ra i vi loi in mi c to bi cc
phn t t phn cc. b. Phn cc lectron cn c gi l phn cc do bin dng xy ra
i vi loi in mi c to bi cc phn t khng t phn cc. c. Phn cc ion xy ra
i in mi tinh th. Khi t trong in trng ngoi, di tc dng ca in trng
ngoi, cc lng cc in s nh hng ( loi phn t t phn cc) hoc trng tm in
tch m b ko dn ( phn t khng t phn cc tr nn c moomen lung cc in nh
hng theo E o ) hoc mang cc ion m v dng s dch chuyn (loi in mi tinh
th). Kt qu l, trn mt gii hn ca cht in mi 183
Chng IX. in mi ng cht v ng hng s xut hin cc in tch lin kt tri du
c mt in mt ' . Cc in tch lin kt ny gy ra mt in trng E ngc chiu EO v
in trng tng hp trong cht in mi gim i ln so vi in trng trong chn
khng: E ='
Eo
; trong = 1 +
c gi l hng s in mi. c trng cho mc phn cc ca phn t cht in mi, ngi
ta a ra vc t phn cc in mi Pe . Hnh chiu ca Pe ln php tuyn n ca khi
in mi bng ': Pn = ' . Gia vc t in cm D = o E v Pe c mi lin h: D = o
E + Pe . i vi cht in mi khng ng cht v ng hng, vc t Pe khng t l vi E
. Trong trng hp ny xc nh D , ta phi dng cng thc va nu cho D , ch
khng dng c cng thc D = o E Ta xt thm cht in mi sc nht. c im ni bt
ca loi in mi ny l khng phi l hng s m ph thuc vo in trng ngoi E v
nhit . Trong mt khong nhit no c th ln ti 10 000. Tnh cht ny c dng
tng in dung cho t in. Cui cng, ngi ta tm thy mt s tinh th nh
Tuamalin, mui sc nht, thch anh c tnh cht l khng cn in trng ngoi, m
ch cn lm bin dng theo nhng phng c bit th cc tinh th ny cng b phn cc
( hiu ng p in thun), hoc t mt hiu th hai mt bn thch anh th thch anh
s b nn hay dn. l hiu ng p in nghch. Cc ng dng ny c nhiu ng dng
trong k thut. III. CU HI N TP 1. Hin tng phn cc in mi v hin tng in
hng ca vt dn khc nhau nhng im no? Phn tch. 2. Nu s khc nhau gia phn
t t phn cc v phn t khng t phn cc. 3. Th no l cht in mi ng cht v ng
hng. 4. Phn bit ba loi phn cc. Nu s khc nhau gia chng v vn dng chng
gii thch s phn cc ca cht in mi ng cht v ng hng. 5. nh ngha vc t phn
cc in mi, tm mi lin h gia vc t phn cc in mi v mt in tch lin kt mt.
6. Chng minh rng cng in trng trong cht in gim i ln so vi trong chn
khng. 7. Nu nhng c tnh ca in mi sc nht. 8. Trnh by hiu ng p in
thun, p in nghch v nu nhng ng dng ca n. 9. Tm mi lin h gia gia D v
Pe
184
Chng IX. in mi
IV.BI TP
Th d 1:Mt t in phng c cc bn cc vi din tch S = 115cm2 v cch nhau
mt khong d = 1,24cm. Mt hiu in th U = 85,5V c t vo gia hai bn t in.
Sau ngt n ra khi hiu in th trn v mt tm in mi dy b = 0,78cm v c hng
s in mi = 2,61 c a vo gia cc bn cc ca t in (xem hnh bn). Tnh: a) in
dung C0 ca t trc khi tm in mi c a vo. b) in tch t do xut hin trn cc
bn cc. b c) in trng E0 trong khe gia cc bn t v tm in mi. d) in trng
E trong tm in mi. e) Hiu in th gia cc bn t sau khi a tm in mi vo.
f) in dung khi c tm in mi Bi gii: d a) Tnh C0: Trc khi a tm in mi
vo, y l t in khng kh ( 1) nn:
8,85.1012.115.104 = = 8, 21 pF C0 = d 1, 24.102b) Tnh q = C0U =
8,21.11-12. 85,5 = 7,02 . 10-10C in tch t do ny khng i khi a tm in
mi vo t in. c) Tnh E0 theo cng thc E0 =
0S
q 7, 02.1010 = = = 6900 (V / m) 0 0 S 8,85.1012.115.104
d) Tnh E theo cng thc (9-5) E 6900 E= 0 = = 2640 (V / m ) 2, 61
e) Tnh U = U1 + U2, trong U1 l hiu in th trn khe gia cc bn t v tm
in mi; Cn U2 l hiu in th gia hai mt gii hn ca tm in mi. Ta c U1 =
E0 (d b), U2 = Eb nn: U = E0 (d b) + Eb = E0d (E0 E)b = 6.900.
1,24. 10-2 (6900 2640).0,78.10-2 = 52,3V. f) Tnh C theo cng thc
C =
q 7, 02.1010 = = 1,34.1011 F U 52,3
Th d 2: 185
Chng IX. in mi Cho hai mt phng kim loi A. B song song tch in u,
t cch nhau mt khong D= 1cm, ln lt c mt in mt A = (2 / 3).10 9 C /
cm 2 v B = (1 / 3).10 9 C / cm 2 . Hng s in mi ca lp mi trng c di
d= 5mm gia chng l = 2 . Xc nh hiu in th gia hai mt . Bi gii: V hai
mt phng mang in tch cng du nn vc t cng in trng do hai mt phng mang
in gy ra c hng ngc nhau. Cng in trng tng hp trong lp mi B trng c
hng s in mi v b dy d c ga tr bng E d = A 2 o Cng in trng tng hp
trong khong khng gian cn li gi hai mt phng mang in trn c tr s bng:
B B A Eo = A 2 o Hiu in th ng vi lp in mi b dy d c tr s bng: ( B )d
U O = Ed d = A 2 o Hiu in th ng vi lp khng kh cn li c b dy D- d
bng: ( B )( D d ) U O = EO ( D d ) = A 2 o
+ + +
dD
+ + +
Hnh btvd2
Hiu in th gia hai mt phng mang in A, B song song trn c tr s bng:
( A B )( D d ) ( A B )d U = UO +Ud = + 2 o 2 o
( A B ) D d + d = 2 o
=1413 V
Th d 3: C hai mt phng song song v hn mang in u tri du mt in mt
bng nhau. Ngi ta lp y gia hai mt phng mt lp thy tinh dy 3 mm ( =
7). Hiu in th gia hai mt phng trn l 1000 V. Xc nh mt in tch lin kt
trn mt cht in mi. Bi gii: V hai mt phng mang in l v hn v mt in u
nen cc vc t D v u vung gc vi hai mt phng. Ta c Dn= D, En= E, = D O
E .U vo biu thc trn ta c: d U ' = O ( 1) E = O ( 1) d Thay o , , U,
d bng nhng tr s ca chng ta c:
Thay D= O E v E =
186
Chng IX. in mi
' = 8,86.10 12.6000 / 3.10 3 C / m 2 = 1,77.10 5 C / m 2BI TP T
GII 9.1 Mt t in phng c cha in mi ( = 6 ) khong cch gia hai bn l 0,4
cm, hiu in th gia hai bn l 1200 V. Tnh: a. Cng in trng trong cht in
mi. b. Mt in mt trn hai bn t in. c. Mt in mt trn cht in mi. U HD: E
= ; = o E ; = o ( 1) E d 9.2. Cho mt t in phng, mi trng gia hai bn
ban u l khng kh ( 1 = 1 ), din tch mi bn l 0,01m2, khong cch gia
hai bn l 0,5 cm, hai bn c ni vi mt hiu in th 300 V. Sau b ngun i ri
lp y khong khng gian gia hai bn bng mt cht in mi c 2 = 3. a. Tnh
hiu in th gia hai bn t in sau khi lp y in mi. b. Tnh in tch trn mi
bn. HD: a. V lp khong khng gian gia hai bn bng in mi sau khi b ngun
nn in tch trn cc bn trc v sau khi lp khng thay i ( q= const), do mt
in mt trn cc bn q = cng khng i. s U V E = nn trc khi lp y in mi ta
c o 2U 2 = d = O d Rt ra: U 2 =
1U 1 = 100V 2
b. q= C1U1= 5,8.10-9 C 9.3. Gia hai bn ca mt t in phng cch nhau
5mm, ngi ta thit lp hiu in th U= 150 V. K st mt bn t in c mt bn s
mng song song hng s in mi , b dy 3mm. Tm cng in trng trong khng kh
v trong lp s gia hai bn t in. HD: Gi 1 v 2 ln lt l hng s in mi ca
lp khng kh v in mi gia hai bn. B dy ca lp ny bng d1 v d2. Gi U1 v
U2 ln lt l hiu in th gia cc mt song song ca lp khng kh v in mi.
Theo u bi, ta c hai t mc ni tip vi nhau. Gi E1 v E2 ln E E ; E 2= ;
v lt l cng in trng trong lp khng kh v lp s. Ta vit c E 1 = 1 2
U = E1d1+E2d2. T cc ng thc trn rt ra: 2U E = 60kV / m ; E 2 = 1
1 = 10kV / m . E1 = 2 d1 1 + d 2 2187
Chng IX. in mi
9.4. Cho mt t in phng vi cc bn cch nhau 5mm v din tch mi bn l
100 cm2. Hiu in th gia hai bn l 300V. Sau khi ngt t khi ngun, ngi
ta lp y khong khng gian gia hai bn bng bnit. a. Tm hiu in th gia
hai bn sau khi lp y bnit. b. Tm in dung ca hai bn sua khi lp y
bnit. c. Tm mt in mt trc v sau khi lp y bnit. Cho bit hng s in mi
ca bnit = 2,6 . HD: Trong trng hp bi ton q1 = q2 , trong q1 v q2 l
in tch trn bn t in trc v sau khi lp y in mi ( v ct t khi ngun). Nh
vy q= const v mt in mt trn cc bn = a. V E =q = const . S
U nn trc khi lp y in mi d = U 1 o 1 v sau khi lp y in = o d
mi d = U 2 o 2 . V = const v d = const nn U 1 1 = U 2 2 v U 2 =
b. C1 = o Sd = 1,77.10 11 F v C 2 =
U 1 1
2
= 115V .
o 2 Sd
= 4,6.10 11 F .
q CU = = 5,31.10 7 C / m 2 . S S 9.5. Hai bn t in phng cch nhau
mt on d= 1cm. Hiu in th U gia hai bn bng 300V. Khong khng gian gia
hai bn t c lp y bng mt bn thy tinh song phng c 1 = 6 ;
c. 1 = 2=
dy d1 = 0,5 cm v mt bn parafin song phng c = 2 , dy d2 = 0,5 cm.
Tm: a. Cng in trng trong mi lp in mi. b. Hiu in th gia hai mt ca mi
lp. c. in dung ca t in, cho bit din tch ca mi bn t bng S =100 cm2
d. Mt in tch trn mi bn t. HD: Gi E1, U1, 1 , d1 v E2, U2, 2 , d 2 l
in trng, hiu th gia hai mt ca lp in mi, hng s in mi v b dy ca cc
lp. Ta c: 1 E1 = 2U 2 v E1 d1 + E2 d 2 = U . T cc ng thc ny ta rt
ra: 2U E = 1,5.10 4 V / m ; E = 1 1 = 4,5.10 4 V / m a. E1 = 1 d 2
+ 2 d1 2 b. U1 =75V, U2 = 225V c. in dung C ca h l: C = Vy C =
S S C1C 2 ; trong C1 = o ; C 2 = o (C1 + C 2 ) d1 d2
o 1 2 S = 2,66.10 11 F . 2 d1 + 1d 2188
Chng IX. in mi d. in tch trn mi bn t bng: q = S = C1U1=C2U2=CU.
Do : CU = = 8.10 7 C / m 2 . S 9.6 Tm mt khi nng lng ca in trng ti
mt im: a. Cch 2 cm mt mt qu cu dn in tch in c bn knh R = 1cm. b. St
mt mt phng v hn tch in u. c. Cch 2 cm mt dy dn tch in di v hn. Cho
bit mt in mt trn qu cu v mt phng v hn bng 1,67.10-5C/m2 v mt in di
trn dy tch in bng 1,67.10-7 C/m. Cho hng s in mi l 2. 1 HD:- Mt nng
lng in trng = o E 2 (1) 2 - in trng gy bi qu cu mang in u ti mt im
cch qu cu mt on x q 1 :E = thay q= 4R 2 v kt hp vi (1) ta c 2 4 0 (
R + x) W =
2R4 = 9,7.10 2 J / m 3 4 2 o ( R + x) 2 W = = 1,97 J / m 3 2 o 8
o
- in trng gy bi mt mt phng v hn: E =
- in trng gy bi mt dy dn tch in vi mt in di ti mt im cch dy mt
on x
2 E= W = = 0,05 J / m 3 2 2 o x 8 ox9.7. Hai t in phng, mi t c
in dung C= 10-6 F c mc ni tip vi nhau. Tm s thay i in dung ca h nu
lp y mt trong hai t in bng mt cht in mi c hng s in mi = 2 . C HD:
in dung ca h hai t in mc ni tip: C1 = 2 C Sau khi lp y t in th hai
bng cht in mi ta c: C 2 = +1 Nh vy s thay i in dung ca h l: ( 1)C C
= C 2 C1 = = 1,7,10 7 F . 2( + 1) 9.8. Gia hai bn ca mt t in phng
cch nhau mt on d= 3mm, ngi ta thit lp mt hiu in th U= 1000V. Sau ct
t khi ngun v lp y t in bng mt cht in mi = 7 . Tm mt in tch lin kt
xut hin trn mt in mi.
189
Chng IX. in mi
HD: in trng trong t in E =
' . Gi 1 v 2 l hng s in mi ca khng kh v O
ca mt cht in mi no th trc khi lp y in mi, d = U 1 o 1 , v sau
khi lp y in mi 2 , ta c: d = U 2 o 2 . Nu t vn ni vi ngun th U1=U2=
U. Nu t ct khi ngun th = const , d = const v ta c U 1 o 1 = U 2 o 2
, tc l hiu in th gia hai bn t s gim U 2 =
U 1 1
2
. Gi ' l mt in mt xut hin trn in mi, ta vit c:
' = Pn = D o E = o 2 E o E = o ( 2 1) E == o ( 2 1)12 3
o ( 2 1)U 2d
1U 1 8,85.10 .6.10 = = 2,53.10 6 C / m 2 3 2d 7.3.10
9.9. Mt t in phng c cc bn cch nhau mt on d=2,0 mm c tch in ti
hiu in th U= 200 V. Ngi ta lp khong khng gian gia hai bn t bng mt
lp thy tinh c = 6 . Tm mt in tch t do trn cc bn t in v mt in tch
lin kt ' trn mt bn thy tinh. HD: Khi lp y t in bng mt cht in mi (
thy tinh) c = 6 th in trng gia hai U bn gim i ln. Nhng in trng gia
hai bn l in trng u E = ; do ta c d U U = 6,3.10 6 C / m 2 E= = . T
rt ra: = = o o d d Vc t phn cc P cng chiu vi E , hng vung gc t bn
dng sang bn m ca t in, do cng vung gc vi bn thy tinh. V th P = Pn =
' . Ta bit: D = P + o E , trong D l vc t in cm D = E . Do :
d 9.10. Cho mt t in phng, din tch mi bn l S, khong cch gia hai
bn do. Ngi ta a vo trong t in mt bn in mi c b dy d < do. Tm in
dung ca t khi . HD: Trong trng hp ny, ta c hai t mc ni tip vi nhau.
in dung C ca h bng: d d do CC o S C= 1 2 = C1 + C 2 o S o S d + d d
d o
' = P = o E o E =
o ( 1)U
= 4,4.10 6 C / m 2 .
o S o S
190
Chng IX. in miNhng do do d, ngha l nu ta t ln mt in mi ( b dy d)
mt l rt mng c b dy d'= do- d rt nh th in dung ca t in c lp in mi d
ban u s khng i, ngha l C vn S bng o d 9.11. Trong mt t in phng,
khong cch gia cc bn l d, ngi ta t mt tm in mi dy d1< d song song
vi cc bn t in. Tm in dung t in trn. Cho bit hng s in mi ca in mi l
, din tch ca tm bng din tch S ca bn t. S S S HD: H t gm ba t mc ni
tip: C1 = o ; C 2 = o ; C 3 = o , trong d2 v d3 l d d2 d3 khong cch
gia cc mt ca tm in mi v cc bn t in, r rng d2 + d3 = d- d1. in dung
ca t c tnh t: o S 1 1 1 1 C= = + + d + d1 (1 ) C C1 C 2 C 3 9.12.
Mt in tch q c phn b u trong khp th tch ca mt qu cu bn knh R. Tnh:
a. Nng lng in trng bn trong qu cu. b. Nng lng in trng bn ngoi qu
cu. HD: a. in trng bn trong qu cu ti nhng im cch tm qu cu mt khong
r bng: qr E1 = 4 o R 3
1 q2 Vy nng lng ti nhng im bn trong qu cu: W1 = o E12 .4r 2 dr =
2 4 o .10 R 0b. in trng ti nhng im bn ngoi qu cu: E1 =q 4 o r 2
R
1 q2 2 2 Vy nng lng ti nhng im bn ngoi qu cu: W2 = o E 2 .4r dr
= 2 8 o .R 0
191
