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

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