Materi Fisika Listrik 3 Kapasitansi [PDF]

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Pengertian Kapasitor • Dua penghantar (plat) berdekatan yang diberi muatan sama tetapi berlawanan jenis disebut kapasitor. • Sifat menyimpan energi listrik / muatan listrik. • Kapasitas suatu kapasitor (C) ditentukan oleh perbandingan antara besar muatan Q dari salah satu penghantarnya dengan beda potensial V antara kedua penghantar itu.



Kegunaan Kapasitor • Untuk U k menghindari hi d i terjadinya j di lloncatan li listrik ik pada d rangkaian2 yang mengandung kumparan bila tiba2 diputuskan arusnya. • Rangkaian yang dipakai untuk menghidupkan mesin mobil • Untuk memilih panjang gelombang yang ditangkap oleh pesawat penerima radio. Bentuk kapasitor • Kapasitor bentuk keping sejajar • Kapasitor bentuk bola sepusat • Kapasitor bentuk silinder



MacamMacam -Macam Kapasitor



Capacitor: A Device to Store Energy in the Form of an Electric Field



Kapasitansi Kapasitansi C adalah jumlah muatan yang disimpan dibagi dengan tegangan



Q = CV Satuan untuk kapasitansi Coulomb/Volt = Farad



Capacitance the h amount off charge h stored divided by the voltage is the capacitance C



+- - - - - - - - - - - -



++++++++



Q = CV unit of capacitance Coulomb/Volt = Farad



Kapasitor Pelat Paralel Dua pelat paralel yang luas • luas A • jarak antar pelat d • kerapatan muatan sama ttp berbeda tanda σ



C = Q/∆V = σA/(σd/ε0)



d



C = ε0 A/d Untuk menaikkan nilai kapasitansi membesarkan b k lluasan



+σ



–σ



memperkecil celah



Bila ukuran ke dua pelat cukup besar dibanding jarak celah d



E



Medan diantara ke dua pelat mempunyai nilai seragam



E = σ/ε0 Beda potensial diantara k d ke dua pelat l



+σ



–σ



∆V = Ed = σd/ε0



C = Q/V = Q/(Q/ 4 πε0 R) = 4 πε0 R Kapasitansi bergantung pada bentuk geometri dari permukaan.



Capacitance p Depends p on… • Separation of the plates (d) [d in m] 1 C∝ d † Plate Area (A)



[A in m2]



C∝A † Dielectric Constant ((ε))



C ∝ε



[[ε in C2//N-m2]



Show Demo Model, calculate its capacitance , and show how to charge g it up p with a battery. y Circular parallel plate capacitor r



r = 10 cm



r



A = πr2 = π(.1)2 A = .03 m 2



s



S = 1 mm = .001 m



C=



ε0A S



C = (10 −11 )



.03 Coulomb .001 Volt



}Farad



C = 3 ×10 −10 F C = 300 pF



p = pico = 10-12



1. Consider two plates separated by d=1.5 cm , where the electric field between them is 100 V/m, V/m and the charge on the plates is 30.0 µC. What is the capacitance?



The Capacitance is: 2.0 10-5 2.



F Or 20 microfarad.



What is the capacitance of a capacitor if the charge is 0 075 C held at V = 400 V 0.075 C = Q/V =0.075 C/400 V = 1.875 X 10-4 f = 187.5 µF



Dielectrics K • Capacitors are usually constructed with an insulating material between the plates • This helps prevent a breakdown or arcing between the plates at higher voltages • The plates can be placed closer together without danger of touching reducing d • The capacitance is increased by a factor K (dielectric constant) • The formula for capacitance is now



A C = Kε 0 d



Permitivitas Dielektrik



ε = Kε0



K adalah konstanta dielektrik tergantung g gp pada material dielektrik umumnya nilai K = 1 sampai 10



Bila dielektrik sepenuhnya mengisi volume diantara pelat-pelat :



⇒ E berkurang oleh faktor K



E = σ// ε = σ/K /K ε0



⇒ V berkurang oleh faktor K



V = σd/K ε0



⇒ C meningkat oleh faktor K



C = Kε0 A/d



Material



K



Vacuum Gl Glass Mica Mylar P ffi Wax Paraffin W Plexiglas Polyethylene P l i l chloride Polyvinyl hl id Porcelain Teflon Germanium Rubber



1.00000 5 10 5-10 3-6 3.1 2 1 – 2.5 2.1 25 3.40 2.25 3 18 3.18 6.0 – 8.0 2.1 16 2.5 – 3.0



Water Glycerin Liquid ammonia(-78°C) Benzene Air(1 atm) Air(100 atm)



80.44 80 42.5 25 2 284 2.284 1.00059 1.0548



Question: The energy content of a charged capacitor resides in its (a) plates (b) potential difference (c) charge Answer: d (d) electric field Question: The plates of a parallel-plate capacitor of capacitance C are brought together to one-third their original separation. The capacitance is now ((a)) C/9 (b) C/3 (c) 3C (d) 9C Answer: c



Example p •



Two rectangular sheets of copper foil 16 X 20 cm are separated by a thin layer of paraffin wax 0 0.2 2 mm thick thick. Calculate the capacitance if the dielectric constant for the wax is 2.4.



Kε o A d A = ( 16 cm )( 20 cm ) = 320 cm 2 = 0.032 m 2



C=



d = 0.2 mm = 2 X 10 −4 m ( 8.854 X 10 −12 C 2 / Nm 2 )( 2.4 )( 0.032 m 2 ) C= 2 X 10 −4 m C = 3.4 X 10 −9 f = 3400 pf



Energy Storage in Capacitors • The energy stored in a capacitor is just the work necessary to separate the charges in p the first place. • As you keep moving charges, you have a greater force to overcome because the g potential keeps increasing • The total work is jjust the total charge g times the average voltage or W = QV/2.



Energy Storage in Capacitors Total charge stored on capacitor Q



W = ∫ V0dq



V=q/C



Q



W = ∫ (q/C)dq ( /C)d = Q /2C 2



0



1 Energy = QV Q 2 Q = CV



W = CV = 12 QV 1 2



2



1 2 1Q E Energy = U = CV = 2 2 C



2



where is the energy located? W = C V2/2 = C (E d)2/2 = ((ε0 A / d)) E2 d2 / 2 = (ε0 E2/2)(A d) = (energy density)(volume)



ε0 E /2 2



is the energy density



the energy is where the electric field is between the plates Misal kapasitor 1.2 µF diberi muatan hingga berpotensial 3 kV, berapa energi yang tersimpan didalamnya?



W = 12 CV 2 = 12 QV



W = 5.4 Joule



Hubungan paralel kapasitor c Parallel: V bernilai sama untuk ke dua kapasitor. Kapasitansi tinggal di jumlahkan. jumlahkan



Ceq = Qtotal/V Ceq = Q1/V1 + Q2/V2 V1 = V2 = V Ceq = C1 + C2



Hubungan Paralel Kapasitor



Q1 = C1V ;



Q2 = C2V ;



Q3 = C3V ;



Q = C pV ;



C p = C1 + C2 + C3



Kapasitor yang dihubungkan paralel, tegangan antara ujung2 kapas ujung kapasitor tor adalah sama, sebesar V.



Hubungan seri kapasitor



Seri : V tidak sama pada masing2 kapasitor. Tetapi Q sama (mengapa?) 1/ Ceqq = V1/ Q1 + V2/ Q2 Q1 = Q2 = Q 1/ Ceq = 1/ C1 + 1/ C2



Hubungan seri kapasitor Q Vab = ; C1



Q Vbc = ; C2



Q Vcd = ; C3



Q Vad = Cs



1 1 1 1 = + + Cs C1 C2 C3 Kapasitor yang dihubungkan seri akan mempunyai muatan yang sama.



Q = Q1 = Q2 = Q3



Dua kapasitor dihubungkan seri yang dihubungkan dengan sumber 1000 V. Tentukan: a. C-subsitusi (gabungan) b. Muatan masing-masing kapasitor c. Beda potensial pada ujung masing-masing kapasitor d. Energi yang tersimpan dalam susunan kapasitor Penyelesaian: a.



1 1 1 1 1 = + = + Csub C1 C2 3 pF 6 pF



C sub = 2 pF



b.



q1 = q2 → q = Csub.V = (2 x 10-12 F) (1000 V) = 2 nC



c.



q1 2 x10 − 9 C = = 667 V V1 = − 12 3 x10 F C1



d.



Energi g dalam



(



W1 =



1 1 q 1V 1 = 2 x 10 2 2



W2 =



1 1 q 2V 2 = 2 x 10 2 2



(



−9



C −9



)(667 C



q2 2 x10 − 9 C = = 333 V V2 = 6 x10 −12 F C2



V ) = 6 . 7 x 10



)(333



−7



V ) = 3 . 3 x 10



W sub = (6.7 + 3.3 ) x 10 −7 J = 10 x10 -7 J



J −7



J



Heart Muscle Cells • The cell wall acts as a capacitor since charges are separated on the interior and exterior surfaces. The voltage is in the millivolt range. Molecules in the cell wall are polarized. The charge on the outside of the wall is on the order of 10-88 Coulomb. When the heart beats, beats the wall depolarizes in a wave as shown in this picture. Here the wave proceeds from l f to right. left i h After Af depolarization d l i i is i complete, the cell wall repolarizes. This cycle gives rise to an electrical signal that can be detected externally with an EKG (elektrokardiogram) machine.



EKG Tracing



The P wave is Th i contraction t ti off the th upper chambers h b or atria. t i The Th QRS is the ventricles. The wave proceeds from left to right and toward the front, then down to the left and toward the rear of the heart. The T wave is the re-polarization of the muscle in preparation for the next beat. Heart defects show up as variations in the wave pattern.