Bab 4 Medan Listrik Potensial Litrik dan Kapasitor

(1)

Medan Listrik, Potensial

Listrik, dan Kapasitor

A. Muatan Listrik

B. Medan Listrik

C. Energi Potensial

Listrik dan

Potensial Listrik

D. Kapasitor

:-(+?52D6?DED6=29>6>A6=2;2B:>6?86?2:=:CDB:<':CDB:<CE529 252C6;2<;282DB2I2:?:=29:B'652<2?A6D:B>6BEA2<2?4@?D@9<636B2522? =:CDB:<<2B6?2A6D:B252=2992C:=A6=6A2C2?>E2D2?=:CDB:<5:2G2?

2?I2<C6<2=:2A=:<2C:=:CDB:<52=2><69:5EA2?C692B: 92B:.6=6F:C: B25:@<@>AED6B52?2=2D 2=2D6=6<DB@?:<=2:?>6?88E?2<2?=:CDB:<C63282: CE>36B 6?6B8:?I2 -6D6=29 5:D6>E<2??I2 =:CDB:< 32?I2< 5:D6>E<2? D6<?@=@8: D6<?@=@8:32BE-63282:4@?D@9=:92D=29328:2?52=2>C63E29 B25:@:52=2>?I2?5252A2D>6=:92D36B>242> >242><@>A@?6? =:CDB:<2A2D<29?52>6?I63ED<2??2>2 ?2>2<@>A@?6?=:CDB:<I2?8 D6B52A2D5:52=2>B25:@

:2?D2B2C6<:2?32?I2<<@>A@?6?I2?8D6B52A2D5:52=2>B25:@ ?52A2CD:>6?6>E<2?<2A2C:D@BA2<29<2A2C:D@B:DEA2<68E?22? <2A2C:D@B282:>2?242B2<6B;2C63E29<2A2C:D@B

?52A2CD::?8:?>6?86D29E:;2G232?52B:A6BD2?I22? A6BD2?I22? D6BC63ED/?DE<:DED6>E<2?;2G232??I256?82?>6>A6=2;2B:323:?:

memformulasikan gaya listrik, kuat medan listrik, fluks, potensial listrik, energi potensial listrik, serta penerapannya pada keping sejajar.

Setelah mempelajari bab ini, Anda harus mampu:

menerapkan konsep kelistrikan dan kemagnetan dalam berbagai penyelesaian masalah dan produk teknologi.

Hasil yang harus Anda capai:

Berbagai jenis kapasitor yang berfungsi untuk menyimpan muatan listrik.

Bab

4

(2)

A. Muatan Listrik

+B@C6C5EA=:<2C:?2C<29@=69>6C:?7@D@<@A:>6BEA2<2?C2=29C2DE 4@?D@9 2A=:<2C: <@?C6A =:CDB:< CD2D:C 52=2> <69:5EA2? C692B: 92B: +6>36?DE<2?C2=:?2?92C:=7@D@<@A:<2B6?2A6?6>A6=2?C6B3E<2D2ED@?6B I2?836B>E2D2??682D:7A252A6B>E<22?<6BD2C@?D@9=2:?86;2=2=:CDB:< CD2D:CI2?8=63:9C656B92?2252=29>6?6>A6=?I2B2>3ED2D2EA@D@?82? <6BD2CA252A6?882B:CA=2CD:<I2?8D6=295:8@C@<<2:?G@=

&65E286;2=2!:C:<2D6BC63ED>6?E?;E<<2?329G25E23E2936?5252A2D C2=:?8>6?2B:<282:>2?2A2B229=:7:C:<2>6?;6=2C<2?86;2=2D6BC63ED +6B92D:<2? <65E2A6B:CD:G2 A252&2'&6 52?&2'&6 &6D:<2C63E2932D2?8<2B6D5:8@C@<56?82?<2:?G@=<6>E5:2?5:56<2D<2? A25232D2?8<242I2?85:8@C@<<2:?CEDB2D6B?I2D2<65E232D2?8D6BC63ED D2B:< >6?2B:<-632=:<?I2C63E2932D2?8<2B6DI2?85:8@C@<56?82?<2:? G@=52?5:56<2D<2?56?82?32D2?8<2B6DI2?8=2:?D6B?I2D2<65E232D2?8 <2B6DD@=2< >6?@=2</?DE<>6?;6=2C<2?86;2=2D6BC63ED52A2D5:2?882A 329G2A6?88@C@<2?<2:?G@=52?<2:?CEDB2A25232D2?8>6>36B:<2? >E2D2?=:CDB:<A25232D2?8D6BC63ED

6B52C2B<2?86;2=27:C:<2D6C63ED;6=2C329G2>E2D2?A252<242 52?>E2D2?A25232D2?8<2B6D36B3652;6?:C?I2*3/&2.3 6&301.3

O>6?I63ED<2?329G2>E2D2?=:CDB:<A25232D2?8<242C63282: >E2D2?A@C:D:752AE?>E2D2?A25232D2?8<2B6D5:C63ED>E2D2??682D:7 6?52D6BCECE?2D2CB:3E2?329<2?;ED22?2D@>D@>D6BCECE?2D2C AB@D@??6EDB@?52?6=6<DB@?+B@D@?52??6EDB@?D6B52A2D5:52=2>:?D: 2D@>C652?8<2?6=6<DB@?D6B52A2D5:<E=:D2D@>%E>=29AB@D@?52?6=6<DB@? 52=2>C63E292D@>252=29C2>2$DE=29C6323?I22D@>36BC:72D?6DB2= -63282:4@?D@92D@>9:5B@86?>6?82?5E?8C2DEAB@D@?52?C2DE6=6<DB@? D@>36B>E2D2?A@C:D:7252=292D@>I2?8>6>:=:<:;E>=29AB@D@?=63:9 32?I2<52B:;E>=296=6<DB@?C652?8<2?2D@>36B>E2D2??682D:7252=29 2D@>I2?8>6>:=:<:;E>=29AB@D@?=63:9C65:<:D52B:A252;E>=296=6<DB@? 6?5236B>E2D2??6DB2=D6BCECE?2D2C2D@> 2D@>I2?8D:52<36B >E2D2??6DB2=D@>?6DB2=252=292D@>I2?8>6>:=:<:;E>=29AB@D@? C2>256?82?;E>=296=6<DB@?>:C2=?I22D@>9:5B@86?C6A6BD:D6B=:92D A252&2'&6

1. Interaksi Elektrostatis antara Dua Muatan Listrik

%6?:C:?D6B2<C:6=6<DB@CD2D:C252>242>I2:DE

2 D2B:< >6?2B:<2?D2B2>E2D2? >E2D2?D:52<C6;6?:C 3 D@=2< >6?@=2<2?D2B2>E2D2? >E2D2?C6;6?:C

-&61*79,978.349142'O>6?8E<EB36C2B?I2D2B:< 2?52?D@=2<2?=:CDB:<C642B2<E2?D:D2D:7$2;E82>6?I:>AE=<2?9E<E> I2?8 >6?82DEB D2B:<2? 52? D@=2<2? =:CDB:< D6BC63ED #E<E> D6BC63ED 5:<6?2=56?82?1'1)+1(+)#E<E>@E=@>3>6?I2D2<2?329G2 Gambar 4.1

Mesin fotokopi

Gambar 4.2

Setelah batang karet digosok kain wol dan batang kaca digosok kain sutra kedua batang tersebut tarik-menarik.

batang karet batang karet Sumber: Conceptual Physics, 1998

"*'*1922*25*1&/&6.0437*5*)&3.786.0!48*37.&1.786.0)&3&5&7.8460*6/&0&31&-74&174&1'*6.098)&1&2 '9091&8.-&3

Tes Kompetensi Awal

A2<29C6D:2A36?52>6>:=:<:>E2D2?=:CDB:< A2C2;2<29A6?IECE?52B:C63E292D@>%6=2C<2?

A6?IECE? A6?IECE??I2D6BC63ED

A2I2?8?52<6D29E:D6?D2?8<2A2C:D@BA2 C2;2<297E?8C:?I2282:>2?2AB:?C:A52C2B<6B;2?I2 A2I2?85:>2<CE556?82?>652?=:CDB:<

batang karet batang kaca

Gambar 4.3

Setelah kedua batang karet digosok kain wol dan didekatkan, kedua batang karet tolak-menolak.

Gambar 4.4

Pada atom hidrogen terdapat satu proton dan satu elektron.

(3)

50.%')!*.%'010+(')!*+('*0. 11$)10*(%/0.%' !/.*5!.* %*#(1.1/ !*#*$/%('(%!/.'! 1)10*0!./!10 * !.* %*#0!.(%' !*#*'1 .0&.'*0.'! 1)10*0!./!10

+6B92D:<2?&2'&652?&2'&6-642B2>2D6>2D:C82I2 D2B:< >6?2B:<2D2ED@=2< >6?@=2<82I2@E=@>352=2>F2<E>52A2D 5:DE=:CC63282:36B:<ED

2D2EF2<E> '- -_. O

"2I2@E=@>3I2?8D6B;25:52=2>CE2DE>65:E>2D2E3292?>6>:=:<: A6BC2>22?82I2@E=@>3C63282:36B:<ED

3292?

%:<2+6BC2>22?O252?+6BC2>22?O35:823E?8<2?5:52A2D

F2<E> 3292?

_O2

-69:?8825:52A2D9E3E?82?82I2@E=@>3A2523292?52?82I2@E=@>3 A252F2<E>C63282:36B:<ED

3292? F2<E>

&6D6B2?82?

82I2@E=@>3)

- >E2D2?=:CDB:<

<@?CD2?D25:6=6<DB:<'P)> _{A6B>:D:F:D2CBE2?892>A2}PO_)>

_r A6B>:D:F:D2CB6=2D:73292?

. ;2B2<2?D2B>E2D2?>

E2<6A:?8=@82>I2?8D6B3E2D52B:3292?C2>25:36B:>E2D2?C2>236C2B

2 6B2A2<29>E2D2?5:C6D:2A<6A:?8;:<25:<6D29E:82I2@E=@>3C636C2B)52? ;2B2<2?D2B<6A:?8=@82>>

3 %:<2<65E2<6A:?836B2525:52=2>3292?>65:E>56?82?_.36B2A2<29

36C2B82I2@E=@>3?I2 &:&'

:<6D29E:

)

2 '- -_. 56?82?--

--. _- .

' '

- . _' > _)> ) PO

Gambar 4.5

Interaksi muatan listrik berlainan jenis akan tarik-menarik

Gambar 4.6

Interaksi muatan listrik sejenis akan tolak-menolak

Contoh

4.1

F F

q₂ q₁

F F

q₂ q₁

F F

q₂ q₁

Tugas Anda 4.1

Anda masih ingat alat yang dinamakan dengan elektroskop? Apakah fungsi alat tersebut? Bagaimanakah cara kerjanya? Coba Anda cari informasi dari internet atau buku-buku referensi.

(4)

2. Resultan Gaya Coulomb

+6B92D:<2?&2'&6 "2>32B D6BC63ED 52A2D 5:56C<B:AC:<2? C63282:36B:<ED252=2982I2@E=@>3I2?85:2=2>:>E2D2?-2<:32D A6?82BE9>E2D2?-252=2982I2@E=@>3I2?85:2=2>:>E2D2?-

2<:32DA6?82BE9>E2D2?-252=2982I2@E=@>3I2?85:2=2>:>E2D2?

-2<:32DA6?82BE9>E2D2?-252=2982I2@E=@>3I2?85:2=2>: >E2D2?-2<:32DA6?82BE9>E2D2?-

,6CE=D2?82I2@E=@>35:CE2DED:D:<5:BE>EC<2?C63282:36B:<ED

6B2BD:B6CE=D2?82I2@E=@>3I2?85:2=2>:>E2D2?-A252&2'&6

252=29

2=2>92=:?:

52?

- -

' '

. .

Gambar 4.7

Resultan gaya Coulomb

.:823E29A2BD:<6=36B252A252C2DE82B:C=EBECC6D:2AA2BD:<6=36B>E2D2?

52? ?D2BA2BD:<6=36B;2B2<C2>2I2:DE4>

2 .6?DE<2?36C2B2B2982I2A252A2BD:<6=36B>E2D2?

3 :>2?2A2BD:<6= 5:=6D2<<2?282B82I2@E=@>3A252A2BD:<6=D6BC63ED

?@= &:&'

:<6D29E:- - -

2 '6D2<>E2D2?--52?-C6A6BD:82>32B36B:<ED

%2B2<2?D2B2<65E2A2BD:<6=C2>236C2BI2:DE..4>52AE?

>6BEA2<2?82I2@E=@>392C:=:?D6B2<C:D2B:< >6?2B:<-52?-C652?8<2?

>6BEA2<2?82I2@E=@>392C:=:?D6B2<C:D2B:<>6?2B:<-52?-6?82? >6>AB65:<C:<2?>2<2B6CE=D2?82I2@E=@>3?I2>6>6?E9: A6BC2>22?36B:<ED

Contoh

4.2

r₁₂ = 10 cm

q₁ F12 _q

F₂₃ _q

r₂₃ = 10 cm

- - -' _. ' _.

)> ₎

F₁₂ q₁ q₂ F₂₃ F₂₁ F₃₂ q₃

r₁ r₂

Tokoh

Charles Coulomb

(1736– 1806)

Coulomb adalah ahli Fisika kelahiran Angouleme, Prancis. Dia merupakan orang pertama yang menunjukkan perilaku bahwa muatan listrik saling tolak satu sama lain. Namanya kemudian digunakan sebagai satuan muatan listrik.

Sumber: Jendela IPTEK, 1995

---

%25:36C2B>E2D2?=:CDB:<5:C6D:2A<6A:?8252=29 3 3292? F2<E>

)) %25:36C2B82I2@E=@>3?I2252=29 )

(5)

E23E29A2BD:<6=>2C:?8 >2C:?836B>E2D2? 5:82?DE?8A252D2=:C6A6BD:A252 82>32B%:<2D2=:5:2?882AD:52<36B>2CC2D6?DE<2?36C2B?I2

2 D682?82?D2=:

3 >2CC2C6D:2AA2BD:<6=;:<2#>C

&:&'

:<6D29E:---PO

#>C_'_P_)>

?5252A2D>6?6?DE<2?36C2BD682?82?D2=:52?>2CC2)92?I256?82?>6?:?;2E C6D6?829328:2?52B:C:CD6>

2 -:CD6>5:<2D2<2?52=2><62522?C6D:>32?8;:<2

2D2EC:?L

2B:!*67&2&&3<5:<6D29E:329G2'

-. >2<2

2<2?5:A6B@=69C:?L'

Nm / C C

m 2 2

)

3 552?4@CL)#

))>C)<8

%25:>2CC2C6D:2AA2BD:<6=252=29<8

3 82B?:=2:B6CE=D2?82I2@E=@>35:-C2>256?82??@=>2<236C2B92BEC

C2>256?82?C69:?882A6BC2>22??I2>6?;25:

'- - '- - .

-. . .

. _. _.

(:C2=<2?>E2D2?-5:=6D2<<2?4>52B:>E2D2?->2<2

.4> 52? .O4>

..O444>

%25:282B82I2@=@E>3A252-C2>256?82??@=>E2D2?-5:=6D2<<2? >52B:>E2D2?-

Contoh

4.3

T 30°

T sin 30°

T cos 30°

Kata Kunci

• proton • elektron • neutron • atom netral • gaya Coulomb

T 60°

30 cm q

%:<282I2 82I2I2?836<6B;2A252A2BD:<6= A2BD:<6=D6BC63ED 5:82>32B<2?2<2?D6B=:92DC6A6BD:82>32B36B:<ED

Perhatikan gambar berikut.

Hitunglah resultan gaya F yang bekerja pada muatan q.

r r

– q + q

Tantangan

(6)

B. Medan Listrik

?52 >E?8<:? A6B?29 >6?56?82B 329G2 C6D:2A >E2D2? I2?8 5:=6D2<<2?5:CE2DE526B292<2?>6>:=:<:>652?=:CDB:<5:C6<:D2B?I2 %:<2C63E29>E2D2?E;:-5:=6D2<<2?A252526B29D6BC63ED>E2D2? D6BC63ED2<2?>6?82=2>:82I2@E=@>3&E2D>652?=:CDB:<5:CE2DED:D:< 5:567:?:C:<2?C63282:82I2@E=@>3A6BC2DE2?>E2D2?I2?85:2=2>:@=69 C63E29>E2D2?5:D:D:<D6BC63ED-642B2>2D6>2D:C5:DE=:C<2?C63282: 36B:<ED

56?82? '-- .

>2<2

2D2E

90 cm +q

74° T

q q' 15 cm

Tes Kompetensi

Subbab

A

*6/&0&31&-)&1&2'9091&8.-&3

6B2A2<2936C2B82I2D2B:< >6?2B:<2?D2B2AB@D@?52? 6=6<DB@?52=2>C63E292D@>9:5B@86?I2?836B;2B: ;2B: 2D@>KPO_>

+6B92D:<2?82>32B36B:<ED

%:<25:<6D29E:---O52?> D6?DE<2?B6CE=D2?82I2@E=@>3A252-

.:823E29A2BD:<6=36B25252=2>C2DE82B:C=EBEC>2C:?8 >2C:?8 36B>E2D2? O 52? O ?D2BA2BD:<6=>6>:=:<:;2B2<I2?8C2>2> 2 .6?DE<2?36C2B52?2B2982I2A252A2BD:<6=

36B>E2D2?

3 : >2?2 A2BD:<6= I2?8 36B>E2D2?

5:=6D2<<2?282B82I2@E=@>3A252A2BD:<6= D6BC63ED36B?:=2:?@=

E23E29A2BD:<6=>2C:?8 >2C:?836B>E2D2?

5:82?DE?8A252D2=::C@=2D@BC6A6BD:82>32B36B:<ED

q₁ q₂ q₃

a a

.2=:5:2?882AD:52<36B>2CC2 .6?DE<2?36C2B?I2

2 D682?82?D2=:

3 >2CC2C6D:2AA2BD:<6=;:<2

#>C

282:>2?2<29;2B2<2?D2B25E2A2BD:<6=36B>E2D2? 92BEC5:E329282B82I2D@=2<<65E2?I2>6?;25: 2<2=:C6>E=2 3<2=:C6>E=2

(E2D2?I2?836C2B?I25:=6D2<<2?A25282B:C 9E3E?82?D2B2>E2D2?I2?836C2B?I2O52? >E2D2?I2?836C2B?I2(E2D2?D6B=6D2< 4>52B:52?4>52B:%:<2;2B2<4> D6?DE<2?36C2B?I2B6CE=D2?82I2I2?836<6B;2A252 2 >E2D2?

3 >E2D2?

-63E2936?5236B>2CC2852?36B>E2D2?-

5:82?DE?8<2?A252C6ED2CD2=:B:?82?I2?8>2CC2?I2 52A2D5:232:<2?.6A2D5:C636=29<2?2?36?52A252 ;2B2<4>5:=6D2<<2?-OI2?8>6?I6323<2? A@C:C:36?52>6?;25:C6A6BD:A25282>32B36B:<ED

A C

%:<2

P)>52?#>C

D6?DE<2?=29

236C2BCE5ED52?

(7)

&6D6B2?82?

<E2D>652?=:CDB:<)

- >E2D2?=:CDB:<

. ;2B2<2?D2B>E2D2?>

' P_)>

1. Garis-Garis Gaya

/?DE<>6?882>32B<2?>652?=:CDB:<52A2D;E825:=E<:C<2?52=2> 36?DE<82B:C 82B:C82I2(%*!/+""+.!#E3E?82?2?D2B282B:C 82B:C82I2 52?F6<D@B>652?=:CDB:<252=29C63282:36B:<ED

2 06<D@B<E2D>652?5:CE2DED:D:<A25282B:C82I2>6?I:?88E?882B:C 82I25:D:D:<D6BC63ED

3 2?I2<?I282B:CA6BC2DE2?=E2CA6?2>A2?8I2?8D682<=EBEC56?82? 82B:C 82B:CD6BC63ED252=29C632?5:?856?82?36C2B?I2>652?=: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.

+6B92D:<2?82>32B36B:<ED

-63E296=6<DB@?!OPO₎_PO_<8

36B86B2<D2?A2<646A2D2?2G2=52B:A6=2D36B>E2D2??682D:7 >6?E;EA6=2DA@C:D:7I2?836B;2B2<4>52=2>>652? =:CDB:<9@>@86?)#:DE?8=29

2 A6B46A2D2?I2?85:>:=:<:6=6<DB@?

3 G2<DEI2?85:A6B=E<2?>6?42A2:A6=2DA@C:D:7 4 =2;E6=6<DB@?C22DD:325:A6=2DA@C:D:7

elektron

+2BD:<6=36B>E2D2?52?36B>2CC2>8D6B2AE?83632C52=2>>652?=:CDB:< C6A6BD:A25282>32B36B:<ED

%:<2#>C_{D6?DE<2?=2936C2B?I2<E2D>652?=:CDB:<I2?8>6>6?82BE9:A2BD:<6=}

D6BC63ED &:&' :<6D29E:

)>8O_<8

#>C

-PO

+6B92D:<2?82>32B36B:<ED

3 2

Contoh

4.4

2=2><62522?C6D:>32?8

3 -)#

<8 >C ₎

Contoh

4.5

E v

(8)

2. Resultan Kuat Medan Listrik

(652? =:CDB:< >6BEA2<2? 36C2B2? F6<D@B *=69 <2B6?2 :DE A6?;E>=292??I2 >6?8:<ED: 2DEB2? A6?;E>=292? F6<D@B ?52 52A2D >6?882>32BF6<D@B F6<D@B>652?=:CDB:<5:C6<:D2B>E2D2?CD2D:CI2?8 >6?E?;E<<2? 36C2B 52? 2B29 >652? =:CDB:< A252 D:D:< D:D:< 5: C6<:D2B >E2D2?D6BC63ED,6CE=D2?36C2B<E2D>652?5:D:D:<,252=29C63282: 36B:<ED

&6D6B2?82?

<E2D>652?=:CDB:<5:D:D:<,2<:32D>E2D2?- <E2D>652?=:CDB:<5:D:D:<,2<:32D>E2D2?-

(652? =:CDB:< >6BEA2<2? 36C2B2? F6<D@B *=69 <2B6?2 :DE E?DE< >6?89:DE?8B6CE=D2?52B:>652?=:CDB:<52A2D5:=2<E<2?56?82?42B2 >6D@562?2=:C:C>6?88E?2<2?F6<D@BC2DE2?2D2E56?82?>6?88E?2<2? >6D@568B27:<(6D@568B27:<52A2D5:=2<E<2?56?82?CI2B2DC6D:2A>652? =:CDB:<5:<6D29E:2B29F6<D@B?I2

Gambar 4.9

Titik p dipengaruhi oleh muatan

q₁ dan q₂.

&:&'

*=69<2B6?2>2CC26=6<DB@?C2?82D<64:=82I236B2D)#52A2D5:232:<2?D6B9252A82I2 @E=@>3-

:<6D29E:

- !OPO

)PO_<8

4>>

)

2 2B:#E<E>$$)6GD@?5:A6B@=69

) - )

) _>C

)

D2?52?682D:7>6?E?;E<<2?A6B46A2D2?I2?85:2=2>:6=6<DB@?

3 12<DED6>AE96=6<DB@?5:A6B@=6952B:A6BC2>22?"'I2:DE/200

56?82?2/

P>C0

0 PO_C6<@?

4 '2;E6=6<DB@?C22D>6?I6?DE9A6=2DA@C:D:75:A6B@=6956?82?A6BC2>22?

2₀ ₂

P_>C_>

2₀ P_>C

+6B92D:<2?82>32B36B:<ED

.6?DE<2?

2 <E2D>652?=:CDB:<5:D:D:<

3 82I2A252>E2D2?OPO_5:D:D:<

q₁ q₂

p E₁ E₂

r₁ r₂

Contoh

4.6

5 cm 5 cm

0,2 C 0,05C

(9)

&:&'

2 +6B92D:<2?82>32B36B:<ED

(E2D2?E;:3:2C2?I25:2?882A36B>E2D2?A@C:D:7

<E2D>652?=:CDB:<2<:32D- <E2D>652?=:CDB:<2<:32D- 52? - ' ' . . , - ' ' . .

*=69<2B6?2...>2<2

)> > ' -.

)> ) >

%25:<E2D>652?=:CDB:<5:D:D:<252=29P₎

3 (E2D2?OPO_{>6?82=2>:82I2C63282:36B:<ED}

_,_,P-P_)OPO_O)

.2?52?682D:7>6?E?;E<<2?,2B29?I2<6<:B:

+252D:D:< D:D:<CE5ED52?C63E29A6BC68:>2C:?8 >2C:?85:=6D2<<2?C63E29 A2BD:<6=36B>E2D2?-82B<E2D>652?=:CDB:<5:D:D:<?@=D6?DE<2?36C2B >E2D2?I2?892BEC5:=6D2<<2?5:D:D:<

&:&'

*=69<2B6?2---52?.>2<2

'_.-

&E2D>652?5:D:D:<@=69>E2D2?5:52?

252=29

' - '

_. _.

82B92BECC2>236C2BD6D2A:36B=2G2?2?2B2956?82?

/?DE<:DE-92BEC36B>E2D2??682D:7

O - -' ' . -' . -' . -. _.-

- -?682D:7

%25:36C2B>E2D2?I2?892BEC5:=6D2<<2?5:D:D:<252=29 -

D C B

E_AD E_BD _E

q₁

r₁ = 5 cm

0,2 C 0,05C

P E1 E2 q2

r₁ = 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. q 2 d. q 2 e. 2q 2

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_E2

E_BD = E 2kq₂ 2

q E_AC = E_BD

2 2 2

k q kq q AC

2 2 2

q q

₂ 2 2 2

q q q q q_C = 2q 2

(10)

3. Hukum Gauss

#E<E>"2ECC5:52C2B<2?A252<@?C6A"(1'/!=E<C252=29<E2?D:D2CI2?8 >6?882>32B<2?36B2A232?I2<F6<D@B>652?82B:C 82B:C82I2I2?8>6?6>3EC CE2DEA6B>E<22?52=2>2B29D682<=EBEC+6B92D:<2?&2'&6

%:<2D6B52A2D82B:C 82B:C82I252B:CE2DE>652?=:CDB:<9@>@86?I2?8 >6?6>3ECD682<=EBECCE2DE3:52?8C6=E2C;E>=2982B:C>652?I2?8 >6?6>3ECD682<=EBEC3:52?8D6BC63EDC2>256?82?A6B<2=:2?52? +6B<2=:2?2?D2B252?:?:5:?2>2<2?"(1'/(%/0.%' -642B2>2D6>2D:C 5:DE=:C<2?C63282:36B:<ED

&6D6B2?82?

7=E<C=:CDB:<)>_2D2EG636B

<E2D>652?=:CDB:<)

=E2C3:52?8I2?85:D6>3EC>652?=:CDB:<>

%:<282B:C 82B:C82I2D6BC63ED>6?6>3EC3:52?8D:52<C642B2D682< =EBEC7=E<C=:CDB:<?I2252=29

4@C

56?82?252=29CE5ED2?D2B2F6<D@B>652?52?=E2CA6B>E<22?I2?8 5:D6>3EC>652?=:CDB:<

2B:<@?C6A7=E<C=:CDB:<:?:=29"2ECC>6?86>E<2<2?9E<E>?I2 I2?85:?I2D2<2?C63282:36B:<ED

71)($#.%/#55*#'!(1. .%/101,!.)1'*0!.0101,/!* %*# !*#*&1)($)10*(%/0.%'5*# %(%*#'1,%+(!$,!.)1'*0!.0101,0!./!108

-642B2>2D6>2D:C5:DE=:C

A6B>E<22?D6BDEDEA

Gambar 4.10

Garis-garis gaya yang menembus bidang permukaan.

Gambar 4.11

Garis gaya yang menembus suatu permukaan membentuk sudut.

#:DE?8=297=E<C=:CDB:<A252CE2DE3:52?8A6BC68:I2?836BE<EB2?P4>;:<2 <E2D>652?=:CDB:<9@>@86?C636C2B)52?2B29?I2

2 C6;2;2B3:52?8

3 >6>36?DE<CE5EDLD6B9252A3:52?8 4 D682<=EBECD6B9252A3:52?8

&:&' :<6D29E:

'E2C3:52?8A6BC68:4>P4>4>_PO_>

&E2D>652?=:CDB:<)

!=E<C=:CDB:<52A2D?52A6B92D:<2?A25282>32B36B:<ED

2 3 4

n n

E 37°

Luas = A

Contoh

4.8

2 /?DE<CE5EDL 4@C

(11)

4. Perhitungan Medan Listrik dengan Menggunakan Hukum

Gauss

a. Medan Listrik pada Keping Sejajar

(652?=:CDB:<5:2?D2B2A6=2DC6;2;2B52A2D5:9:DE?856?82?>E529 >6?88E?2<2?#E<E>"2ECCE23E29A6=2D<6A:?8I2?8>6>:=:<:=E2C

>2C:?8 >2C:?85:36B:>E2D2?C2>2D6BC632B>6B2D2D6D2A:36B=2G2?2? ;6?:CI2:DE-52?O-C6A6BD:A252&2'&6,2A2D>E2D2?

D:2A <6A:?8 5:567:?:C:<2? C63282: >E2D2?- A6B C2DE2? =E2C -642B2 >2D6>2D:C5:DE=:C<2?C63282:36B:<ED

Gambar 4.12

Medan listrik antara dua keping sejajar dengan rapat

muatandan.

&E2D>652?=:CDB:<A252A6=2D<@?5E<D@B5:D6?DE<2?36B52C2B<2? <@?C6A #E<E> "2ECC 2B2?I2 56?82? >6>3E2D CE2DE A6B>E<22? D6BDEDEAC6A6BD:C:=:?56BE?DE<>6>E529<2?A6B9:DE?82?+6B92D:<2?

&2'&6

6B52C2B<2?!*67&2&&3<7=E<C=:CDB:<A252C:=:?56BD6BDEDEA D6BC63ED252=29

C:=:?56BD6BDEDEA

4@CL4@CL4@CL

*=69<2B6?2>2<2

C:=:?56BD6BDEDEA

6B52C2B<2?!*67&2&&3< 5:52A2D<2?A6BC2>22?

C:=:?56BD6BDEDEA

C69:?882

*=69 <2B6?2

B2A2D >E2D2? >2<2 <E2D >652? =:CDB:< I2?8

5:D:>3E=<2?@=69C2DEA6=2D<@?5E<D@B5:?I2D2<2?56?82?A6BC2>22?

6?82?56>:<:2?36C2B?I2<E2D>652?=:CDB:<I2?85:D:>3E=<2? @=695E2A6=2D<@?5E<D@B5:?I2D2<2?56?82?A6BC2>22?

&6D6B2?82?

B2A2D>E2D2?>

A6B>:D:F:D2CBE2?892>A2PO)>

3 /?DE<CE5EDL 4@C

L)PO_>₁₃

4 /?DE<CE5EDL 4@C

L)PO_>₁₃

Gambar 4.13

Perhitungan kuat medan listrik

E pada pelat konduktor menggunakan permukaan tertutup (silinder) berdasarkan Hukum Gauss.

E₁

A₁ E₂ A₂

E₃

A₃

Kata Kunci

• medan listrik • garis-garis gaya • fluks

• fluks listrik • keping sejajar • bola konduktor • permukaan Gauss

(12)

b. Kuat Medan Listrik pada Bola Konduktor Berongga

+6B92D:<2?&2'&6%:<2<652=2><@?5E<D@B3@=236B@?882 I2?836B;2B: ;2B:5:36B:C6;E>=29>E2D2?A@C:D:72D2E>E2D2??682D:7 >E2D2?D6BC63ED2<2?D6BC632B>6B2D292?I25:A6B>E<22?3@=252AE? 5:52=2>3@=2D:52<D6B52A2D>E2D2?=:CDB:<6B52C2B<2?#E<E>"2ECC 52A2D5:D6?DE<2?36C2B>652?=:CDB:<5:52=2>>2EAE?5:=E2B3@=2I2?8 36C2B?I2

2D2E

:328:2?52=2>3@=256?82?.36C2B?I2>652?=:CDB:<#2= D6BC63ED5:C6323<2?36C2B?I2>E2D2?I2?85:=:?8<EA:A6B>E<22?"2ECC$

-52AE?E?DE<A6B>E<22?"2ECC$$56?82?.36C2B?I2>E2D2? =:CDB:<I2?85:=:?8<EA:A6B>E<22?"2ECC$$C2>256?82?;E>=29>E2D2? =:CDB:<A2523@=2D6BC63ED6?82?56>:<:2?>652?=:CDB:<5:A6B>E<22? "2ECC$$252=29

- -

. .

&E2D>652?=:CDB:<5:=E2B3@=252A2D5:A6B@=6956?82?>6?82?882A3@=2 C63282:>E2D2?=:CDB:<I2?8D6B=6D2<5:AEC2D3@=2%25:C642B2<6C6=EBE92? >652?=:CDB:<5:C6<:D2B3@=236B@?882252=29

M 5:52=2>3@=2<2B6?2- M 5:A6B>E<22?3@=2 ' -

M 5:=E2BA6B>E<22?3@=2 '- .

Gambar 4.14

Bola konduktor berongga yang memiliki jari-jari R.

r = jarak titik ke pusat bola.

Gambar 4.15

Grafik E terhadap r untuk bola konduktor berongga.

R a

0 _r₌_R _r E = 0

Tes Kompetensi

Subbab

B

*6/&0&31&-)&1&2'9091&8.-&3

-63E296=6<DB@?5:D6>32<<2?56?82?<646A2D2?2G2= P_{>CC62B2956?82?<E2D>652?=:CDB:<I2?8}

36C2B?I2P_).6?DE<2?=29

2 <2A2?6=6<DB@?2<2?36B96?D: 3 ;2B2<I2?85:D6>AE9?I2 +6B92D:<2?82>32B36B:<ED

-63E296=6<DB@?!OPO₎_PO_<8

5:=6A2C<2?D2?A2<646A2D2?2G2=52B:C:C:A6=2D<6A:?8 36B>E2D2??682D:752?5:A6B46A2D>6?E;EA6=2DA@C:D:7 %:<2;2B2<2?D2B2A6=2D4>52?>652?=:CDB:< )D6?DE<2?=29

2 36C2BA6B46A2D2?6=6<DB@?

3 G2<DEI2?85:A6B=E<2?6=6<DB@?E?DE<>6?42A2: A6=2DA@C:D:7

4 =2;E6=6<DB@?C22DD:325:A6=2DA@C:D:7

E23E29A2BD:<6=36B>E2D2?D6B=6D2<A252C2DE82B:C =EBECC6A6BD:A25282>32B36B:<ED

permukaan Gauss II permukaan Gauss I

%:<2;2B2<2?D2B2A2BD:<6=4>5:>2?2<29=6D2<D:D:< I2?8<E2D>652?>28?6D:<?I2?@=

+252<66>A2DCE5EDC63E29A6BC68:5:=6D2<<2?6>A2D 3E29>E2D2?I2?8C2>2C6A6BD:82>32B36B:<ED

.6?DE<2?<E2D>652?=:CDB:<5:AEC2DA6BC68:;:<2 2 <66>A2D>E2D2?:DEA@C:D:7

3 D2?52A@C:D:752??682D:7<66>A2D>E2D2?:DE 36BC6=2?8 C6=:?8

q k

E =

2 q k

E =

R r>R r<R

14 cm

–24q

–4q

20 cm

4 C 4C

4 C 4C d

F E

(13)

#:DE?87=E<C=:CDB:<I2?8>6?6>3EC3:52?8A6BC68: C:C:4>;:<2<E2D>652?=:CDB:<C636C2B) I2?82B29?I2

2 C6;2;2B56?82?3:52?8 3 D682<=EBEC3:52?8

4 >6>36?DE<CE5EDLD6B9252A3:52?8

-63E296=6<DB@?5:=6D2<<2?5:AEC2DC63E29<6A:?8 =@82> 36B>E2D2? ?682D:7 =6<DB@? D6BC63ED5: A6B46A2D>6?E;E<6A:?8=@82>A@C:D:75:56<2D?I2 C636C2BP_>C_{%:<2<65E2<6A:?8>6>:=:<:}

B2A2D >E2D2? I2?8 C2>2 D6?DE<2? ?:=2: B2A2D >E2D2??I2)6PO<8-6PO

PO_)>

C. Energi Potensial Listrik dan Potensial Listrik

1. Energi Potensial Listrik

+6B92D:<2?&2'&6-63E29>E2D2?E;:-5:56<2D<2?56?82? >E2D2?-<:32D?I2D6B;25::?D6B2<C:2?D2B2>E2D2?-52?-36BEA2 82I24@E=@>3I2?82B29?I2D@=2< >6?@=2</?DE<36BA:?52952B:A@C:C:

.<6A@C:C:.>E2D2?-92BEC>6=2<E<2?EC2926C2B?I2EC292I2?8

92BEC5:=2<E<2?252=29C632?5:?856?82?36C2B?I282I2@E=@>352? A6BA:?5292??I2

.2?52?682D:7>6?E?;E<<2?EC292>6=2G2?82I2@E=@>3%:<2EC292 I2?85:=2<E<2?->6=2=E:A6BA:?5292?I2?8C2?82D<64:=>2<2EC292-

52A2D5:A6B@=6956?82?42B2

O .

. .

_. .

. '

-. -.

-6CE2:56?82?567:?:C:?I2EC292252=29AB@C6CDB2?C76B6?6B8:2D2E 36C2B?I2 A6BE3292? 6?6B8: 2D2E 52=2> 36?DE< >2D6>2D:C?I2 C6A6BD: 36B:<ED

, , ,

2A2D<:D2C:>AE=<2?329G2A6BC2>22?6?6B8:A@D6?C:2=5:CE2DE D:D:<252=29

, '

&6D6B2?82?

EC292=:CDB:<;@E=6

82I2@E=@>3)

< <@?CD2?D2@E=@>3P_)>

- >E2D2?E;:

- >E2D2?CE>36B

. ;2B2<2?D2B2-52?->

+ – –

q₂ q₁ _F q₁

c F

(2) (1)

r₁ r₂

Gambar 4.16

Interaksi antara muatan q₁ dan q₂.

(14)

2. Potensial Listrik

+@D6?C:2==:CDB:<252=2936C2B?I26?6B8:A@D6?C:2==:CDB:<A6BC2DE2? >E2D2?+252&2'&6A@D6?C:2==:CDB:<I2?85:>:=:<:@=69-252=29 C63282:36B:<ED

- - '

_- ' _{- .} _. _O

-642B2E>E>A6BC2>22?A@D6?C:2==:CDB:<5:CE2DED:D:<I2?836B;2B2<.

52B:>E2D2?CE>36B252=29

%:<2>E2D2?=:CDB:<I2?8>6?82<:32D<2?>E?4E=?I2A@D6?C:2==:CDB:< ;E>=29?I2=63:952B:C2DEA@D6?C:2==:CDB:<5:C63E29D:D:<>6BEA2<2? ;E>=292=;232BA@D6?C:2=D6B9252AC6D:2A>E2D2?=:CDB:<6C2B?I2>E2D2? A@D6?C:2=5:D:D:<,I2?85:C6323<2?@=69>E2D2?D:D:<---_*252=29

, *

. O

+6B92D:<2?&2'&6%:<292?I2252>E2D2?A@D6?C:2==:CDB:< 5:D:D:<+252=29

, - -

. . .

3. Hubungan Usaha dan Beda Potensial Listrik

+6B92D:<2?&2'&6/C292I2?85:=2<E<2?E?DE<>6>:?529<2? >E2D2?-52B:A@C:C:.<6A@C:C:.52B:2B29>E2D2?252=29

, , ,

' - ' -

. .

' '

-. .

- O

56?82?O252=293652A@D6?C:2==:CDB:<2?D2B2D:D:<52?D:D:< Gambar 4.17

Potensial listrik oleh 3 muatan,

q₁, q₂, q₃ di titik P.

P _r

3 +

q₃

–

q₂

q₁

r₂ r₁

Gambar 4.18

Muatan q, bergerak dari titik 1 ke titik 2 dipengaruhi muatan Q.

r₂

r₁

-63E29>E2D2?A@C:D:7-PO_{5:86B2<<2?>6?E;EC63E29:?D:2D@>I2?8}

36B>E2D2?%2B2<A:C292G2=<65E2A2BD:<6=D6BC63ED252=29PO_>52?;2B2<

A:C292<9:B?I2252=29PO_{>%:<2EC292I2?85:A6B=E<2?E?DE<>6>:?529<2?}

PO_{%D6?DE<2?>E2D2?:?D:2D@>D6BC63ED}

&:&' :<6D29E:

- PO _.

PO>

. PO> PO%

Contoh

4.9

(15)

-68:D:82C:<E C:<E5:56?82?4>52?4>-63E29>E2D2? =:CDB:<-OO_{2<2?5:A:?529<2?52B:D:D:<}_<6D:D:<_{I2?8D6B=6D2<A252}

A6BD6?8292?%:<2>E2D2?-O_52?>E2D2?

OO2?882A<6925:B2?

-D:52<36BA6?82BE9D6B9252AA@D6?C:2=5:D6?DE<2?=29 2 A@D6?C:2=5:@=29-52?-

3 A@D6?C:2=5:@=29-52?-

4 EC292I2?85:A6B=E<2?E?DE<>6>:?529<2?>E2D2?-52B:<6 &:&'

:<6D29E:-O-OO

-OO_4>>

2 +@D6?C:2==:CDB:<5:2<:32D-52?-252=29

_' - _'-

P_)>

> >

F@=D

3 +@D6?C:2==:CDB:<5:2<:32D>E2D2?-52?-

*=69<2B6?2-6-52?>2<2OC69:?882

O 4 /C292E?DE<>6>:?529<2?>E2D2?252=29

-OOO_O0PO_;@E=6

Tantangan

untuk Anda

Hitunglah usaha yang diperlukan untuk memindahkan muatan positif yang besarnya 10 C dari suatu titik yang potensialnya 10 V ke suatu titik yang potensialnya 60 V.

56?82?!*67&2&&3<5:A6B@=69

. .

PO_%P_PO

%25:>E2D2?:?D:2D@>252=29PO

Contoh

4.10

A B

10 cm 6 cm

4 cm 4 cm

4. Hukum Kekekalan Energi Mekanik dalam Medan Listrik

#E<E><6<6<2=2?6?6B8:>6<2?:<AE?36B=2<EA25286B2<A2BD:<6= C6A6BD:86B2<AB@D@?52?6=6<DB@?5:52=2>>652?=:CDB:<#2=D6BC63ED 36B=2<E<2B6?2>652?=:CDB:<>6BEA2<2?>652?<@?C6BF2D:7

?6B8:D@D2=C63E29A2BD:<6=56?82?>2CC2)52?>E2D2?-I2?8 36B86B2<52=2>>652?=:CDB:<252=29

,','

2D2E

- )2 - )2 _O

(6?8:?82D6?6B8:A@D6?C:2==:CDB:<,-52?6?6B8:<:?6D:<'

)2 ;:<2<646A2D2?2G2=A2BD:<6=2!*67&2&&3<>6?;25:

- )2 O

!*67&2&&3<>6?E?;E<<2?A6BE3292?6?6B8:A@D6?C:2=>6?;25: 6?6B8:<:?6D:<

Hukum kekekalan energi mekanik partikel dalam medan listrik berlaku jika pada partikel tersebut tidak ada gaya lain yang bekerja selain gaya Coulomb.

(16)

+6B92D:<2?82>32B36B:<ED

A B

652A@D6?C:2=5:2?D2B25E2A6=2DC6;2;2BA25282>32BD6BC63ED 252=290-63E29AB@D@?2G2=?I25:<6A:?8%:<25: 2?D2B2<65E2A6=2D92>A2E52B29:DE?8<646A2D2?AB@D@? C636=E>>6?I6?DE9<6A:?8(2CC2AB@D@?)PO_<8

>E2D2?AB@D@?-PO

&:&'

(6=2=E:#E<E>&6<6<2=2? ?6B8:(6<2?:<52=2>>652? =:CDB:<5:A6B@=69

?6B8:>6<2?:<5:6?6B8:>6<2?:<5:

- )2 - )2 2 2 -

2 2 ₎

6B52C2B<2?36C2B2?I2?85:<6D29E:A252C@2=52A2D5:A6B@=69

_{>C >C}

&646A2D2?AB@D@?C636=E>>6?I6?DE9A6=2D252=29_>C

5. Potensial di antara Dua Keping Sejajar

%:<25E23E29<6A:?8C6;2;2B5:9E3E?8<2?56?82?CE>36BD682?82? 32D6B2:>2<2<65E2<6A:?82<2?>6>:=:<:>E2D2?I2?8C2>2D6D2A: 36B=2G2?2? ;6?:C /C292 I2?8 5:=2<E<2? 82I2 =:CDB:< - E?DE< >6>:?529<2?>E2D2? >E2D2?C6;2E9 252=29C636C2B

52AE?52=2>=:CDB:<CD2D:C36C2B?I2EC292252=29

6?82?>6?8823E?8<2?<65E2A6BC2>22?EC292D6BC63ED5:A6B@=69

- -

O O

&6D6B2?82?

;2B2<2?D2B2<65E2<6A:?8>

652A@D6?C:2=O52A2D5:?I2D2<2?C63282:3652A@D6?C:2=

2?D2B2<65E2<6A:?8C6;2;2B Gambar 4.19

Dua buah keping sejajar dan terpisah sejauh d diberi muatan yang sama.

Contoh

4.11 Kata Kunci

• energi potensial listrik • potensial listrik • volt

A B

(17)

-63E296=6<DB@?!5:D6>32<<2?>2CE<5:2?D2B25E2<6A:?8C6;2;2B56?82?<646A2D2? 2G2=2C6A6BD:A25282>32B36B:<ED

%:<23652A@D6?C:2=<65E2<6A:?8252=2952?;2B2< 2?D2B2<6A:?8 D6?DE<2?

2 G2<DEI2?85:A6B=E<2?6=6<DB@?9:?882 >6?I6?DE9<6A:?8C636=292D2C

3 ;2B2<>6?52D2B4I2?85:D6>AE96=6<DB@? ;:<2A6?82BE9A6B46A2D2?8B2F:D2C:5:232:<2? &:&'

2B:#E<E>$$)6GD@?52?!*67&2&&3<5:A6B@=69

) - )

- ) ₎ 56?82?2B29<62D2C

2 -6CE2:A6BC2>22?"'329G2

0 /C652?8<2?- ₎ >2<2E?DE</ 2<2?5:A6B@=69

)

0 _- _- )

3 2B:A6BC2>22?42056?82?45:A6B@=69

42_-)

) -

Tes Kompetensi

Subbab

C

*6/&0&31&-)&1&2'9091&8.-&3 +6B92D:<2?82>32B36B:<ED

r = 50 × 10–10_m

%:<25:<6D29E:.PO_>52?_-_PO

52?C63E29AB@D@?D6B=6D2<5:D:D:<D6?DE<2?=296?6B8: A@D6?C:2=AB@D@?

E2>E2D2?D:D:<-O*52?-*D6BA:C29 A252;2B2<4>.6?DE<2?A@D6?C:2=5:D:D:<D6?829 82B:C9E3E?82?D2B2<65E2>E2D2?D6BC63ED

>A2D3E29>E2D2?=:CDB:<---52?-D6B=6D2< A252D:D:<CE5EDA6BC68:I2?8A2?;2?8C:C:?I24> C6A6BD:A25282>32B36B:<ED

Contoh

4.12

q d x

V 1

.6?DE<2?A@D6?C:2=D:D:<5:AEC2DA6BC68::DE;:<2 2 ----

3 --O

--O

E23E29>E2D2?D:D:<A252<@@B5:?2D2BD6C:EC 56?82?<@@B5:?2DC63282:36B:<ED-5:

D:D:<52?-O5:D:D:<O.6?DE<2?

A@D6?C:2==:CDB:<5:D:D:<

-63E296=6<DB@?5:D6>32<<2?56?82?<646A2D2? _{>C>2CE<<652=2>5E2<6A:?8C6A6BD:82>32B}

%:<2<E2D>652?2?D2B25E2<6A:?8P₎

D6?DE<2? ;2B2< >6?52D2B4 C2>A2: 6=6<DB@? >6?E>3E<<6A:?8I2?82D2C)6PO<8

-6PO52?#>C

q₄ q₃

q₁ q₂

x = ?

y = 1 cm V

Tantangan

untuk Anda

Sebuah elektron dengan massa 9,11 × 10–31_{kg dan muatan listrik} –1,6 × 10–19_{C, lepas dari katode} menuju anode yang jaraknya 2 cm. Jika kecepatan awal elektron = 0 dan beda potensial antara anode dan katode = 200 V, hitunglah kecepatan elektron ketika sampai di anode.

(18)

D. Kapasitor

1. Muatan pada Kapasitor

+6B?29<29?52>6>6B92D:<2?<@>A@?6? <@>A@?6?I2?82525: 52=2>2=2D6=6<DB@?:<C6A6BD:D6=6F:C:52?B25:@&2A2C:D@B2D2E<@?56?C2D@B 252=29C2=29C2DE<@>A@?6?6=6<DB@?:<I2?82525:52=2>?I2+6B92D:<2?

&2'&6 <2A2C:D@B2D2E<@?56?C2D@B252=295E23E29A6?892?D2B A6=2D<@?5E<D@BI2?85:A:C29<2?@=69CE2DE%/+(0+.2D2EJ2D %!(!'0.%'

E?DE<>6>A6B@=69>E2D2?I2?8C2>2D6D2A:36B=2G2?2?;6?:C&2A2C:D@B 36B7E?8C:E?DE<>6?I:>A2?6?6B8:A@D6?C:2==:CDB:<

?5252A2D>6>32I2?8<2?C63E29EC292E?DE<>6>36B:<2?>E2D2? A252<2A2C:D@B56?82?42B2<EDE3A@C:D:732D6B2:>6?2B:<6=6<DB@?A252 <6A:?8C636=29<:B:9:?882<6A:?8D6BC63ED>6?;25:36B>E2D2?A@C:D:7 52? >6?5@B@?8 6=6<DB@? 6=6<DB@? D6BC63ED <6 <6A:?8 C636=29 <2?2? >6?;25:36B>E2D2??682D:79:?882<65E2<6A:?8>6>:=:<:>E2D2?C2>2 92?I2;6?:C?I2I2?836B3652+B@C6CA6BA:?5292?6=6<DB@?52B:<6A:?8 <:B:<6<6A:?8<2?2?>6=2=E:6?6B8:&:>:252B:32D6B2:I2?836B=2?8CE?8 D6BEC >6?6BECC2>A2:3652A@D6?C:2=2?D2B2<6A:?8C2>256?82?3652 A@D6?C:2=I2?85:>:=:<:32D6B2:-2>A2:2<9:B?I2<2A2C:D@BD6=29A6?E9 >E2D2?52?D:52<52A2D5::C:=28:

&2A2C:D@B5:C:>3@=<2?56?82? &2A2C:D@B5:8E?2<2?E?DE< 2 >6?I:>A2?>E2D2?2D2E6?6B8:=:CDB:<

3 C63282:C2=29C2DE<@>A@?6?52=2>B2?8<2:2?A6?2=2I2?836B8E?2 E?DE<>6>:=:97B6<E6?C:A252A6C2G2DB25:@52?

4 >6?46829=@?42D2?=:CDB:<A252B2?8<2:2? B2?8<2:2?I2?8>6?82?5E?8 <E>A2B2?;:<2D:32 D:322BEC=:CDB:<5:AEDEC<2?

e–

– +

Gambar 4.20

Proses pengisian muatan pada kapasitor.

keping konduktor isolator

keping konduktor

Gambar 4.21

(a) Jenis kapasitor; (b) Skema kapasitor.

2. Kapasitas Kapasitor

?52>E?8<:?A6B?29>6=:92D<2A2C:D@BD6BED2>252=2>A6B2=2D2? 6=6<DB@?:< %:<2 C63E29 <2A2C:D@B 5:9E3E?8<2? 56?82? CE>36B 3652 A@D6?C:2=I2?836BE329 E329>2<2>E2D2?I2?8D6BC:>A2?5:52=2>?I2 AE?36BE329 E329

<@?CD2?

* * - - -

+6B32?5:?82?- 52? D6BC63ED 5:?2>2<2? <2A2C:D2C <2A2C:D@B I2?8 92B82?I2<@?CD2?E?DE<D:2A<2A2C:D@BC69:?88252=2>36?DE<>2D6>2D:C 5:DE=:CC63282:36B:<ED

(19)

- O &6D6B2?82?

- >E2D2?

3652A@D6?C:2=0

<2A2C:D2?C:02D2E!

&2A2C:D2C<2A2C:D@BD:52<5:A6?82BE9:@=69A6BE3292?>E2D2?52? A@D6?C:2==:CDB:<

52D:82;6?:C<2A2C:D@BI2?832?I2<5:8E?2<2?I2:DE

2 ,/%0+.'!.0/&6BD2CA252<2A2C:D@B:?:36B7E?8C:C63282:A6?I6<2D5: 2?D2B2<65E2A6=2D=@82>

3 ,/%0+.2.%!(&2A2C:D@B:?:5:8E?2<2?52=2>B2?8<2:2?A6?2=2 A252A6C2G2DB25:@

4 ,/%0+.!(!'0.+(%0!(+&2A2C:D@B;6?:C:?:>6>:=:<:<2A2C:D2?C:A2=:?8 D:?88:I2:DEC2>A2:56?82?A!

Gambar 4.22

(a) Kapasitor kertas; (b) Kapasitor variabel; (c) Kapasitor elektrolit. kertas

lempeng logam

keping bergerak

keping tetap kapasitor elektrolit

-63E29<2A2C:D@B5::C:@=6932D6B2:F@=DC69:?88236B>E2D2?.6?DE<2?=29 2 <2A2C:D2C<2A2C:D@BD6BC63ED

3 >E2D2?I2?8D6BC:>A2?52=2><2A2C:D@B;:<25:>E2D:32D6B2:F@=D &:&'

:<6D29E:F@=D-PO

2 &2A2C:D2C<2A2C:D@B5:9:DE?856?82?!*67&2&&3<

0 !

3 (E2D2?I2?8D6BC:>A2?5:A6B@=6956?82?A6BC2>22?

-PO_!0

3. Kapasitor Keping Sejajar

+252.(-&*1&6&)&;>6=2<E<2?A6?6=:D:2?D6?D2?8A6?82BE9 A6?8:C:2?BE2?85:2?D2B2A6=2D A6=2D<2A2C:D@B56?82?>6?88E?2<2?3292? 5:6=6<DB:<&6&)&;>6?88E?2<2?5E2<2A2C:D@BI2?8:56?D:<:C2=29 C2DE<2A2C:D@B5:D6>A2D<2?CE2DE3292?5:6=6<DB:<C652?8<2?<2A2C:D@B =2:??I236B:C:E52B2A252D6<2?2??@B>2=&6>E5:2?<65E2<2A2C:D@B D6BC63ED5:36B:A@D6?C:2==:CDB:<I2?8C2>236C2B?I2C6A6BD:5:DE?;E<<2? @=69&2'&6(6=2=E:6<CA6B:>6?D6BC63ED&6&)&;>6>A6B@=6992C:= CA6<D2<E=6B(E2D2?5:<2A2C:D@BI2?8>6?82?5E?85:6=6<DB:<;2E9=63:9 36C2B52B:A252>E2D2?5:<2A2C:D@BI2?8>6?82?5E?8E52B2

Contoh

4.13 Tugas Anda 4.2

Berdasarkan kepolarannya, dikenal dua jenis kapasitor, yaitu kapasitor polar dan kapasitor non-polar. Coba Anda cari informasi mengenai perbedaan di antara keduanya.

(20)

#&'*1

-:72D C:72D52B:636B2A2:6=6<DB:<

&-&3 4378&38&.*1*086.0 *09&8&3.*1*086.0_0%22

02<E>

/52B2

:B O

&6BD2C

(:<2>6B2956=:>2 ,E33I>:42

+@BC6=6?

&G2BC2I2?85:=63EB

"6=2CA:B6H

2<6=:D

+@=:6D=6?

>36B

.67=@?

)6@AB6?

(:?I2<DB2?C7@B>2D@B .:D2?:E>5:@<C:52

+@=:CD:B6?

*=69<2B6?2>E2D2?-=63:936C2BE?DE<I2?8C2>2A252<2A2C:D@B I2?8>6?82?5E?85:6=6<DB:<6B52C2B<2?A6BC2>22? - 5:A6B@=69 92C:=329G2<2A2C:D2CC63E29<2A2C:D@B2<2? 36BD2>329 36C2B 2<:32D A6?6>A2D2?3292?5:6=6<DB:<5:2?D2B2A6=2D A6=2D<2A2C:D@B

-:72D C:72D:?:252=29<:B2 <:B2A252D6>A6B2DEB<2>2B52?E?DE<<@?5:C: <@?5:C: C656>:<:2?BEA2C69:?882>652?=:CDB:<5:52=2>5:6=6<DB:<D:52<36BE329 56?82?G2<DE

$?:=298B25:6?A@D6?C:2=>2<C:>E>I2?8>E?8<:?D6B52A2D5:52=2>5:6=6<DB:< D2?A2D6B;25:?I2<68282=2?=:CDB:<!(!0.%(.!' +3*:6=6<DB:<C6B:?8<2=: 5:D6>A2D<2?5:2?D2B2A=2D A=2DA6?892?D2BE?DE<>6?8:B:><2?CE2DEA6B36522? A@D6?C:2=I2?8=63:9D:?88:E?DE<5:A2<2:<2?5:2?D2B2A=2D A=2DD6BC63ED52B:A252 A6B36522?A@D6?C:2=I2?8>E?8<:?5:52A2D56?82?>6?88E?2<2?E52B2C63282: 5:6=6<DB:<

+6B92D:<2?&2'&6 <2A2C:D@B<6A:?8 C6;2;2BD6B5:B:2D2C5E2 <6A:?8=@82>I2?8D6BA2C2?8C6;2;2BA252;2B2<A:C29 >6D6BI2?8;2E9 =63:9<64:=52B:=E2C<6A:?8>_{&65E2<6A:?8D6BC63ED5:36B:>E2D2?}

-I2?8C2>236C2BD6D2A:D:52<C6;6?:C

%:<25:2?D2B2<6A:?8C6;2;2B252=29E52B2F2<E><2A2C:D2?C:<2A2C:D@B >6>6?E9:A6BC2>22?36B:<ED

&6D6B2?82?

=E2C<6A:?8>

;2B2<2?D2B<6A:?8>

A6B>:D:F:D2CF2<E>E52B2PO_)>

Gambar 4.24

Kapasitor keping sejajar dihubungkan dengan sumber tegangan V.

Gambar 4.23

(a) Baterai B menyebabkan perbedaan potensial yang sama pada setiap kapasitor; kapasitor dengan bahan dielektrik memiliki muatan yang lebih banyak. (b) Kedua kapasitor mengangkut muatan yang sama. Kapasitor dengan bahan dielektrik memiliki perbedaan potensial lebih rendah, seperti ditunjukkan oleh pembacaan alat ukur. bahan dielektrik udara

–q

V A

(21)

4. Dielektrik pada Kapasitor

:2?D2B25E2A6=2D<6A:?8A252<2A2C:D@B3:2C2?I25:C:C:A:5:A2C2?8 CE2DE3292?:C@=2D@BI2?85:C63ED %!(!'0.%'292?:?:5:8E?2<2?<2B6?2 52A2D>6>A6B36C2B?:=2: <2A2C:D2C<2A2C:D@B@?D@93292?5:6=6<DB:< 252=29<242>:<2<6BD2C52?<2B6D

292?5:6=6<DB:<>6>:=:<:92B82A6B>:D:F:D2CI2?836B365256?82? 92B82 A6B>:D:F:D2C F2<E> 6B52C2B<2?!*67&2&&3 < 52A2D 5:<6D29E:329G2A6B>:D:F:D2C3292? _. C69:?882<2A2C:D2C<2A2C:D@B I2?8>6?88E?2<2?3292?5:6=6<DB:<252=29

2D2E _. O

6?82?>6>6B92D:<2?!*67&2&&3<52?!*67&2&&3<

5:52A2D9E3E?82?<2A2C:D@BD2?A25:6=6<DB:<52?<2A2C:D@B56?82?3292? 5:6=6<DB:<252=29C63282:36B:<ED

-63E29<2A2C:D@B<6A:?8C6;2;2B=E2CD:2A<6A:?84>_{52?D6BA:C29>>C2DE}

C2>2=2:?652A@D6?C:2=5:2?D2B2<6A:?8F@=D&6>E5:2?<2A2C:D@B5:=6A2C 52B:CE>36BD682?82?52?BE2?85:2?D2B2<6A:?85::C:56?82?5:6=6<DB:<I2?8>6>:=:<: A6B>:D:F:D2CB6=2D:73292?

#:DE?8=29

2 <2A2C:D2C>E=2 >E=2 3 >E2D2?-A252D:2A<6A:?8

4 <E2D>652?=:CDB:<C636=E>5:36B:3292?5:6=6<DB:< 5 <2A2C:D2C<2A2C:D@BC6D6=295:C:C:A:3292?5:6=6<DB:< 6 A6B>:D:F:D2C5:6=6<DB:<I2?85:8E?2<2?

7 3652A@D6?C:2=<6A:?8C6D6=295:36B:3292?5:6=6<DB:< 8 <E2D>652?=:CDB:<C6D6=295:36B:>E2D2?5:6=6<DB:< &:&'

:<6D29E:

_PO_)>

>>PO_>

4>_PO_>

>_> !

3 -PO!0PO

0 0> >

.PO!PO!

6 . PO)>PO)>

Contoh

4.14

Jika permitivitas dielektrik disisipkan di antara keping kapasitor, kapasitas dan beda potensialnya berubah, tetapi muatannya (q) tetap.

Ingatlah

Tugas Anda 4.3

Mengapa dielektrik tidak menggunakan bahan konduktor? Coba tanya pada guru Anda atau cari informasi dari internet.

(22)

5. Kapasitor Bentuk Bola

+6B92D:<2?&2'&6E2<E=:D<@?5E<D@B36B36?DE<3@=2I2?8 5:A:C29<2?BE2?8F2<E>&E=:D328:2?=E2B36B>E2D2??682D:756?82? ;2B: ;2B:=E2BC652?8<2?<E=:DI2?8C636=2952=2>36B>E2D2?A@C:D:7

56?82?;2B: ;2B:+@D6?C:2=A2523@=2328:2?52=2>5:D:>3E=<2? @=69>E2D2?A@C:D:7-52?>E2D2?3@=2328:2?=E2BO-52AE?A@D6?C:2=

52B:3@=2<@?5E<D@B328:2?=E2B36B?:=2:?@=<2B6?25:D2?29<2? &:D2 <6D29E: 329G2 <E2D >652? 52? A@D6?C:2= =:CDB:< @=69 3@=2 <@?5E<D@B5:D:D:<52?>6>:=:<:A6BC2>22?

52?0

-' '

6?82? 56>:<:2? 3652 A@D6?C:2= 2?D2B2 <65E2 3@=2 <@?5E<D@B 36B>E2D2?D6BC63ED252=29

>6BEA2<2?D6D2A2?<6C632?5:?82?E?DE<BE2?892>A2' P_)>O_{6?82?56>:<:2?<2A2C:D2C}_{52B:<2A2C:D@B3@=2252=29}

_'- '

C69:?882

E?DE<<E=:D3@=2<@?5E<D@B36B@?8825:>2?25:A6B@=69

' O

6. Susunan Kapasitor

a. Susunan Seri

2=2>AB2<D:<?52C6B:?8<2=:A6B=E>6?8823E?8<2?<2A2C:D@BE?DE< >6?52A2D<2??:=2:<2A2C:D2CI2?85::?8:?<2?2B2A6?8823E?82?<2A2C:D@B I2?8C2?82D52C2B252=2942B2A6?8823E?82?56?82?CECE?2?C6B:52? A2B2=6=

Gambar 4.25

Kapasitor bola konsentris dengan V₂ dibumikan

R₁

R₂

bola 2

bola 1

7 - _! 0

5:A6B@=69--<2B6?29E3E?82?56?82?32D6B2:5:AEDECC69:?882D:52<>6>:=:<: ;2=2?=2:?

8 _>0 0>

Generator Van de Graaff adalah alat untuk mengumpulkan dan menyimpan muatan listrik statik dalam jumlah yang sangat besar. Rambut wanita ini akan saling tolak-menolak setelah memegang generator Van de Graaff dan ujung-ujungnya berdiri karena termuati oleh muatan yang sama.

Van de Graff is a tool to collect and store statics electric charge

enormously in quantity. The woman’s hair could be repulsive each other after held the Van de Graff generator and a hair would be repulsives caused by loading the common charges.

Informasi

untuk Anda

Sumber:Physics for You,2001

(23)

+6B92D:<2?&2'&6 2=2> CECE?2? C6B: <2A2C:D@B 36B=2<E 2DEB2?36B:<ED

(E2D2? A252 C6D:2A <2A2C:D@B C2>2 I2:DE C2>2 56?82? >E2D2? <2A2C:D@BA6?882?D:

-_D@D---- O 652A@D6?C:2=A252E;E?8 E;E?8<2A2C:D@BA6?882?D:C2>256?82?

;E>=29D@D2=3652A@D6?C:2=E;E?8 E;E?8C6D:2A<2A2C:D@B

_D@D O 2B:!*67&2&&3<5:<6D29E:329G2

- 2D2E

- -

C652?8<2?_D@D

D@D

%:<292B82 92B8252?5:>2CE<<2?<652=2>!*67&2&&3 <2<2?5:A6B@=69A6BC2>22?

D@D

6C2B<2A2C:D2CA6?882?D:E?DE<*3E29<2A2C:D2D@BI2?85:CECE? C6B:252=29

D@D

Gambar 4.26

Susunan seri kapasitor

C₁ C₂ C₃

V₁ V₂ V₃

+ – + – + –

.:823E29<2A2C:D@BI2?8<2A2C:D2C?I2>2C:?8 >2C:?8!!52?!5:CECE?C6B: 52?5:9E3E?8<2?<6CE>36BD682?82?0#:DE?8=29

2 <2A2C:D2CA6?882?D:

3 >E2D2?-52?3652A@D6?C:2=C6D:2A<2A2C:D@B &:&'

:<6D29E:!!!D@D0

2 &2A2C:D2C<2A2C:D@BA6?882?D:

_D@D

D@D

! ! ! !

3 (E2D2?<2A2C:D@B5:A6B@=6956?82?>6?88E?2<2?!*67&2&&3<

-_D@D

D@DD@D!0

-_D@D

C69:?882D682?82?C6D:2A<2A2C:D@B252=29

₀ ₀ ₀

! ! !

- -

Contoh

4.15

• Kapasitas total dua buah kapasitor seri adalah

1 2 1 2

C C C

C C

• Perbandingan potensial pada kapasitas seri adalah

1 2 3

1 1 1 1

: : : ... : n : : : ... :

V V V V

C C C C

Ingatlah

Tugas Anda 4.4

Ketika Anda sedang merangkai peralatan elektronik, kadang-kadang sulit menemukan nilai kapasitas kapasitor yang tepat dengan kebutuhan, maka teknik penyusunan kapasitor bermanfaat sekali dalam hal ini. Jika Anda hanya memiliki kapasitor bernilai besar, susunan kapasitor apakah yang akan Anda gunakan untuk mendapatkan kapasitor bernilai kecil?

b. Susunan Paralel

2=2>CECE?2?A2B2=6=<2A2C:D@B36B=2<E92= 92=36B:<ED 652A@D6?C:2=C6D:2A<2A2C:D@B252=29C2>2

(24)

(E2D2?D@D2=<2A2C:D@BA6?882?D:C2>256?82?;E>=29D@D2=>E2D2?

-_D@D--- O +6B92D:<2?&2'&6 (E2D2?A252>2C:?8 >2C:?8<2A2C:D@B 252=29--52?-C652?8<2?-_D@D2=%:<2 92B82 92B82---52?-5:>2CE<<2?<652=2>!*67&2&&3< 2<2?5:A6B@=69

_D@D

6C2B<2A2C:D2CA6?882?D:E?DE<*3E29<2A2C:D@BI2?85:CECE?A2B2=6= 252=29

D@D* O

.:823E29<2A2C:D@BI2?8<2A2C:D2C?I2>2C:?8 >2C:?8!!52?!5:CECE? A2B2=6=52?5:9E3E?8<2?<6CE>36BD682?82?F@=D#:DE?8=29

2 <2A2C:D2CD@D2=_D@D

3 3652A@D6?C:2=C6D:2A<2A2C:D@B 4 >E2D2?C6D:2A<2A2C:D@B 5 >E2D2?D@D2=<2A2C:D@B &:&'

:<6D29E:!!!_D@D0 2 &2A2C:D2CD@D2=?I2252=29_D@D

!!!!

3 +252CECE?2?A2B2=6=36?52A@D6?C:2=5:C6D:2A<2A2C:D@BC2>2;E8256?82? A@D6?C:2=D@D2=?I2C69:?882_D@D

4 >E2D2?5:C6D:2A<2A2C:D@B252=29

!P0

5 (E2D2?D@D2=<2A2C:D@B252=29-_D@D

7. Energi yang Tersimpan dalam Kapasitor

&2A2C:D@B;E8236B7E?8C:E?DE<>6?I:>A2?6?6B8:A@D6?C:2=I2?8 5:36B:<2?52B:CE>36BA@D6?C:2==:CDB:< ?6B8:A@D6?C:2=D6BC63ED5:8E?2<2? <6D:<2B2?8<2:2?=:CDB:<>E=2:36<6B;2-6;E>=29>E2D2?<652=2><2A2C:D@B D6BEC36BD2>329C6=2>2AB@C6CA6?8:C:2?D6BEC36B=2?8CE?836BC2>22? 56?82?A6BD2>3292?A@D6?C:2=A252<65E2<6A:?8C6;2;2B9:?882A6?E9

C636C2B -

Contoh

4.16 Tantangan

untuk Anda

Perhatikan gambar rangkaian kapasitor berikut.

Tentukan:

a. kapasitas pengganti antara titik

x dan titik y,

b. beda potensial antara titik x dan

y jika muatan pada kapasitor 5F adalah 60C.

x C₂ C₁

3 F

4 F

5 F

3 F

Gambar 4.28

Grafik perubahan potensial (V) sebagai fungsi perubahan muatan selama proses pengisian berlangsung.

V (V)

q(C)

V C

Gambar 4.27

Susunan paralel kapasitor

C₁ C₂ C₃

V₁ V₂ _V

V +

–

+ –

dq Q

&6?2:<2?D@D2=6?6B8:A@D6?C:2=A252<2A2C:D@B_,252=29A6?;E>=292? 2D2E:?D68B2=>E2D2?-!O_{52B:?@=C2>A2:}_A6B92D:<2?_&2'&6

6?82?>6>6B92D:<2?!*67&2&&3<6?6B8:A@D6?C:2=I2?8D6BC:>A2? 52=2><2A2C:D@B52A2D5:A6B@=69C63282:36B:<ED

, -

(25)

E23E29<2A2C:D@B>2C:?8 >2C:?8<2A2C:D2C?I252?5:CECE?C642B2C6B: <6>E5:2?5:9E3E?8<2?56?82?D682?82?0.6?DE<2?

2 <2A2C:D2CD@D2=

3 6?6B8:I2?8D6BC:>A2?52=2>C:CD6> &:&'

:<6D29E:0

D@D2=

_{! 0 %}

Contoh

4.17

?52D6=29>6?86D29E:86?6B2D@BF2?56B"B22752B:3+462&7.938903)&=2D D6BC63ED>6?892C:=<2?>E2D2?C6;6?:C52=2>;E>=2936C2B.E82C?52252=29 >6?42B::?7@B>2C:>6?86?2:42B2<6B;22=2DD6BC63ED5:C6BD2:56?82?82>32B+292>: 7E?8C: 7E?8C:C6D:2A328:2?52=2>2=2DD6BC63ED%:<2D6=29C6=6C2:AB6C6?D2C:<2?5: 56A2?<6=2C

Mari Mencari Tahu

Kata Kunci

• kapasitas kapasitor • kapasitor kertas • kapasitor variabel • kapasitor elektrolit • farad

• kondensator • isolator • dielektrik • katode • anode

-63E29<2A2C:D@B!5:9E3E?8<2?56?82?C63E29 CE>36BD682?82?09:?882D6B:C:A6?E9&6>E5:2? <EDE3 <EDE3?I25:9E3E?8<2?<6<2A2C:D@B!I2?8 D:52<36B>E2D2?#:DE?8=29

2 >E2D2?D@D2=

3 3652A@D6?C:2=A252E;E?8 E;E?8C6D:2A<2A2C:D@B 4 6?6B8:2<9:BD@D2=

5 A6?8EB2?82? 6?6B8: <6D:<2 <2A2C:D@B C2=:?8 36B9E3E?82?

-63E29<2A2C:D@B<6A:?8C6;2;2B36B36?DE<=:?8<2B2? 56?82?B25:EC4>52?;2B2<2?D2B2<6A:?8>> %:<25:9E3E?8<2?<632D6B2:09:DE?8=29 2 36C2B<2A2C:D2C?I2

3 36C2B>E2D2?A252<6A:?8<2A2C:D@B

4 6?6B8:A@D6?C:2=I2?8D6BC:>A2?52=2><2A2C:D@B

Tes Kompetensi

Subbab

D

*6/&0&31&-)&1&2'9091&8.-&3

.:823E29<2A2C:D@BI2?8<2A2C:D2C?I2>2C:?8 >2C:?8 !!52?!5:CECE?C6B:=2=E5:9E3E?8<2?<6 CE>36BD682?82?F@=D#:DE?8=29

2<2A2C:D2C<2A2C:D@BA6?882?D:

3>E2D2?-52?3652A@D6?C:2=C6D:2A<2A2C:D@B .:823E29<2A2C:D@B56?82?<2A2C:D2C>2C:?8 >2C:?8

!!52?!5:9E3E?8<2?C642B2A2B2=6=<65E2 E;E?8?I25:9E3E?8<2?56?82?CE>36BD682?82? F@=D#:DE?8=29

2 <2A2C:D2CD@D2=

3 3652A@D6?C:2=C6D:2A<2A2C:D@B 4 >E2D2?C6D:2A<2A2C:D@B 5 >E2D2?D@D2=<2A2C:D@B

.6?DE<2?<2A2C:D2CA6?882?D:2?D2B252?52B: >E2D2?D@D2=?I252B:<2A2C:D@BI2?85:A2C2?8C6A6BD: A25282>32B36B:<ED;:<25:9E3E?8<2?<6CE>36B

D682?82?F@=D 5 mm

10 cm

2 4 V

2 3 4

4 F

24 F

6 F

2 F

3 F

2 F 4F

2 F

a b

5 F

12 F 4F

(1)

E23E29<2A2C:D@B>2C:?8 >2C:?8<2A2C:D2C?I252?5:CECE?C642B2C6B: <6>E5:2?5:9E3E?8<2?56?82?D682?82?0.6?DE<2?

2 <2A2C:D2CD@D2=

3 6?6B8:I2?8D6BC:>A2?52=2>C:CD6>

&:&'

:<6D29E:0

D@D2=

_{! 0 %}

Contoh

4.17 Mari Mencari Tahu

Kata Kunci

• kapasitas kapasitor • kapasitor kertas • kapasitor variabel • kapasitor elektrolit • farad

• kondensator • isolator • dielektrik • katode • anode

-63E29<2A2C:D@B!5:9E3E?8<2?56?82?C63E29 CE>36BD682?82?09:?882D6B:C:A6?E9&6>E5:2? <EDE3 <EDE3?I25:9E3E?8<2?<6<2A2C:D@B!I2?8 D:52<36B>E2D2?#:DE?8=29

2 >E2D2?D@D2=

3 3652A@D6?C:2=A252E;E?8 E;E?8C6D:2A<2A2C:D@B 4 6?6B8:2<9:BD@D2=

5 A6?8EB2?82? 6?6B8: <6D:<2 <2A2C:D@B C2=:?8 36B9E3E?82?

-63E29<2A2C:D@B<6A:?8C6;2;2B36B36?DE<=:?8<2B2? 56?82?B25:EC4>52?;2B2<2?D2B2<6A:?8>> %:<25:9E3E?8<2?<632D6B2:09:DE?8=29 2 36C2B<2A2C:D2C?I2

3 36C2B>E2D2?A252<6A:?8<2A2C:D@B

4 6?6B8:A@D6?C:2=I2?8D6BC:>A2?52=2><2A2C:D@B

Tes Kompetensi

Subbab

D

*6/&0&31&-)&1&2'9091&8.-&3

.:823E29<2A2C:D@BI2?8<2A2C:D2C?I2>2C:?8 >2C:?8 !!52?!5:CECE?C6B:=2=E5:9E3E?8<2?<6 CE>36BD682?82?F@=D#:DE?8=29

2<2A2C:D2C<2A2C:D@BA6?882?D:

3>E2D2?-52?3652A@D6?C:2=C6D:2A<2A2C:D@B .:823E29<2A2C:D@B56?82?<2A2C:D2C>2C:?8 >2C:?8

!!52?!5:9E3E?8<2?C642B2A2B2=6=<65E2 E;E?8?I25:9E3E?8<2?56?82?CE>36BD682?82? F@=D#:DE?8=29

2 <2A2C:D2CD@D2=

3 3652A@D6?C:2=C6D:2A<2A2C:D@B 4 >E2D2?C6D:2A<2A2C:D@B 5 >E2D2?D@D2=<2A2C:D@B

.6?DE<2?<2A2C:D2CA6?882?D:2?D2B252?52B: >E2D2?D@D2=?I252B:<2A2C:D@BI2?85:A2C2?8C6A6BD: A25282>32B36B:<ED;:<25:9E3E?8<2?<6CE>36B

D682?82?F@=D 5 mm

10 cm

2 4 V

2 3 4

4 F

24 F

6 F

2 F

3 F

2 F 4F

2 F

a b

5 F

12 F 4F

(2)

Rangkuman

(E2D2?=:CDB:<D6B5:B:2D2CAB@D@?I2?836B>E2D2? A@C:D:76=6<DB@?I2?836B>E2D2??682D:752??6EDB@? I2?8D:52<36B>E2D2?

%:<25E2>E2D2?I2?836B=2:?2?;6?:C36B:?D6B2<C: 2<2?D6B;25:82I2D2B:< >6?2B:<2?D2B2<65E2>E2D2? D6BC63ED-632=:<?I2;:<25E2>E2D2?=:CDB:<C6;6?:C 36B:?D6B2<C:2<2?D6B;25:82I2D@=2< >6?@=2< #E<E>@E=@>3>6?I2D2<2?329G236C2B82I2D2B:<

>6?2B:<2D2ED@=2< >6?@=2<2?D2B25E2>E2D2?=:CDB:< 36B32?5:?8=EBEC56?82?92C:=<2=:36C2B<65E2 >E2D2?52?36B32?5:?8D6B32=:<56?82?<E25B2D;2B2< 2?D2B2<65E2>E2D2?D6BC63ED

56?82?'<@?CD2?D2@E=@>3

&E2D>652?=:CDB:<5:CE2DED:D:<5:567:?:C:<2?C63282: 82I2@E=@>3A6BC2DE2?>E2D2?I2?85:2=2>:@=69 C63E29>E2D2?5:D:D:<D6BC63ED

#E<E>"2ECC>6?I2D2<2?329G2;E>=2982B:C82I2 I2?8<6=E2B52B:CE2DEA6B>E<22?D6BDEDEAC632?5:?8 56?82?;E>=29>E2D2?=:CDB:<I2?85:=:?8<EA:@=69 A6B>E<22?D6BDEDEAD6BC63ED

A6B>E<22?D6BDEDEA

&E2D>652?=:CDB:<I2?85:D:>3E=<2?@=69C63E29 A6=2D<@?5E<D@B5:?I2D2<2?56?82?A6BC2>22?

56?82?.B2A2D>E2D2?

&E2D>652?=:CDB:<5:C6<:D2B3@=236B@?882252=29 2 5:52=2>3@=2<2B6?2-

3 5:A6B>E<22?3@=2 ' -

4 5:=E2BA6B>E<22?3@=2 ' - .

652A@D6?C:2==:CDB:<5:D:D:<%<2B6?2A6?82BE9>E2D2?

5:?I2D2<2?56?82?A6BC2>22?

' .

/C292E?DE<>6>:?529<2?C63E29>E2D2?=:CDB:<-

52B:C2DED:D:<<6D:D:<=2:??I25:?I2D2<2?56?82?

-O

&2A2C:D@B252=29<@>A@?6?=:CDB:<I2?836B7E?8C: E?DE<>6?I:>A2?>E2D2?52?6?6B8:A@D6?C:2==:CDB:< &2A2C:D@B>6BEA2<2?5E2A6=2D<@?5E<D@BI2?8 5:A:?529<2?@=69CE2DE:C@=2D@B2D2EJ2D5:6=6<DB:< &2A2C:D2C<2A2C:D@B252=29<6>2>AE2?CE2DE<2A2C:D@B

E?DE<>6>A6B@=692D2E>6?I:>A2?>E2D2?=:CDB:<

56?82? <2A2C:D2?C:!2B25

- >E2D2?

3652A@D6?C:2=0@=D

6C2B?I2<2A2C:D2C<2A2C:D@B56?82?3292?5:6=6<DB:< 5:?I2D2<2?56?82?A6BC2>22?

56?82?.A6B32?5:?82?2?D2B2A6B>:D:F:D2C3292?

52?A6B>:D:F:D2CF2<E>

%:<2*3E29<2A2C:D@B5:CECE?C6B:36C2B>E2D2?A252 C6D:2A<2A2C:D@B252=29C2>2

-_D@D---

-652A@D6?C:2=A252E;E?8 E;E?8<2A2C:D@BA6?882?D: C2>256?82?;E>=29D@D2=3652A@D6?C:2=<2A2C:D@B

_D@D_*

6C2B<2A2C:D2C<2A2C:D@BA6?882?D:E?DE<*3E29 <2A2C:D@BI2?85:CECE?C6B:252=29

D@D

%:<2*3E29<2A2C:D@B5:CECE?A2B2=6=36C2B3652 A@D6?C:2=A252C6D:2A<2A2C:D@B252=29C2>2

_D@D

6C2B>E2D2?D@D2=<2A2C:D@BA6?882?D:C2>256?82? 36C2B>E2D2?D@D2=<2A2C:D@B

-D@D---

6C2B<2A2C:D2C<2A2C:D@BA6?882?D:E?DE<*3E29 <2A2C:D@BI2?85:CECE?A2B2=6=252=29

D@D*

6C2B 6?6B8: =:CDB:< I2?8 D6BC:>A2? 52=2> <2A2C:D@B252=29

(3)

Setelah mempelajari bab ini, Anda tentu menjadi paham tentang listrik statis, medan listrik, potensial listrik, energi potensial listrik, kapasitor dan jenis-jenisnya, serta bagaimana cara kapasitor bekerja. Dari semua materi pada bab ini, bagian mana yang menurut Anda sulit untuk dipahami? Coba Anda diskusikan dengan teman atau guru Fisika Anda.

Refleksi

Pemasangan kapasitor pada sakelar-sakelar listrik yang memiliki daya tinggi dapat menghilangkan terjadinya percikan bunga api. Pemanfaatan ini berguna sekali untuk keamanan. Coba Anda tuliskan manfaat lain penggunaan kapasitor.

9&8&3.786.0

>6?:>3E=<2? 36BEA2

Peta Konsep

4@?D@9 A6BC2>22?<E2D>652??I2

(652?':CDB:<

>6?I:>A2? >6?I6323<2?

652+@D6?C:2=':CDB:<

_. ?6B8:+@D6?C:2=':CDB:<_,-_.

(E2D2?.:D:< (E2D2?.6B5:CDB:3EC:

M @=2&@?5E<D@B M &2A2C:D@B

(4)

Tes Kompetensi

Bab 4

*6/&0&31&-5&)&'9091&8.-&3

!.1.-1&-7&1&-7&89/&:&'&3;&3,5&1.3,8*5&8

@=252?>2C:?8 >2C:?8>E2D2?52? %2B2<2?D2B3@=2252=29>%:<2<65E23@=2D6BC63ED 36B2525:E52B2>2<2

2 6?6B8:A@D6?C:2=3@=2P_;@E=6

3 6?6B8:A@D6?C:2=3@=2P_;@E=6

4 _,3@=2_,3@=2 5 ,3@=2,3@=2

6 6?6B8:A@D6?C:2=3@=2P_;@E=6

:<6D29E:3E2936?52D:D:<I2?836B>E2D2?I2:DE 52?%:<2>6?2B:<>6?@=2<52? >6?2B:<C652?8<2?36B>E2D2??682D:7A6B?I2 D22?I2?836?2B252=29

2 >E2D2?A@C:D:7>E2D2??682D:7 3 >E2D2?A@C:D:7>E2D2?A@C:D:7 4 >E2D2??682D:7>E2D2?A@C:D:7 5 >E2D2??682D:7>E2D2??682D:7 6 >E2D2?A@C:D:7>E2D2??682D:7

-2DE2?>652?=:CDB:<52A2D5:?I2D2<2?52=2>N

2 ) _{5 %)}

3 0> 6 ) 4 )>

E23E29<6A:?8C6;2;2B5:36B:>E2D2?I2?8C2>236C2B 52?36B=2G2?2?;6?:C&E2D>652?=:CDB:<5:2?D2B25E2 <6A:?8D6BC63EDN

2 36B32?5:?8=EBEC56?82?B2A2D>E2D2??I2 3 36B32?5:?8D6B32=:<56?82?B2A2D>E2D2??I2 4 36B32?5:?8 D6B32=:< 56?82? <E25B2D ;2B2<

2?D2B<6A:?8

5 36B32?5:?8=EBEC56?82?;2B2<<65E2<6A:?8 6 2B29?I2>6?E;E<6<6A:?8I2?836B>E2D2?A@C:D:7 -63E296=6<DB@?52=2>CE2DE>652?=:CDB:<D:52<>6?8

2=2>:82I2;:<26=6<DB@?D6BC63ED

2 36B86B2<52=2>2B29D682<=EBECD6B9252A>652? =:CDB:<

3 36B86B2<C6;2;2B56?82?2B29>652?=:CDB:< 4 36B86B2<56?82?2B29C6>32B2?8

5 36B86B2<>6=:?8<2B 6 D:52<36B86B2<

E2>E2D2?D:D:<C6;6?:C52?C2>236C2B--

36B252A252;2B2<4>C2DEC2>2=2:?%:<2

'P_)>_{36C2B82I2D@=2<I2?85:2=2>:}

<65E2>E2D2?D6BC63ED252=29 2 )

3 P₎

4 PO₎

5 PO₎

+6B92D:<2?82>32B36B:<ED

+B@D@?I2?836B86B2<52B:<6A:?8<6<6A:?8C6A6BD: D2>A2<A25282>32BD6BC63ED>6>:=:<:<646A2D2?P _{>C%:<2;2B2<2?D2B2<6A:?852=2>F2<E>}

4>>2CC2AB@D@?O_{<8>E2D2?AB@D@?P}O _{3652A@D6?C:2=<6A:?8C6;2;2BD6BC63ED252=29}

2 0 5 0

3 0 6 0

4 0

$!# E23E2936?5236B>E2D2?O-52?-36B;2B2<.C2DE C2>2=2:?%:<2;2B2<.5:E329 E329>2<28B27:< 9E3E?82?82I2@E=@>356?82?.252=29 2 F

r 3 F

r 4 F

r 5 F

r 6

A B

(5)

-63E29<2A2C:D@BF5:9E3E?8<2?56?82?CE>36B

D682?82?9:?88252A2D>6?I:>A2?6?6B8:C636C2B ;@E=66C2B>E2D2?I2?8D6BC:>A2?52=2><2A2C:D@B D6BC63ED252=294@E=@>3

2 PO _{5 P}O

3 PO _{6 P}O

4 PO

+6B92D:<2?82>32B36B:<ED

A B

+2q – q

%:<2,252=29=6D2<D:D:<I2?8B6CE=D2?<E2D>652? =:CDB:<?I2?@=>2<2,252=29

2 D6A2D5:D:D:<* 3 2?D2B2D:D:<52?* 4 2?D2B2D:D:<*52?

5 A252A6BA2?;2?82?56?82?, , 6 A252A6BA2?;2?82?56?82?, , +252<2A2C:D@BI2?836B<2A2C:D2?C:5:36B:<2?>E2D2?

C636C2B-C69:?8823652A@D6?C:2=?I26C2B6?6B8: 52=2><2A2C:D@B

2 - ₅

3 6 -

+6B92D:<2?82>32B36B:<ED

%:<2<2A2C:D@B<6D:82<2A2C:D2?C:?I2C2>2<2A2C:D@B A6?882?D:2?D2B2E;E?8252?3252=29!72B25

2 5

3 6

E23E29<2A2C:D@B>2C:?8 >2C:?8>6>:=:<:<2A2C:D2C 52?5:CECE?C6B:&2A2C:D@BA6?882?D:?I2 2 5

3 6

-63E29 <2A2C:D@B A6=2D C6;2;2B D2?A2 5:6=6<DB:< 36B<2A2C:D2C%:<23292?<2A2C:D@BD6BC63ED5::C: 56?82?5:6=6<DB:<>:<2I2?8>6>:=:<:<@?CD2?D2 5:6=6<DB:<'<2A2C:D2?C:?I2C6<2B2?8>6?;25: 2

+6B32?5:?82?82I2@E=@>352?82I28B2F:D2C:2?D2B2 3E296=6<DB@?52=2>BE2?8F2<E>252=29N 2 P₎

3 P₎

4 P₎

5 P₎

6 C2>236C2B

/C292E?DE<>6>32G2C63E296=6<DB@?52B:<EDE3 32D6B2:F@=D<6<EDE3?682D:7?I2252=29 2 PO_;@E=6

3 PO_;@E=6

4 PO_;@E=6

5 PO_;@E=6

6 PO_;@E=6

&2A2C:D@B5:36B:>E2D2?9:?882<:=@F@=D6C2B 6?6B8:I2?8D6BC:>A2?52=2><2A2C:D@B252=29 2 ;@E=6

3 ;@E=6 4 ;@E=6 5 ;@E=6 6 ;@E=6

(6?EBED>@56=@9BD6?D2?82D@>9:5B@86?6=6<DB@? -6!>6?86=:=:?8:AB@D@?-656?82?;2B: ;2B: PO_%:<2_!_PO_{>2<282I2@E=@>3}

2?D2B26=6<DB@?52?AB@D@?A2522D@>9:5B@86?C636C2B ?6GD@?

2 PO

3 PO

4 PO

5 PO

6 PO

-63E296=6<DB@?>6?86=:=:?8:AB@D@?A252;2B2<K %:<2>2CC26=6<DB@?PO_{<8>2<2<646A2D2?}

6=6<DB@?>6?86=:=:?8:AB@D@?252=29>C 2 P

3 P

4 P

5 P

6 P

2=2>4@E=@>3D6B52A2D3E296=6<DB@?

2 P

3 P

4 P

5 P

6 P

a b

C₁

C₂ F

(6)

+6B92D:<2?82>32B36B:<ED

.6?DE<2?

2 <E2D>652?5:D:D:<

3 82I2I2?85:2=2>:>E2D2? +6B92D:<2?82>32B36B:<ED

.6?DE<2?

2 <2A2C:D2?C:A6?882?D:_D@D2=2?D2B2D:D:<252?3 ;:<2C6D:2A<2A2C:D@B>6>:=:<:<2A2C:D2C3_F

3 >E2D2?A252C6D:2A<2A2C:D@B;:<2D:D:<252?3 5:36B:3652A@D6?C:2=0@=D

4 6?6B8:A252C6D:2A<2A2C:D@B;:<252?/5:36B: 3652A@D6?C:2=052?52?.5:36B:3652 A@D6?C:2=0

-63E29<2A2C:D@B<2A2C:D2?C:?I2;:<25:6=6<DB:<?I2 F2<E>52?;:<25:6=6<DB:<?I2A@BC6=6?#:DE?8=29 <@?CD2?D25:6=6<DB:<E>A@BC6=6?

+6B92D:<2?82>32B36B:<ED

%:<2>E2D2?6=6<DB@?PO_52?>2CC2

6=6<DB@?PO_{<85:=6A2CD2?A2<646A2D2?2G2=}

D6A2D5:D:D:<D6?DE<2?=29

2 G2<DEI2?85:3EDE9<2?6=6<DB@?C2>A2:5:D6A: 3 <646A2D2?C22DD:325:D6A:

-63E296=6<DB@?-!PO₎_PO_<8

5:D6>32<<2?56?82?<646A2D2?P_>C52=2>

2B29C6;2;2B<E2D>652?=:CDB:<)6B2A2;2B2< I2?85:D6>AE96=6<DB@?C636=E>36B96?D:

@=2<64:=36B>E2D2??O??52? O?5:=6D2<<2?5:D:D:< D:D:<CE5EDC63E29A6BC68:

&:&'1&-5*68&3;&&3'*6.098.3.)*3,&38*5&8

-63E29>E2D2?E;:PO_{36B;2B2<>>52B:}

>E2D2?CE>36BOPO_{6B2A2EC292I2?8}

92BEC5:=2<E<2?A252>E2D2?E;:282BD6BA:C29A252 ;2B2<>>52B:>E2D2?D6BC63ED

E2<6A:?8=@82>I2?8C6;2;2B52?;2B2<?I24> C2DE52B:I2?8=2:?5:36B:>E2D2?=:CDB:<I2?836B=2G2?2? 9:?8823652A@D6?C:2=0

6B2A2 82I2 I2?8 5:2=2>: C63E29 6=6<DB@? I2?8 5:=6D2<<2?5:2?D2B2<6A:?8

.6?DE<2?<2A2C:D2CA6?882?D:2?D2B252?A252 CECE?2? CECE?2?<2A2C:D@B36B:<ED:?:

&2A2C:D@B5:9E3E?8<2? 56?82? CE>36B D682?82?0C6=2?;ED?I2CE>36BD682?82?5:AEDEC 52?<2A2C:D@B<6>E5:2?5:9E3E?8<2?A2B2=6=<6 <2A2C:D@B

2 6B2A2>E2D2?D@D2=<2A2C:D@BD6BC63ED 3 6B2A23652A@D6?C:2=C6D:2A<2A2C:D@B

4 6B2A26?6B8:I2?89:=2?8A252<2A2C:D@B

<6D:<25:9E3E?8<2?56?82?<2A2C:D@B

A B

6 cm 8 cm

4 C 3C

s C₁ C₂ C₃

C₄ C₅

C₆ C₇

B A

V_B – V_A = 200 volt

15 cm

A B

C C

C C C

A B

C C

A B

40 F 12F

13 F

6 F

28 F

16 F

12 F d=10 cm