Chapter 20 Electric Potential and Electric potential Energy

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Milton McCormick
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1 Outline Chapter 2 Electric Potential and Electric potential Energy 2-1 Electric Potential Energy and the Electric Potential 2-2 Energy Conversation 2-3 The Electric Potential of Point Charges 2-4 Equipotential Surfaces and the Electric Field. 2-5 Capacitor and Dielectrics 2-6 Electric Energy Storage

2 2-5 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 C Q = (2 9) V SI units: coulomb/volt=farad, F (1F = 1 C/V) Note: 1 pf = 1-12 F; 1 uf = 1-6 F

3 2-41 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 (2-9) for V: V Q = = = C F C V

4 Parallel-Plate Capacitor Deriving capacitance C: determined by geometrical parameters Figure 2-13 A Parallel-Plate Capacitor

5 Deriving capacitance C From Active Example 19-3, we have E = σ Q (2 1) ε = ε A The magnitude of the potential difference is Δ V = V = Ed = Qd ε A Therefore, Q Q C = = V ( Qd / ε A )

6 Capacitance of a parallel-plate capacitor in Air is C ε A = (2 12) d SI units: coulomb/volt=farad, F The constant Permittivity of free space is Note: ε 1 = = C / N m 4π k The capacitance of a capacitor is determined by its geometrical parameters.

8 Dielectrics How to increase the Capacitance? Figure 2-15 The Effect of a Dielectric on the Electric Field of a Capacitor

9 After a permanent dipole material is inserted into the capacitor, for the same amount of charge Q, the electric field E is decreased as κ E = E κ Where is the dielectric constant, which is greater than 1, and is determined by the material. Now we have E Ed V V = Ed = d = = κ κ κ C = Q = Q = κ Q = κc V ( V / κ ) V

10 Capacitance of a parallel-plate capacitor with Dielectric C κε A = (2 15) d The capacitance of a dielectric capacitor is determined by its geometric parameters (A, d) and its material κ

11 2-45 A parallel-plate capacitor has plates with an area.12 m 2 for each plate, and a separation of.88 mm. The space bewteen the plates is filled with a dielectric whose dielectric constant is 2.. What is the potential difference between the plates when the charge on the capacitor plate is 4.7 μc. Solution: Solve equation 2-9 for V and substitute equation 2-15 for C: V 6 3 ( C)(.88 1 m) ( 2.)( C / N m )(.12 m ) Q Qd = = = = C κε A 19 kv

12 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 2-2 Dielectric strengths Substance Dielectric Strength (V/m) Mica 1x1 6 Air 3x1 6

13 2-6 Electrical Energy Storage Figure 2-17 The Energy Required to Charge a Capacitor

14 The total energy U (average) in a capacitor is 1 U = QVav = QV (2 16) 2 Since Q=CV, we have U = CV (2 17) With V=Q/C, we have U 2 Q = (2 18) 2C

15 2-53 Calculate the work done by a 3.-V battery as it charges a 7.2-μF capacitor in the flash unit of a camera.

16 Summary 1) Capacitance of a parallel-plate capacitor in Air is C ε A = (2 12) d 2) Capacitance of a parallel-plate capacitor with Dielectric κε A C = (2 15) d 3) The total energy U (average) in a capacitor is 1 U = QVav = QV (2 16) 2

17 Exercise 2-3 A capacitor of.75 uf is charged to a voltage 16 V. What is the magnitude of the charge on each plate of capacitor? Solution Q=CV = (.75x1-6 F)(16 V) = 1.2 x1-5 C

18 Example 2-5 All charge up A capacitor is constructed with plates of areas.28 m 2 and separation.55 mm. Find the magnitude of the charge on each plate of this capacitor when the potential difference between the plates is 2.1 V

19 Solution 1) Find the capacitance C ε A C N m m d.55 1 m ( /. )(.28 ) = = = F 2) Find the charge 1 9 Q CV ( = = F)(2.1 V) = C

20 Example 2-6 Even More charge up A capacitor is constructed with plates of areas.28 m 2 and separation.55 mm. It is filled with dielectric with dielectric constant κ. When it is connected to 12.-V battery, each plate has a charges of magnitude 3.26x1-8 C. What is the value of dielectric constant? κ Picture the Problem Example 2-6 Even More Charged Up

21 Solution 1) The capacitance 8 Q ( C) C = = = V 12V 2) Find the dielectric constant κ 9 F C = κε d A κ = Cd ε A 9 3 (3.2 1 F)(.55 1 m) = = ( C / N m )(.28 m ) 6.7

22 Example 2-7 In a typical defibrillator, a 175-uF capacitor is charged until the potential difference is 224V. (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 2-7 The Defibrillator: Adding a Shock to the System

23 Solution Part (a) Q= CV = ( 175x1-6 F)(224 V) =.392 C Part (b) U = 1 2 CV 2 1 ( )(224 ) = F V = J 2

Q24.1 The two conductors a and b are insulated from each other, forming a capacitor. You increase the charge on a to +2Q and increase the charge on b

Q24.1 The two conductors a and b are insulated from each other, forming a capacitor. You increase the charge on a to +2Q and increase the charge on b to 2Q, while keeping the conductors in the same positions.