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 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

‘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 = /2l pe0 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/K-s 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