Download Presentation
## Physics 1220/1320

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Electromagnetism –**part one: electrostatics Physics 1220/1320 Lecture Electricity, chapter 21-26**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**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**Coulomb’s Law**Force on a charge by other charges ~ ~ ~ Significant constants: e = 1.602176462(63) 10-19C i.e. even nC extremely good statistics (SI) 1/4pe0**Principle of superposition of forces:**If more than one charge exerts a force on a target charge, how do the forces combine? Find F1 Luckily, they add as vector sums! Consider charges q1, q2, and Q: F1 on Q acc. to Coulomb’s law 0.29N Component F1x of F1 in x: With cos a = -> F1x= Similarly F1y = What changes when F2(Q) is determined? What changes when q1 is negative?**Electric Fields**How does the force ‘migrate’ to be felt by the other charge? : Concept of fields**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**Electric field lines**For the visualization of electric fields, the concept of field lines is used.**Electric Dipoles**Net force on dipole by uniform E is zero. Product of charge and separation ”dipole moment” q d Torque**Group Task:**Find flux through each surface for q = 30^ and total flux through cube What changes for case b? n1: n2: n3: n4: n5,n6:**Gauss’s Law**Basic message: ‘What is in the box determines what comes out of the box.’ Or: No magic sources.**http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html**http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html http://www.falstad.com/vector3de/**Group Task**2q on inner 4q on outer shell http://www.falstad.com/vector3de/**Line charge:**F(E) = Opp. charged parallel plates EA = Infinite plane sheet of charge: 2EA =**Potential**Difference Potential difference: [V/m]**Calculating velocities from**potential differences Dust particle m= 5 10-9 [kg], charge q0 = 2nC Energy conservation: Ka+Ua = Kb+Ub**Moving charges: Electric Current**• Path of moving charges: circuit • Transporting energy • Current http://math.furman.edu/~dcs/java/rw.html Random walk does not mean ‘no progression’ Random motion fast: 106m/s Drift speed slow: 10-4m/s e- typically moves only few cm Positive current direction:= direction flow + charge**Current through A:= dQ/dt**charge through A per unit time Work done by E on moving charges heat (average vibrational energy increased i.e. temperature) Unit [A] ‘Ampere’ [A] = [C/s] Current and current density do not depend on sign of charge Replace q by /q/ Concentration of charges n [m-3] , all move with vd, in dt moves vddt, volume Avddt, number of particles in volume n Avddt What is charge that flows out of volume?**Resistivity and Resistance**Properties of material matter too: For metals, at T = const. J= nqvd ~ E Proportionality constant r is resistivityr = E/JOhm’s law Reciprocal of r is conductivity Unit r: [Wm] ‘Ohm’ = [(V/m) / (A/m2)] = [Vm/A]**Resistance**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 [W] • vs. R R =rL/A R = V/I V = R I I = V/R**Group Task**E, V, R of a wire Typical wire: copper, rCu = 1.72 x 10-8 Wm 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 = rJ = rI/A = 0.0210 V/m (a) 21 V/m (b) V = EL = 21 mV (a), 21 mV (b), 2.1 V (c) R = V/I = 2.1V/1A = 2.1 W**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) : Vab = E = IR**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, Vab = E – Ir and I = E/(R+r)**Circuit Diagrams**Voltage is always measured in parallel, amps in series**Energy and Power in Circuits**Rate of conversion to electric energy: EI, rate of dissipation I2r – difference = power output source When 2 sources are connected to simple loop: Larger emf delivers to smaller emf**Resistor networks**Careful: opposite to capacitor series/parallel rules!**Group Task:**Find Req and I and V across /through each resistor!**Group task:**Find I25 and I20**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 Req fails.**‘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 ‘**Group task: Find values of I1,I2,and I3!**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.ofno.unknowns) Let’s set R1=R3=R4=1W and R2=R5=2W**Capacitance**E ~ /Q/ Vab ~ /Q/ Double Q: Charge density, E, Vab double too But ratio Q/Vab is constant Capacitance is measure of ability of a capacitor to store energy! (because higher C = higher Q per Vab = higher energy Value of C depends on geometry (distance plates, size plates, and material property of plate material)**Plate Capacitors**E = s/e0 = Q/Ae0 For uniform field E and given plate distance d Vab = E d = 1/e0 (Qd)/A Units: [F] = [C2/Nm] = [C2/J] … typically micro or nano Farad**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.