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

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Presentation on theme: "Physics 1220/1320 Electromagnetism – part one: electrostatics Lecture Electricity, chapter 21-26."— Presentation transcript:

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 with two kinds of active matter: electrons and protons and one kind of neutral matter: neutrons Two important laws: Conservation & quantization of charge from experiment: Like charges repel, unlike charges attract

3 The phenomenon of charge The phenomenon of charge

4 Electric Properties of Matter (I) Materials which conduct electricity well are called ___________ Materials which prohibited the flow of electricity are called _____________ ‘Earth’ or ‘ground’ is a conductor with an infinite reservoir of charge _____________ are in between and can be conveniently ‘switched’ _____________ are ideal conductors without losses

5 Induction : Conductors and Insulators Induction : Conductors and Insulators

6 Coulomb’s Law Force on a charge by other charges ~ ~ Significant constants: e = (63) C i.e. even nC extremely good statistics (SI) 1/4  0

7 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 0.29N Component F 1x of F 1 in x: With cos  F 1x = Similarly F 1y = What changes when F 2 (Q) is determined? What changes when q1 is negative? Find F 1

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


10 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

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

12 Electric Dipoles Net force on dipole by uniform E is zero. Product of charge and separation ”dipole moment” q d Torque

13 Gauss’s Law, Flux


15 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 :

16 Gauss’s Law Basic message: ‘What is in the box determines what comes out of the box.’ Or: No magic sources.

17 Important Applications of Gauss’s Law


19 Group Task 2q on inner 4q on outer shell

20 Line charge:  (E) = Infinite plane sheet of charge: 2EA = Opp. charged parallel plates EA =

21 Electric Potential Energy


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

24 Potential difference: [V/m] Potential Difference

25 Calculating velocities from potential differences Energy conservation: K a +U a = K b +U b Dust particle m= [kg], charge q 0 = 2nC


27 Equipotential Surfaces


29 Potential Gradient

30 Moving charges: Electric Current Path of moving charges: circuit Path of moving charges: circuit Transporting energy Transporting energy Current Current Random walk does not mean ‘no progression’ Random motion fast: 10 6 m/s Drift speed slow: m/s e - typically moves only few cm Positive current direction:= direction flow + charge

31 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/

32 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]

33 Types of resistivity

34 Ask for total current in and potential at ends of conductor: Relate value of current to Potential difference between ends. For uniform J,E ‘resistance’ R = V/I [  ]   vs. R R =  L/A R = V/I V = R I I = V/R Resistance


36 E, V, R of a wire Typical wire: copper,  Cu = 1.72 x  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) 21 V/m (b) V = EL = 21  V (a), 21 mV (b), 2.1 V (c) R = V/I = 2.1V/1A = 2.1  Group Task

37 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

38 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)

39 Circuit Diagrams Voltage is always measured in parallel, amps in series

40 Energy and Power in Circuits Rate of conversion to electric energy: E I, rate of dissipation I 2 r – difference = power output source When 2 sources are connected to simple loop: Larger emf delivers to smaller emf

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

42 Combining Measuring Instruments

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

44 Group task: Find I 25 and I 20

45 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.

46 ‘Charging a battery’ Circuit with more than one loop! Apply both rules. Junction rule, point a: Similarly point b: Loop rule: 3 loops to choose from ‘

47 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!

48 Capacitance E ~ /Q/  V ab ~ /Q/ Double Q: Charge density, E, V ab double too 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)

49 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

50 Capacitor Networks In a series connection, the magnitude of charge on all plates is the same. The potential differences are not the same. In a parallel connection, the potential difference for all individual capacitors is the same. The charges are not the same.

51 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.

52 Capacitor Networks 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: Charge on C1 and C2: Q C1 = Q C2 =

53 Energy Storage in Capacitors V = Q/C W =

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

55 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 Note: Cylindrical capacitor Cylindrical capacitor Find E E r =  0 1/r Find V V ab = Find C Q = What is C dep. On?

56 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

57 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:

58 Induced charge: Polarization – Molecular View Dielectric breakdown

59 Change with dielectric: E 0 =  0  E = (  i )/  0 and = E 0 /K  i =  0 E 0 /K   i  /K  i =  (1-1/K) E =  Empty space: K=1,  0

60 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 Separate variables: Integrate, take exp.:

61 Discharging a capacitor Characteristic time constant  = RC !

62 Ex 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

63 Ex 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 increases c) Power output source a) P R = b) P C = dU/dt = c) P  = E I = d) What is the answer to the questions after ‘a long time’?  all zero Group Task

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