Electric Potential and Capacitance

1. Electric Potential and Capacitance What’s a volt anyway?

2. Why Potential? • Energy makes some problems easier to solve • Energy is a scalar • The concept of a potential is useful to have in our physics toolbox

3. Define Change in Electric Potential Energy • When a positive charge released near the positive plate moves to the right it loses potential energy equal in magnitude to the work done by the field on the charge Positive plate Negative plate b a

4. Electrical Potential Energy • The change in potential energy equals the negative of the work done by the field • Electrical PE changes to kinetic • The positive charge has its greatest PE near positive plate • Only differences in potential energy are measurable

5. Electrical Potential • Potential is potential energy per unit charge • Analogous to field which is force per unit charge • Symbol of potential is V; Va = PEa/q • Only potential differences are measurable; zero point of potential is arbitrary • Vab = Va – Vb = - Wba/q • Wba is work done to move q from b to a

6. Units • Unit of electric potential is the volt • Abbreviation V • 1 V = 1 Joule/Coulomb = 1 J/C • Thus electrical work = qV • Potential difference is called voltage

7. V = 0 Arbitrary • Usually ground is zero point of potential • Sometimes potential chosen zero at infinity • + terminal of 12V battery is said to be at 12V higher potential than – terminal

8. Electric Potential and Potential Energy • DPE = PEb – PEa = qVba • If object with charge q moves through a potential difference Vba its potential energy changes by qVba • Example: What is the gain of electrical PE when 1 C of charge moves between the terminals of a 12 volt battery? • Water Analogy: • Voltage is like water pressure 12 joules

9. Example: Electron in Computer Monitor • An electron is accelerated from rest through a potential difference of 5000 volts • Find its change in potential energy • Find its speed after acceleration Vba = 5000 v = Vb - Va b a

10. Example: Electron in Computer Monitor • An electron is accelerated from rest through a potential difference of 5000 volts • Find its change in potential energy • Find its speed after acceleration • PE = qV = 1.6x 10-19 C x 5 x 103 V = 8 x 10-16 J = ½ m v2 • m = 9.11 x 10-31 kg ; v = 4.2 x 107 m/s

11. Questions on Preceding Example • Does the energy depend on the particle’s mass? • Does the final speed depend on the mass?

12. Electric Potential and Electric Field • Can describe charge distribution in terms field or potential. Consider uniform field: • F = Eq • W = qVba • W = Fd =qE d • Thus Vba = Ed or E = Vba/d • Alternate units for E: volts per meter v/m Instead of? N/C

13. Equipotential Lines • An equipotential surface is perpendicular to the electric field at any point • If not charges would move along the surface • This contradicts the idea of an equipotential

15. Describe the equipotentials… • …between large oppositely charged conducting plates + -

16. Electric Potential Due to Point Charges • V = k Q/r derived from Calculus • Here V = 0 at r = infinity; V represents potential difference between r and infinity

18. Example: Work to force two point charges together • What work is needed to bring a 2mC charge from far away to within 10cm of a 5 mc charge? • Work required = change in potential energy • W = qVba = q{kQ/rb – kQ/ra} = qkQ/rb 0.90 J

19. Potential at an Arbitrary Point Near Several Point Charges • Add the potentials due to each point charge • Use the right sign for the charge • Relax; potential isn’t a vector • What is true about the mid-plane between two equal point charges of opposite sign? Potential everywhere zero

20. Find… • The potential due to a microcoulomb charge at a point one meter away • The potential energy of two microcoulomb charges one meter apart 9000 volts 9 x 10-3 J

21. The Electron Volt • The energy acquired by a particle carrying a charge equal to that of the electron when accelerated through one volt • 1 eV = 1.6 x 10-19 joules • It’s about ENERGY • 1 KeV = 1000 eV = 1.6 x 10-16 joules • 1 MeV = 106 eV • 1 Gev = 109 eV

22. Capacitance • Stores charge • Two conducting plates • NOT touching • May have insulating material between Q = CV

23. Capacitance • Symbol C • Unit: coulombs per volt = farad • 1 pf = 1 picofarad = 10-12 farad • 1nf = 1 nanofarad = 10-9 f • 1mf = 1 microfarad = 10-6 f

24. C is Constant for a Given Capacitor • Does not depend on Q or V • Proportional to area • Inversely proportional to distance between plates • C = e0 A/d • If dielectric like oil or paper between plates use e = Ke0; K is called dielectric constant • e iscalled permittivity. e0 is permittivity of free space

26. Find the Capacitance • A capacitor can hold 5 mC of charge at a potential difference of 100 volts. What is its capacitance

27. Q = CV • C = Q/V = 5 x 10-6 C/ 100V = 5 x 10-8 f = 50000 pf = 0.05 mf

29. Why is the Electric Field inside the dielectric less than outside? Less since fewer field lines

30. Capacitors Photos courtesy Illinois Capacitor, Inc

31. Supercapacitors -

32. Supercapcitors use double layer electrolyte technology, usually with activated carbon and sulfuric acid. The carbon has huge surface area

33. Applications • In automotive ignitions • In strobe lights • In electronic flash • In power supplies • In nearly all electronics

34. Supercapacitor applications • Power motor vehicles for short bursts • Backup power for computers • Operate emergency doors and slides in commercial aircraft • Lower energy density but greater power density than batteries

35. A Capacitor Stores Electric Energy • A battery produces electric energy bit by bit • A capacitor is NOT a type of battery • A battery can be used to charge a capacitor

36. Energy in a Capacitor Holding Charge Q at Voltage V • U = QV/2 = CV2/2 = Q2/2C • Derivation: the work needed to charge a capacitor by bringing charge onto a plate when some is already there (use W =QV) • Initially V = 0 • Average voltage during the charging process is V/2

37. Example • A 20 mf capacitor stores 20 millijoules of energy. What is the voltage across it? • U = CV2/2 • V = (2U/C)1/2 = (2x20 x 10-3J/20x10-6F)1/2 = 44.7 V