Physics 1220/1320 Electromagnetism – part one: electrostatics Lecture Electricity, chapter 21-26
Dec 31, 2015
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
And with two kinds of active matter:electrons and protons
And one kind of neutral matter: neutrons
The phenomenon of charge
ProblemInvisibility
Common problem inphysics: have to believe ininvisible stuff and find waysto demonstrate its existence.
Danger of sth invisible
If we rub electrons onto the plastic, is it feasible to say that we rubprotons on it in the second experiment?
No! If we move protons, we move electrons with them. But what if in thesecond experiment we still moved electrons – in the other direction?
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
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 intothe nucleus?Why does the nucleus not fly apart?Why does the electron not fly apart?
Consequence of QM uncertaintyrelation
More forces, total of fourShort ranged – limit for nucleus sizeUranium almost ready to fly apart
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
Induction : Conductors and Insulators
Induction -
Appears visiblyin conductors
a) Are charges present?
c) Why do like chargescollect at oppositeside?
b) Why are there notmore ‘-’ charges?
d) Why does the metalsphere not stay charged forever?
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) 10-19C i.e. even nC extremely good statistics(SI) 1/4pe0
Modern Physics: why value? how constant?
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
Component F1x of F1 in x:
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
Group task:
Find force ofall combinationsof distances andcharge arrangements
Group task:
Find fields forall combinationsof distances andcharge arrangementsat all charge positions.
Direction E at black point equidistant from chargesis 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 -
Electric field lines
For the visualization of electric fields, the concept of fieldlines is used.
Electric Dipoles
H2O :
O2- (ion)
H1+ H1+
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:
Important Applications of Gauss’s Law
Group Task
2q on inner4q on outershell
http://www.falstad.com/vector3de/
http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.htmlhttp://www.falstad.com/vector3de/
Group TaskFor charges1 = +q, 2 = -q, 3= +2q
Find the flux throughthe surfaces S1-S5
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
Outside sphere: V = k q/r
Surface sphere: V = k q/R
Inside sphere:
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 Resistance
Properties of material matter too:For metals, at T = const. J= nqvd ~ E
Proportionality constant r is resistivity r = E/JOhm’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 I = JA and V =EL with Ohm’s law E=rJ
V/L = rI/AConst. r I ~ V ‘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 = V/m (a) V/m (b)
V = EL = mV (a), mV (b), V (c)
R = V/I = V/ A = W
Group Task
Resistance of hollow cylinder length L, inner and
outer radii a and b
Current flow radially! Not lengthwise!
Cross section is not constant:2paL to 2pbL find resistance ofthin shell, then integrate
Area shell: 2prL with current path drand resistance dR between surfaces
dR= rdr/(2prL) dV = I dR to overcome
And Vtot= SdV
R = int(dR) = /(2r pL) intab(dr/r) = /(2r pL) ln(b/a)
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
(Ideal wires and am-meters haveZero resistance)
No I through voltmeter (infinite R) …i.e. no current at all
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
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:
I = A
[Similarly point b: ]
Loop rule: 3 loopsto choose from
‘1’: decide directionin loop – clockwise(sets signs!)
r = W
Find E from loop 2:
Lets do the loop counterclockwise: E = V
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!
I1 = A, I2 = A, I3 =
Capacitance
E ~ /Q/ Vab ~ /Q/Double Q:
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 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. Or using concept of equivalent capacitance
1/Ceq =
In a parallel connection, the potential difference for all individual capacitors is the same. The charges are not the same.
Or Ceq =
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.
Solution: parallel branch … series branch …
total:
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:
So the overall CEQ = mFCharge on C1 and C2:QC1 =
QC2 = For Vab = 420[V], Vcd=?Vac = Vbd = Vcd=
Energy Storage in Capacitors
V = Q/C
W = 1/C int0Q [q dq] = Q2/2C
Energy Storage in Capacitors
24.24: plate C 920 pF, charge on each plate 2.55 mC
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
Other common geometries
Spherical capacitorNeed Vab for C, need E for Vab:Take Gaussian surface as
sphereand find enclosed charge
Cylindrical capacitor
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 e
Empty space: K=1, =e e0
RC Circuits
Charging a capacitor
From now on instantaneous Quantities I and Vin small fontsvab = i Rvbc = q/CKirchhoff:
As q increases towards Qf, i decreases 0
integral: Take exponential of both sides:
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 26.80/82 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
c) Power output source
a) PR =
b) PC=
c) Pt =
d) What is the answer to the questions after ‘a long time’? all zero
Group Task