Download - Physics 1220/1320
Physics 1220/1320
Electromagnetism –part one:
electrostatics
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
The phenomenon of charge
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
Induction : Conductors and Insulators
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?
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) isdetermined?
What changes when q1is negative?
Find F1
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 distanceand 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 fieldlines is used.
Electric Dipoles
Net force on dipole by uniform E is zero.Product of charge and separation ”dipole moment” q dTorque
Gauss’s Law, Flux
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 comesout of the box.’ Or: No magic sources.
Important Applications of Gauss’s Law
http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.htmlhttp://www.falstad.com/vector3de/
Group Task
2q on inner4q on outershell
http://www.falstad.com/vector3de/
Line charge:F(E) =
Infinite plane sheet of charge:2EA =
Opp. chargedparallel platesEA =
Electric Potential Energy
Electric Potential V units volt: [V] = [J/C]
Potential difference:[V/m]
PotentialDifference
Calculating velocities from potential differences
Energy conservation: Ka+Ua = Kb+Ub
Dust particle m= 5 10-9 [kg], charge q0 = 2nC
Equipotential Surfaces
Potential Gradient
Moving charges: Electric Current
Path of moving charges: circuit
Transporting energy Current
http://math.furman.edu/~dcs/java/rw.htmlRandom walk does not mean ‘no progression’
Random motion fast: 106m/sDrift speed slow: 10-4m/se- typically moves only few cm
Positive current direction:= direction flow + charge
Work done by E on moving charges heat (average vibrational energy increased i.e.
temperature)Current through A:= dQ/dtcharge through A per unit time
Unit [A] ‘Ampere’[A] = [C/s]
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?
Current and current density donot depend on sign of charge Replace q by /q/
Resistivity and ResistanceProperties of material matter too:For metals, at T = const. J= nqvd ~ EProportionality constant r is resistivity r = E/J
Ohm’s law
Reciprocal of r is conductivityUnit r: [Wm] ‘Ohm’ = [(V/m) / (A/m2)] = [Vm/A]
Types of resistivity
Ask for total current in and potential at ends of conductor:Relate value of current toPotential difference between ends.
For uniform J,E
‘resistance’ R = V/I [W]r vs. R
R =rL/AR = V/I V = R I I = V/R
Resistance
E, V, R of a wireTypical 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
Group Task
Electromotive ForceSteady currents require circuits: closed loops of conducting materialotherwise current dies down after short time
Charges which do a full loopmust have unchanged potentialenergy
Resistance always reduces U
A part of circuit is needed whichincreases U again
This is done by the emf.Note that it is NOT a force buta 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!
Combining Measuring Instruments
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 networksIs to look at them as loops and junctions:
This method works evenwhen our rules to reducea circuit to its Req
fails.
‘Charging a battery’Circuit with morethan one loop!
Apply both rules.
Junction rule,point a:
Similarly point b:
Loop rule: 3 loopsto choose from
‘
5 currentsUse junction ruleat a, b and c3 unknown currentsNeed 3 eqn
Loop rule to 3 loops:cad-source cbd-source cab(3 bec.of no.unknowns)
Let’s set R1=R3=R4=1Wand R2=R5=2W
Group task: Find values of I1,I2,and I3!
Capacitance
E ~ /Q/ Vab ~ /Q/Double Q:Charge density, E, Vab
double tooBut ratio Q/Vab is constant
Capacitance is measure of ability of a capacitor to store energy!(because higher C = higher Q per Vab = higher energyValue of C depends on geometry (distance plates, size plates, and materialproperty of plate material)
Plate Capacitors
E = s/e0
= Q/Ae0
For uniform field E and given plate distance dVab = 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.
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 rightHand.
Step 2:Find equivalent for C parallelleft to right.
Step 3:Find equivalent for series.
Capacitor Networks
24.63
C1 = 6.9 mF, C2= 4.6mF
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:QC1 =
QC2 =
Energy Storage in Capacitors
V = Q/C
W =
Energy Storage in Capacitors
24.24: plate C 920 pF, charge on each plate 2.55 mC
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
Other common geometries Spherical capacitorNeed Vab for C, need E for Vab:Take Gaussian surface as
sphereand find enclosed charge
Note:
Cylindrical capacitorFind E Er = l/2pe0 1/r
Find V Vab = Find C Q = What is C dep. On?
Dielectrics
Dielectric constant K
K= C/C0
For constant charge:Q=C0V0=CV
And V = V0/K
Dielectrics are neverperfect insulators: materialleaks
Induced charge: Polarization
If V changes with K, E must decrease too: E = E0/KThis can be visually understood considering that materialsare made up of atoms and molecules:
Induced charge: Polarization – Molecular View
Dielectric breakdown
Change with dielectric:
E0 = s/e0 E = (s-si)/e0 and = E0/Ks-si = e0E0/K si = s -s/K si = s (1-1/K)
E = s/e
Empty space: K=1, e=e0
RC CircuitsCharging a capacitor
From now on instantaneous Quantities I and Vin small fontsvab = i Rvbc = q/CKirchhoff:
As q increases towards Qf, i decreases 0
Separate variables:
Integrate, take exp.:
Discharging a capacitor
Characteristic time constant t = RC !
Ex 26.37 - C= 455 [pF] charged with 65.5 [nC] per plateC then connected to ‘V’ with R= 1.28 [MW]
a) iinitial?b) What is the time constant of RC?
b) i = q/(RC) =
[C/(WF] =! [A]
b) t = RC =
Group Task: Find tcharged fully after 1[hr]? Y/N
Ex 26.82 C= 2.36 [mF] uncharged, then connected in series to R= 4.26 [W] and E=120 [V], r=0
a) Rate at which energy is dissipated at R
b) Rate at which energy is stored in C increases
c) Power output source
a) PR =
b) PC= dU/dt =
c) Pt = E I =
d) What is the answer to the questions after ‘a long time’? all zero
Group Task