Medan Listrik, Potensial Listrik, dan Kapasitor
81
6D6B2?82? E2D652?=:CDB:
- E2D2?=:CDB: . ;2B22?D2BE2D2?
P
1. Garis-Garis Gaya
?DE6?88232B2?652?=:CDB:52A2D;E825:=E:C2?52=2 36?DE82B:C 82B:C82I2++.
E3E?82?2?D2B282B:C 82B:C82I2
52?F6DB652?=:CDB:252=29C63282:36B:ED 2
06DBE2D652?5:CE2DED:D:A25282B:C82I26?I:?88E?882B:C
82I25:D:D:D6BC63ED 3
2?I2?I282B:CA6BC2DE2?=E2CA6?2A2?8I2?8D682=EBEC56?82?
82B:C 82B:CD6BC63ED252=29C632?5:?856?82?36C2B?I2652?=:CDB:
Gambar 4.8
a Garis-garis gaya listrik untuk partikel bermuatan positif;
b Garis-garis gaya listrik untuk partikel bermuatan negatif;
c Garis-garis gaya untuk dua muatan yang sejenis;
d Garis-garis gaya untuk dua muatan yang berbeda jenis.
4
5
+6B92D:2?8232B36B:ED -63E296=6DB?O
P
O
P
O
8 36B86B2D2?A2646A2D2?2G2=52B:A6=2D36BE2D2??682D:7
6?E;EA6=2DAC:D:7I2?836B;2B2452=2652? =:CDB:986?
:DE?8=29
2
A6B46A2D2?I2?85::=::6=6DB? 3
G2DEI2?85:A6B=E2?6?42A2:A6=2DAC:D:7
4
=2;E6=6DB?C22DD:325:A6=2DAC:D:7
elektron
+2BD:6=36BE2D2? 52?36B2CC2
8D6B2AE?83632C52=2652?=:CDB:
C6A6BD:A2528232B36B:ED
:2
C D6?DE2?=2936C2B?I2E2D652?=:CDB:I2?866?82BE9:A2BD:6=
D6BC63ED :
:6D29E: 8
O
8 C
- P
O
+6B92D:2?8232B36B:ED 3
2
Contoh
4.4
q
2=262522?C6D:32?8 3
-
8
C -
E F
w
Contoh
4.5
E v
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82
2. Resultan Kuat Medan Listrik
652? =:CDB: 6BEA22? 36C2B2? F6DB
=69 2B6?2 :DE A6?;E=292??I2 6?8:ED: 2DEB2? A6?;E=292? F6DB
?52 52A2D
6?88232BF6DB F6DB652?=:CDB:5:C6:D2BE2D2?CD2D:CI2?8 6?E?;E2? 36C2B 52? 2B29 652? =:CDB: A252 D:D: D:D: 5: C6:D2B
E2D2?D6BC63ED
,6CE=D2?36C2BE2D652?5:D:D: ,252=29C63282: 36B:ED
O 6D6B2?82?
E2D652?=:CDB:5:D:D:,2:32DE2D2?- E2D652?=:CDB:5:D:D:,2:32DE2D2?-
652? =:CDB: 6BEA22? 36C2B2? F6DB
=69 2B6?2 :DE E?DE 6?89:DE?8B6CE=D2?52B:652?=:CDB:52A2D5:=2E2?56?82?42B2
6D562?2=:C:C6?88E?22?F6DBC2DE2?2D2E56?82?6?88E?22? 6D568B27:
6D568B27:52A2D5:=2E2?56?82?CI2B2DC6D:2A652?
=:CDB:5:6D29E:2B29F6DB?I2
Gambar 4.9
Titik p dipengaruhi oleh muatan q
1
dan q
2
.
: =692B6?22CC26=6DB?C2?82D64:=82I236B2D52A2D5:232:2?D6B9252A82I2
E=3- :6D29E:
- O
P
O
P
O
8 4
2
2B:EE6GD?5:A6B=69
- C
8 -
D2?52?682D:76?E?;E2?A6B46A2D2?I2?85:2=2:6=6DB? 3
12DED6AE96=6DB?5:A6B=6952B:A6BC222?I2:DE2
56?82?2 P
C 0 P
O
C6? 4
2;E6=6DB?C22D6?I6?DE9A6=2DAC:D:75:A6B=6956?82?A6BC222?
2 2
P C
2 P
C
+6B92D:2?8232B36B:ED
.6?DE2? 2
E2D652?=:CDB:5:D:D:
3
82I2A252E2D2?OP
O
5:D:D:
q
1
q
2
p
E
1
E
2
r
1
r
2
Contoh
4.6
5 cm 5 cm
0,2 C 0,05 C
P
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83 :
2
+6B92D:2?8232B36B:ED
E2D2?E;:3:2C2?I25:2?882A36BE2D2?AC:D:7 E2D652?=:CDB:2:32D-
E2D652?=:CDB:2:32D-
52? -
- .
.
,
- -
. .
=692B6?2. .
.22
,
- - .
25:E2D652?=:CDB:5:D:D:252=29P 3
E2D2?OP
O
6?82=2:82I2C63282:36B:ED
, ,
P-P OP
O
O .2?52?682D:76?E?;E2?
,
2B29?I26:B:
+252D:D: D:D:CE5ED
52?C63E29A6BC68:
2C:?8 2C:?85:=6D22?C63E29 A2BD:6=36BE2D2?-
82BE2D652?=:CDB:5:D:D:?=D6?DE2?36C2B
E2D2?I2?892BEC5:=6D22?5:D:D: :
=692B6?2- -
-52?
.22
- .
E2D652?5:D:D:=69E2D2?5:
52? 252=29
- -
. .
82B 92BECC2236C2BD6D2A:36B=2G2?2?2B2956?82?
?DE:DE- 92BEC36BE2D2??682D:7
O
- -
. -
. -
. -
. -
.
- - ?682D:7
25:36C2BE2D2?I2?892BEC5:=6D22?5:D:D:252=29 -
D C
B
A E
AD
E
BD
E
AB
q
1
r
1
= 5 cm 0,2 C
0,05 C P
q
2
E
2
E
1
r
1
= 5 cm
Contoh
4.7
Pembahasan Soal
Pada titik-titik sudut B dan D sebuah bujur sangkar ABCD
masing-masing diletakkan sebuah partikel bermuatan +q. Agar kuat
medan listrik di titik A = nol, maka di titik C harus diletakkan sebuah
partikel bermuatan sebesar .... a.
– q b.
+ q c.
2 q
d. 2
q e.
2 2
q
UMPTN 1991
Pembahasan:
B
+q
E
AD
E
BD
E
AB
A
C D
+q
E
AB
= E
AD
=
2
kq q
= E E
BD
=
2 2
AB AD
E E
=
2
2E E
BD
=
2
2 2
kq E
q E
AC
= E
BD 2
2
2
C
k q
kq q
AC
2 2
2 2
C
q q
q q
2 2
2 2
C
q q
q q
q
C
= 2 2
q
Jawaban: e
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3. Hukum Gauss
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