Ch. 20.5 & 20.6: Capacitors and Energy Storage

Chapter 20 Electric Potential and Electric potential Energy Outline 20-1 Electric Potential Energy and the Electric Potential 20-2 20-3 Energy Conversation The Electric Potential of Point Charges 20-4 Equipotential Surfaces and the Electric Field. 20-5 Capacitor and Dielectrics 20-6 Electric Energy Storage 20-5 Capacitor and Dielectric A capacitor has a capacity to store electric charge and energy, which is determined by its capacitance C, Q = CV Where, Q and V are the charge and voltage, respectively. Definition of Capacitance, C Q C= V (20 − 9) SI units: coulomb/volt=farad, F (1F = 1 C/V) Note: 1 pF = 10-12 F; 1 uF = 10-6 F 20-41 Capacitor It is desired that 5.8 μC of charge be stored on each plate of a 3.2- μC capacitor. What potential difference is required between the plates? Solution: Solve equation (20-9) for V: Q 5.8 × 10−6 C V= = = 1.8 V −6 C 3.2 × 10 F Parallel-Plate Capacitor Deriving capacitance C: determined by geometrical parameters Figure 20-13 A Parallel-Plate Capacitor Deriving capacitance C From Active Example 19-3, we have E= Q σ = ε0 ε0 A (20 − 1) The magnitude of the potential difference is ΔV = V = Ed = Qd ε0 A Therefore, C= Q Q = V (Qd / ε 0 A) Capacitance of a parallel-plate capacitor in Air is C= ε0 A d (20 − 12) SI units: coulomb/volt=farad, F The constant “Permittivity of free space” is ε0 = 1 4π k = 8.85 × 10−12 C 2 / N ⋅ m 2 Note: The capacitance of a capacitor is determined by its geometrical parameters. Dielectrics How to increase the Capacitance? Figure 20-15 The Effect of a Dielectric on the Electric Field of a Capacitor After a permanent dipole material is inserted into the capacitor, for the same amount of charge Q, the electric field E0 is decreased as E= E0 κ Where κ is the dielectric constant, which is greater than 1, and is determined by the material. Now we have V = Ed = C= E0 κ d= E0 d κ = V0 κ Q Q Q = = κ = κ C0 V (V0 / κ ) V0 Capacitance of a parallel-plate capacitor with Dielectric C= κε 0 A d (20 − 15) The capacitance of a dielectric capacitor is determined by • its geometric parameters (A, d) • and its material κ 20-45 A parallel-plate capacitor has plates with an area 0.012 m2 for each plate, and a separation of 0.88 mm. The space bewteen the plates is filled with a dielectric whose dielectric constant is 2.0. What is the potential difference between the plates when the charge on the capacitor plate is 4.7 μC. Solution: Solve equation 20-9 for V and substitute equation 20-15 for C: 4.7 × 10− 6 C )( 0.88 × 10−3 m ) ( Q Qd V= = = = 19 kV −12 2 2 2 C κε 0 A ( 2.0 ) ( 8.85 × 10 C / N ⋅ m )( 0.012 m ) Dielectric Breakdown If the electric field applied to a capacitor is too large, the dielectric material make be damaged, which is referred to as breakdown. The maximum filed a dielectric can withstand without breakdown is called dielectric strength. Table 20-2 Dielectric strengths Substance Dielectric Strength (V/m) Mica 100x106 Air 3x106 20-6 Electrical Energy Storage Figure 20-17 The Energy Required to Charge a Capacitor The total energy U (average) in a capacitor is 1 U = QVav = QV 2 (20 − 16) Since Q=CV, we have 1 U = CV 2 2 (20 − 17) With V=Q/C, we have Q2 U= 2C (20 − 18) 20-53 Calculate the work done by a 3.0-V battery as it charges a 7.2-μF capacitor in the flash unit of a camera. Summary 1) Capacitance of a parallel-plate capacitor in Air is C= ε0 A d (20 − 12) 2) Capacitance of a parallel-plate capacitor with Dielectric C= κε 0 A d (20 − 15) 3) The total energy U (average) in a capacitor is 1 U = QVav = QV 2 (20 − 16) Exercise 20-3 A capacitor of 0.75 uF is charged to a voltage 16 V. What is the magnitude of the charge on each plate of capacitor? Solution Q=CV = (0.75x10-6 F)(16 V) = 1.2 x10-5 C Example 20-5 All charge up A capacitor is constructed with plates of areas 0.0280 m2 and separation 0.550 mm. Find the magnitude of the charge on each plate of this capacitor when the potential difference between the plates is 20.1 V Solution 1) Find the capacitance ε0 A (8.85 ×10−12 C 2 / N .m 2 )(0.0280m 2 ) −10 = = 4.51 × 10 C= F −3 d 0.550 ×10 m 2) Find the charge Q = CV = (4.51×10−10 F )(20.1V ) = 9.06 × 10−9 C Example 20-6 Even More charge up A capacitor is constructed with plates of areas 0.0280 m2 and separation 0.550 mm. It is filled with dielectric with dielectric constant κ. When it is connected to 12.0-V battery, each plate has a charges of magnitude 3.26x10-8 C. What is the value of dielectric constant ? κ Picture the Problem Example 20-6 Even More Charged Up Solution 1) The capacitance Q (3.62 ×10−8 C ) = 3.02 × 10−9 F C= = 12V V 2) Find the dielectric constant C= κ κε 0 A d Cd κ= ε0 A (3.02 × 10−9 F )(0.550 × 10−3 m) = = 6.70 −12 2 2 2 (8.85 × 10 C / N ⋅ m )(0.0280m ) Example 20-7 In a typical defibrillator, a 175-uF capacitor is charged until the potential difference is 2240V. (a) What is the magnitude of the charge on each plate of the final capacitor? (b) find the energy stored in the charged-up defibrillator. Picture the problem Example 20-7 The Defibrillator: Adding a Shock to the System Solution Part (a) Q= CV = ( 175x10-6 F)(2240 V) = 0.392 C Part (b) 1 U = CV 2 2 1 = (175 × 10−6 F ) (2240V ) 2 = 439 J 2