1. Physics 1220/1320 Electromagnetism – part one: electrostatics Lecture Electricity, chapter 21-26

2. Electricity Consider a force like gravity but a billion-billion-billion-billion times stronger And with two kinds of active matter: electrons and protons And one kind of neutral matter: neutrons

3. The phenomenon of charge Problem Invisibility Common problem in physics: have to believe in invisible stuff and find ways to demonstrate its existence. Danger of sth invisible If we rub electrons onto the plastic, is it feasible to say that we rub protons on it in the second experiment? No! If we move protons, we move electrons with them. But what if in the second experiment we still moved electrons – in the other direction?

4. Why matter is usually electrically neutral: Like charges repel, unlike charges attract Mixed + and – are pulled together by enormous attraction These huge forces balance each other almost out so that matter is neutral Two important laws: Conservation & quantization of charge

5. Where do the charges come from? Electrons and protons carry charge. Neutrons don’t. Positive (proton), negative (electron) Consider: Why does the electron not fall into the nucleus? Why does the nucleus not fly apart? Why does the electron not fly apart? Consequence of QM uncertainty relation More forces, total of four Short ranged – limit for nucleus size Uranium almost ready to fly apart

6. Electric Properties of Matter (I) Materials which conduct electricity well are called ______________ Materials which prohibited the flow of electricity are called ________________ ‘_____’ or ‘______’ is a conductor with an infinite reservoir of charge ____________ are in between and can be conveniently ‘switched’ _____________are ideal conductors without losses

7. Induction : Conductors and Insulators Induction : Conductors and Insulators Induction - Appears visibly in conductors a) a)Are charges present? c) Why do like charges collect at opposite side? b) Why are there not more ‘-’ charges? d) Why does the metal sphere not stay charged forever?

8. Coulomb’s Law Concept of point charges Applies strictly in vacuum although in air deviations are small Applies for charges at rest (electrostatics) Force on a charge by other charges ~ ___________ ~ ___________ Significant constants: e = 1.602176462(63) 10 -19 C i.e. even nC extremely good statistics (SI) 1/4  0 Modern Physics: why value? how constant?

9. Principle of superposition of forces: If more than one charge exerts a force on a target charge, how do the forces combine? Luckily, they add as vector sums! Consider charges q 1, q 2, and Q: F 1 on Q acc. to Coulomb’s law Component F 1x of F 1 in x: What changes when F 2 (Q) is determined? What changes when q1 is negative? Find F 1

10. Electric Fields How does the force ‘migrate’ to be felt by the other charge? : Concept of fields

12. Charges –q and 4q are placed as shown. Of the five positions indicated at 1-far left, 2 – ¼ distance, 3 – middle, 4 – ¾ distance and 5 – same distance off to the right, the position at which E is zero is: 1, 2, 3, 4, 5

13. Group task: Find force of all combinations of distances and charge arrangements

14. Group task: Find fields for all combinations of distances and charge arrangements at all charge positions.

15. Direction E at black point equidistant from charges is indicated by a vector. It shows that: a) A and B are +b) A and B are -c) A + B – d) A – B +e) A = 0 B -

16. Electric field lines For the visualization of electric fields, the concept of field lines is used.

17. Electric Dipoles H 2 O : O 2- (ion) H 1+

18. Gauss’s Law, Flux

20. Group Task: ° Find flux through each surface for  = 30° and total flux through cube What changes for case b? n1:n1: n2:n2: n3:n3: n4:n4: n 5,n 6 :

21. Gauss’s Law Basic message:

22. Important Applications of Gauss’s Law

23. Group Task 2q on inner 4q on outer shell http://www.falstad.com/vector3de/

24. http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html http://www.falstad.com/vector3de/

25. Group Task For charges 1 = +q, 2 = -q, 3= +2q Find the flux through the surfaces S1-S5

26. Electric Potential Energy

28. Electric Potential V units volt: [V] = [J/C]

29. Potential difference: [V/m] Potential Difference

30. Calculating velocities from potential differences Energy conservation: K a +U a = K b +U b Dust particle m= 5 10 -9 [kg], charge q 0 = 2nC

31. Outside sphere: V = k q/r Surface sphere: V = k q/R Inside sphere:

32. Equipotential Surfaces

34. Potential Gradient

35. Moving charges: Electric Current Path of moving charges: circuit Path of moving charges: circuit Transporting energy Transporting energy Current Current http://math.furman.edu/~dcs/java/rw.html Random walk does not mean ‘no progression’ Random motion fast: 10 6 m/s Drift speed slow: 10 -4 m/s e - typically moves only few cm Positive current direction:= direction flow + charge

36. Work done by E on moving charges  heat (average vibrational energy increased i.e. temperature) Current through A:= dQ/dt charge through A per unit time Unit [A] ‘Ampere’ [A] = [C/s] Concentration of charges n [m -3 ], all move with v d, in dt moves v d dt, volume Av d dt, number of particles in volume n Av d dt What is charge that flows out of volume? Current and current density do not depend on sign of charge  Replace q by /q/

37. Resistivity and Resistance Properties of material matter too: For metals, at T = const. J= nqv d ~ E Proportionality constant  is resistivity  = E/J Proportionality constant  is resistivity  = E/JOhm’s law Reciprocal of  is conductivity Unit  : [  m] ‘Ohm’ = [(V/m) / (A/m 2 )] = [Vm/A]

38. Types of resistivity

39. Ask for total current in and potential at ends of conductor: Relate value of current to Potential difference between ends. For uniform J,E I = JA and V =EL with Ohm’s law E=  J   V/L =  I/A Const.   I ~ V ‘resistance’ R = V/I [  ]   vs. R R =  L/A R = V/I V = R I I = V/R Resistance

41. E, V, R of a wire Typical wire: copper,  Cu = 1.72 x 10 -8  m cross sectional area of 1mm diameter wire is 8.2x10 -7 m -2 current a) 1A b) 1kA for points a) 1mm b) 1m c) 100m apart E =  J =  I/A = V/m (a) V/m (b) V = EL =  V (a), mV (b), V (c) R = V/I = V/ A =  Group Task

42. Resistance of hollow cylinder length L, inner and outer radii a and b Current flow radially! Not lengthwise! Cross section is not constant: 2  aL to 2  bL  find resistance of thin shell, then integrate Area shell: 2  rL with current path dr and resistance dR between surfaces dR=  dr/(2  rL)  dV = I dR to overcome And V tot =  dV R = int(dR) =  L) int ab (dr/r) =  L) ln(b/a )

43. Electromotive Force Steady currents require circuits: closed loops of conducting material otherwise current dies down after short time Charges which do a full loop must have unchanged potential energy Resistance always reduces U A part of circuit is needed which increases U again This is done by the emf. Note that it is NOT a force but a potential! First, we consider ideal sources (emf) : V ab = E = IR

44. I is not used up while flowing from + to – I is the same everywhere in the circuit Emf can be battery (chemical), photovoltaic (sun energy/chemical), from other circuit (electrical), every unit which can create em energy EMF sources usually possess Internal Resistance. Then, V ab = E – Ir and I = E /(R+r)

45. Circuit Diagrams (Ideal wires and am-meters have Zero resistance) No I through voltmeter (infinite R) … i.e. no current at all Voltage is always measured in parallel, amps in series

46. Energy and Power in Circuits Rate of conversion to electric energy: E I, rate of dissipation I 2 r – difference = power output source

47. Resistor networks Careful: opposite to capacitor series/parallel rules!

48. Combining Measuring Instruments

49. Group Task: Find R eq and I and V across /through each resistor!

50. Group task: Find I 25 and I 20

51. Kirchhoff’s Rules A more general approach to analyze resistor networks Is to look at them as loops and junctions: This method works even when our rules to reduce a circuit to its R eq fails.

52. ‘Charging a battery’ Circuit with more than one loop! Apply both rules. Junction rule, point a: II = A [Similarly point b: ] Loop rule: 3 loops to choose from ‘1’: decide direction in loop – clockwise (sets signs!)   r =  Find E from loop 2: Lets do the loop counterclockwise:  E = V

53. 5 currents Use junction rule at a, b and c 3 unknown currents Need 3 eqn Loop rule to 3 loops: cad-source cbd-source cab (3 bec.of no.unknowns) Let’s set R 1 =R 3 =R 4 =1  And R 2 =R 5 =2  Group task: Find values of I1,I2,and I3! I 1 = A, I 2 = A, I 3 =

54. Capacitance E ~ /Q/  V ab ~ /Q/ Double Q: But ratio Q/V ab is constant Capacitance is measure of ability of a capacitor to store energy! (because higher C = higher Q per V ab = higher energy Value of C depends on geometry (distance plates, size plates, and material property of plate material)

55. Plate Capacitors E =  0 = Q/A  0 For uniform field E and given plate distance d V ab = E d = 1/  0 (Qd)/A Units: [F] = [C 2 /Nm] = [C 2 /J] … typically micro or nano Farad

56. Capacitor Networks In a series connection, the magnitude of charge on all plates is the same. The potential differences are not the same. Or using concept of equivalent capacitance 1/C eq = In a parallel connection, the potential difference for all individual capacitors is the same. The charges are not the same. Or C eq =

57. Equivalent capacitance is used to simplify networks. Group task: What is the equivalent capacitance of the circuit below? Step 1: Find equivalent for C series at right Hand. Step 2: Find equivalent for C parallel left to right. Step 3: Find equivalent for series. Solution: parallel branch … series branch …  total:

58. Capacitor Networks 24.63 C1 = 6.9  F, C2= 4.6  F Reducing the furthest right leg (branch): C=( Combines parallel with nearest C2: C = Leaving a situation identical to what we have just worked out: So the overall C EQ =  F Charge on C1 and C2: Q C1 = Q C2 = For V ab = 420[V], V cd =? V ac = V bd = V cd =

59. Energy Storage in Capacitors V = Q/C W = 1/C int 0Q [q dq] = Q 2 /2C

60. Energy Storage in Capacitors 24.24: plate C 920 pF, charge on each plate 2.55  C a) a)V between plates: V = b)For constant charge, if d is doubled, what will V be? c) How much work to double d? If d is doubled, … Work equals amount of extra energy needed which is mJ

61. Other common geometries Spherical capacitor Spherical capacitor Need V ab for C, need E for V ab : Take Gaussian surface as sphere and find enclosed charge Cylindrical capacitor Cylindrical capacitor

62. Dielectrics Dielectric constant K K= C/C 0 For constant charge: Q=C 0 V 0 =CV And V = V 0 /K Dielectrics are never perfect insulators: material leaks

63. Induced charge: Polarization If V changes with K, E must decrease too: E = E 0 /K This can be visually understood considering that materials are made up of atoms and molecules:

64. Induced charge: Polarization – Molecular View Dielectric breakdown

65. Change with dielectric: E 0 =  0  E =  Empty space: K=1,  0

66. RC Circuits Charging a capacitor From now on instantaneous Quantities I and V in small fonts v ab = i R v bc = q/C Kirchhoff:   As q increases towards Q f, i decreases  0   integral: Take exponential of both sides:

67. Discharging a capacitor Characteristic time constant  = RC !

68. Ex 26.37 - C= 455 [pF] charged with 65.5 [nC] per plate C then connected to ‘V’ with R= 1.28 [M  ] a) a)i initial ? b) What is the time constant of RC? a) a)i = q/(RC) [C/(  F] =! [A] b)  = RC = Group Task: Find  charged fully after 1[hr]? Y/N

69. Ex 26.80/82 C= 2.36 [  F] uncharged, then connected in series to R= 4.26 [  ] and E =120 [V], r=0 a) a)Rate at which energy is dissipated at R b) Rate at which energy is stored in C c) Power output source a) P R = b) P C = c) P  = d) What is the answer to the questions after ‘a long time’?  all zero Group Task