< PreviousMedan Listrik, Potensial Listrik, dan Kapasitor81&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:C82I252?F6<D@B>652?=:CDB:<252=29C63282:36B:<ED 2 06<D@B<E2D>652?5:CE2DED:D:<A25282B:C82I2>6?I:?88E?882B:C82I25: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 untukpartikel bermuatan positif;(b) Garis-garis gaya listrik untukpartikel bermuatan negatif;(c) Garis-garis gaya untuk duamuatan yang sejenis;(d) Garis-garis gaya untuk duamuatan yang berbeda jenis.45+6B92D:<2?82>32B36B:<ED -63E296=6<DB@?!O P O )P O <836B86B2<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=292 A6B46A2D2?I2?85:>:=:<:6=6<DB@?3 G2<DEI2?85:A6B=E<2?>6?42A2:A6=2DA@C:D:74 =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# > CD6?DE<2?=2936C2B?I2<E2D>652?=:CDB:<I2?8>6>6?82BE9:A2BD:<6=D6BC63ED &:&' :<6D29E:) >8 O<8# > C-P O +6B92D:<2?82>32B36B:<ED 32Contoh 4.4q 2=2><62522?C6D:>32?83-)# <8 > C) )#-EFwContoh 4.5EvMudah dan Aktif Belajar Fisika untuk Kelas XII822.Resultan Kuat Medan Listrik(652?=:CDB:<>6BEA2<2?36C2B2?F6<D@B *=69<2B6?2:DEA6?;E>=292??I2>6?8:<ED:2DEB2?A6?;E>=292?F6<D@B ?5252A2D>6?882>32BF6<D@B F6<D@B>652?=:CDB:<5:C6<:D2B>E2D2?CD2D:CI2?8>6?E?;E<<2?36C2B52?2B29>652?=:CDB:<A252D:D:< D:D:<5:C6<:D2B>E2D2?D6BC63ED ,6CE=D2?36C2B<E2D>652?5:D:D:<,252=29C63282:36B:<ED *O&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:DEE?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 4.9Titik p dipengaruhi oleh muatanq1 dan q2. &:&'*=69<2B6?2>2CC26=6<DB@?C2?82D<64:=82I236B2D)#52A2D5:232:<2?D6B9252A82I2 @E=@>3 - :<6D29E:-!O P O )P O <8 4> > ) 2 2B:#E<E>$$)6GD@?5:A6B@=69) -) ) > C <8- )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/2 0 056?82?2 / 0 > P > C00P OC6<@? 4 '2;E6=6<DB@?C22D>6?I6?DE9A6=2DA@C:D:75:A6B@=6956?82?A6BC2>22?20 2 / P > C >20P > C+6B92D:<2?82>32B36B:<ED .6?DE<2?2 <E2D>652?=:CDB:<5:D:D:<3 82I2A252>E2D2?OP O 5:D:D:< q1q2pE1E2r1r2Contoh 4.65 cm5 cm0,2C0,05CPMedan Listrik, Potensial Listrik, dan Kapasitor83 &:&'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?OP O >6?82=2>:82I2C63282:36B:<ED ,,P-P ) OP O O ).2?52?682D:7>6?E?;E<<2?,2B29?I2<6<:B: +252D:D:< D:D:<CE5ED 52?C63E29A6BC68: >2C:?8 >2C:?85:=6D2<<2?C63E29A2BD:<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 '-'-.. 82B 92BECC2>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- DCBAEADEBDEABq1r1 = 5 cm0,2C0,05CPq2E2E1r1 = 5 cmContoh 4.7Pembahasan SoalPada titik-titik sudut B dan Dsebuah bujur sangkar ABCDmasing-masing diletakkan sebuahpartikel bermuatan +q. Agar kuatmedan listrik di titik A = nol, makadi titik C harus diletakkan sebuahpartikel bermuatan sebesar ....a.–qb.+qc.2qd.2qe.22qUMPTN 1991Pembahasan:B(+q)EADEBDEABACD(+q)EAB = EAD = 2kqq = EEBD = 22ABADEE = 22EEBD = 222kqEqEAC = EBD222CkqkqqAC2222Cqqqq 2222Cqqqq qC = 22qJawaban: eMudah dan Aktif Belajar Fisika untuk Kelas XII843.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>3ECCE2DEA6B>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:C5:DE=:C<2?C63282:36B:<ED O&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=294@C O56?82? 252=29CE5ED2?D2B2F6<D@B>652?52?=E2CA6B>E<22?I2?85:D6>3EC>652?=:CDB:< 2B:<@?C6A7=E<C=:CDB:<:?:=29"2ECC>6?86>E<2<2?9E<E>?I2I2?85:?I2D2<2?C63282:36B:<ED 71)( $# .%/# 5 5 *#'!(1 . .%/1 01,!.)1' *0!.0101,/! * %*# !*# *&1)( $)1 0 *(%/0.%'5 *# %(%*#'1,%+(!$,!.)1' *0!.0101,0!./! 108 -642B2>2D6>2D:C5:DE=:CA6B>E<22?D6BDEDEA - O Gambar 4.10Garis-garis gaya yangmenembus bidang permukaan.Gambar 4.11Garis gaya yang menembussuatu permukaanmembentuk sudut .#:DE?8=297=E<C=:CDB:<A252CE2DE3:52?8A6BC68:I2?836BE<EB2? P 4>;:<2<E2D>652?=:CDB:<9@>@86?C636C2B ) 52?2B29?I22 C6;2;2B3:52?83 >6>36?DE<CE5EDLD6B9252A3:52?84 D682<=EBECD6B9252A3:52?8 &:&' :<6D29E:'E2C3:52?8A6BC68: 4>P 4> 4>P O>&E2D>652?=:CDB:< ) !=E<C=:CDB:<52A2D ?52A6B92D:<2?A25282>32B36B:<ED 2 3 4 nEEnnE37°Luas = AE nEContoh 4.8 2 /?DE<CE5ED L4@C L ) P O> Medan Listrik, Potensial Listrik, dan Kapasitor854.Perhitungan Medan Listrik dengan Menggunakan HukumGaussa.Medan Listrik pada Keping Sejajar(652?=:CDB:<5:2?D2B2A6=2DC6;2;2B52A2D5:9:DE?856?82?>E529>6?88E?2<2?#E<E>"2ECC E23E29A6=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:?85:567:?:C:<2?C63282:>E2D2?-A6BC2DE2?=E2C -642B2>2D6>2D:C5:DE=:C<2?C63282:36B:<ED Gambar 4.12Medan listrik antara dua kepingsejajar dengan rapatmuatan dan .- O &E2D>652?=:CDB:<A252A6=2D<@?5E<D@B5:D6?DE<2?36B52C2B<2?<@?C6A#E<E>"2ECC 2B2?I256?82?>6>3E2DCE2DEA6B>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:=:?56BD6BDEDEAD6BC63ED252=29 C:=:?56BD6BDEDEA 4@C L 4@C L 4@C L*=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?85:D:>3E=<2?@=69C2DEA6=2D<@?5E<D@B5:?I2D2<2?56?82?A6BC2>22? O 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? O &6D6B2?82? B2A2D>E2D2? > A6B>:D:F:D2CBE2?892>A2P O )>3 /?DE<CE5ED L4@C L ) P O> 13 4 /?DE<CE5ED L4@C L ) P O> 13 Gambar 4.13Perhitungan kuat medan listrikE pada pelat konduktormenggunakan permukaantertutup (silinder) berdasarkanHukum Gauss.E1A1E2A2E3A3E Kata Kunci•medan listrik•garis-garis gaya•fluks•fluks listrik•keping sejajar•bola konduktor•permukaan GaussMudah dan Aktif Belajar Fisika untuk Kelas XII86b.Kuat Medan Listrik pada Bola Konduktor Berongga+6B92D:<2?&2'&6 %:<2<652=2><@?5E<D@B3@=236B@?882I2?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@=2 52AE?5:52=2>3@=2D:52<D6B52A2D>E2D2?=:CDB:< 6B52C2B<2?#E<E>"2ECC52A2D5:D6?DE<2?36C2B>652?=:CDB:<5:52=2>>2EAE?5:=E2B3@=2I2?836C2B?I2 - 2D2E - O :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$$C2>256?82?;E>=29>E2D2?=:CDB:<A2523@=2D6BC63ED 6?82?56>:<:2?>652?=:CDB:<5:A6B>E<22?"2ECC$$252=29 ---'...&E2D>652?=:CDB:<5:=E2B3@=252A2D5:A6B@=6956?82?>6?82?882A3@=2C63282:>E2D2?=:CDB:<I2?8D6B=6D2<5:AEC2D3@=2 %25:C642B2<6C6=EBE92?>652?=:CDB:<5:C6<:D2B3@=236B@?882252=29M5:52=2>3@=2 <2B6?2- M5:A6B>E<22?3@=2-'M5:=E2BA6B>E<22?3@=2-'.Gambar 4.14Bola konduktor beronggayang memiliki jari-jari R.r = jarak titik ke pusat bola.Gambar 4.15Grafik E terhadap r untukbola konduktor berongga.RaE0r = RrE = 0Tes 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?836C2B?I2P ) .6?DE<2?=292 <2A2?6=6<DB@?2<2?36B96?D:3 ;2B2<I2?85:D6>AE9?I2 +6B92D:<2?82>32B36B:<ED -63E296=6<DB@?!O P O )P O <85:=6A2C<2?D2?A2<646A2D2?2G2=52B:C:C:A6=2D<6A:?836B>E2D2??682D:752?5:A6B46A2D>6?E;EA6=2DA@C:D:7 %:<2;2B2<2?D2B2A6=2D4>52?>652?=:CDB:<) D6?DE<2?=292 36C2BA6B46A2D2?6=6<DB@?3 G2<DEI2?85:A6B=E<2?6=6<DB@?E?DE<>6?42A2:A6=2DA@C:D:74 =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 IIpermukaan 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>A2D3E29>E2D2?I2?8C2>2C6A6BD:82>32B36B:<ED .6?DE<2?<E2D>652?=:CDB:<5:AEC2DA6BC68:;:<22 <66>A2D>E2D2?:DEA@C:D:73 D2?52A@C:D:752??682D:7<66>A2D>E2D2?:DE36BC6=2?8 C6=:?8 2qkrE =2qkRE =Rr>Rr<R14 cm–24q–4q20 cm4C4C4C4CdFEMedan Listrik, Potensial Listrik, dan Kapasitor87 #:DE?87=E<C=:CDB:<I2?8>6?6>3EC3:52?8A6BC68:C:C: 4>;:<2<E2D>652?=:CDB:<C636C2B ) I2?82B29?I22 C6;2;2B56?82?3:52?83 D682<=EBEC3:52?84 >6>36?DE<CE5ED LD6B9252A3:52?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?I2C636C2B P > C %:<2<65E2<6A:?8>6>:=:<:B2A2D>E2D2?I2?8C2>2D6?DE<2??:=2:B2A2D>E2D2??I2)6 P O <8-6 P O 0P O )> C.Energi Potensial Listrik dan Potensial Listrik1.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?-36BEA282I24@E=@>3I2?82B29?I2D@=2< >6?@=2< /?DE<36BA:?52952B:A@C:C:. <6A@C:C:.>E2D2?- 92BEC>6=2<E<2?EC292 6C2B?I2EC292I2?892BEC5:=2<E<2?252=29C632?5:?856?82?36C2B?I282I2 @E=@>352?A6BA:?5292??I2 O .2?52?682D:7>6?E?;E<<2?EC292>6=2G2?82I2 @E=@>3 %:<2EC292I2?85:=2<E<2?- >6=2=E:A6BA:?5292?I2?8C2?82D<64:=>2<2EC292- 52A2D5:A6B@=6956?82?42B2 O . . ..'-- .. .'-- .. '--..O -6CE2:56?82?567:?:C:?I2EC292252=29AB@C6CDB2?C76B6?6B8:2D2E36C2B?I2A6BE3292?6?6B8:2D2E52=2>36?DE<>2D6>2D:C?I2C6A6BD:36B:<ED ,,, O 2A2D<:D2C:>AE=<2?329G2A6BC2>22?6?6B8:A@D6?C:2=5:CE2DED:D:<252=29 ,'--.O &6D6B2?82? EC292=:CDB:<;@E=682I2 @E=@>3)<<@?CD2?D2 @E=@>3P )> - >E2D2?E;: ->E2D2?CE>36B .;2B2<2?D2B2- 52?->+––q2q1q1FcF(2)(1)r1r2Gambar 4.16Interaksi antara muatan q1 dan q2.rMudah dan Aktif Belajar Fisika untuk Kelas XII882.Potensial Listrik+@D6?C:2==:CDB:<252=2936C2B?I26?6B8:A@D6?C:2==:CDB:<A6BC2DE2?>E2D2? +252&2'&6 A@D6?C:2==:CDB:<I2?85:>:=:<:@=69- 252=29C63282:36B:<ED ,--'-'--..O -642B2E>E>A6BC2>22?A@D6?C:2==:CDB:<5:CE2DED:D:<I2?836B;2B2<.52B:>E2D2?CE>36B252=29'.O %:<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 ,---'...O3.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 ,,, '-'- .. '' -.. -O56?82?O 252=293652A@D6?C:2==:CDB:<2?D2B2D:D:< 52?D:D:< Gambar 4.17Potensial listrik oleh 3 muatan,q1, q2, q3 di titik P.Pr3+q3–q2+q1r2r1Gambar 4.18Muatan q, bergerak dari titik 1 ketitik 2 dipengaruhi muatan Q.r2r121-63E29>E2D2?A@C:D:7- P O 5:86B2<<2?>6?E;EC63E29:?D:2D@>I2?836B>E2D2? %2B2<A:C292G2=<65E2A2BD:<6=D6BC63ED252=29P O >52?;2B2<A:C292<9:B?I2252=29 P O > %:<2EC292I2?85:A6B=E<2?E?DE<>6>:?529<2? P O %D6?DE<2?>E2D2?:?D:2D@>D6BC63ED &:&' :<6D29E:- P O . P O >. P O > P O % Contoh 4.9QqMedan Listrik, Potensial Listrik, dan Kapasitor89-68:D:82 C:<E C:<E5: 56?82? 4>52? 4> -63E29>E2D2?=:CDB:<-O O 2<2?5:A:?529<2?52B:D:D:<<6D:D:<I2?8D6B=6D2<A252A6BD6?8292? %:<2>E2D2?- O 52?>E2D2?- O O 2?882A<6925:B2?-D:52<36BA6?82BE9D6B9252AA@D6?C:2=5:D6?DE<2?=292 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 - O O -O O 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<2 O C69:?882 O 4 /C292E?DE<>6>:?529<2?>E2D2?252=29 -OO O O0P O;@E=6 Tantanganuntuk AndaHitunglah usaha yang diperlukanuntuk memindahkan muatan positifyang besarnya 10 C dari suatu titikyang potensialnya 10 V ke suatu titikyang potensialnya 60 V.56?82?!*67&2&&3 < 5:A6B@=69 '- .. P O %P P O >P O %25:>E2D2?:?D:2D@>252=29P O Contoh 4.10ABCD10 cm6 cm4 cm4 cm4.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=D6BC63ED36B=2<E<2B6?2>652?=:CDB:<>6BEA2<2?>652?<@?C6BF2D:7 ?6B8:D@D2=C63E29A2BD:<6=56?82?>2CC2)52?>E2D2?-I2?836B86B2<52=2>>652?=:CDB:<252=29, ' ,'2D2E -)2-)2O(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: -)2O!*67&2&&3 < >6?E?;E<<2?A6BE3292?6?6B8:A@D6?C:2=>6?;25:6?6B8:<:?6D:< Hukum kekekalan energi mekanikpartikel dalam medan listrik berlakujika pada partikel tersebut tidak adagaya lain yang bekerja selain gayaCoulomb.IngatlahMudah dan Aktif Belajar Fisika untuk Kelas XII90+6B92D:<2?82>32B36B:<ED VAB 652A@D6?C:2=5:2?D2B25E2A6=2DC6;2;2BA25282>32BD6BC63ED252=29 0 -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@?) P O<8>E2D2?AB@D@?- P O &:&'(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-)222- -22) 6B52C2B<2?36C2B2?I2?85:<6D29E:A252C@2=52A2D5:A6B@=69 0 <8 2 > C > C 2&646A2D2?AB@D@?C636=E>>6?I6?DE9A6=2D 252=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 /C292I2?85:=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><E2D>652?=:CDB:<0 > 652A@D6?C:2=O 52A2D5:?I2D2<2?C63282:3652A@D6?C:2=2?D2B2<65E2<6A:?8C6;2;2B Gambar 4.19Dua buah keping sejajar danterpisah sejauh d diberimuatan yang sama.Contoh 4.11Kata Kunci•energi potensial listrik•potensial listrik•voltABdEVNext >