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  _____________ ‘switched’ are in between and can be conveniently  _____________ 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 -19 C i.e. even nC extremely good statistics (SI) 1/4 pe 0

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 q 1 , q 2 , and Q:

Find F

1 F 1 on Q acc. to Coulomb’s law 0.29N

Component F 1x of F 1 in x: With cos a = Similarly F 1y = -> F 1x = What changes when F 2 (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

Gauss’s Law, Flux

Group Task: Find flux through each surface for

q

and total flux through cube = 30^

What changes for case b?

n 1 : n 2 : n 3 : n 4 : n 5 ,n 6 :

Gauss’s Law

Basic message: ‘What is in the box determines what comes out of the box.’ Or: No magic sources.

Important Applications of Gauss’s Law

http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html

http://www.falstad.com/vector3de/

2q on inner 4q on outer shell

Group Task

http://www.falstad.com/vector3de/

Line charge:

F

(E) =

Opp. charged parallel plates EA = Infinite plane sheet of charge: 2EA =

Electric Potential Energy

Electric Potential V units volt: [V] = [J/C]

Potential difference: [V/m]

Potential Difference

Calculating velocities from potential differences

Dust particle m= 5 10 -9 [kg], charge q 0 = 2nC

Energy conservation: K a +U a = K b +U b

Equipotential Surfaces

Potential Gradient

Moving charges: Electric Current

   Path of moving charges: circuit Transporting energy http://math.furman.edu/~dcs/java/rw.html

Random walk does not mean ‘no progression’ Current Random motion fast: 10 6 m/s Drift speed slow: e 10 -4 m/s 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/dt charge through A per unit time 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 v d , in dt moves v d dt, volume Av d dt, number of particles in volume n Av d dt What is charge that flows out of volume?

Resistivity and Resistance

Properties of material matter too: For metals, at T = const. J= nqv d Proportionality constant r ~ E is resistivity r = E/J

Ohm’s law

Reciprocal of r is conductivity Unit r : [ W m] ‘Ohm’ = [(V/m) / (A/m 2 )] = [Vm/A]

Types of resistivity

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 ] r vs. R R = r L/A R = V/I V = R I I = V/R

Group Task

E, V, R of a wire

Typical wire: copper,

r

Cu = 1.72 x 10 -8

W

m 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 =

r

J =

r

I/A = 0.0210 V/m (a) 21 V/m (b) V = EL = 21

m

V (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) : V ab = 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, V ab = 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: E I, rate of dissipation I 2 r – 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 R

eq

and I and V across /through each resistor!

Group task: Find I

25

and I

20

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 R eq fails.

‘Charging a battery’

‘ Loop rule: 3 loops to choose from Circuit with more than one loop!

Apply both rules.

Junction rule, point a: Similarly point b:

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.of no.unknowns) Let’s set R 1 =R 3 =R 4 =1 W and R 2 =R 5 =2 W

E ~ /Q/  V ab ~ /Q/ Double Q: Charge density, E, V ab double too But ratio Q/V ab is constant

Capacitance

Capacitance is measure of ability of a capacitor to store energy!

(because higher C = higher Q per V ab = higher energy Value of C depends on geometry (distance plates, size plates, and material property of plate material)

Plate Capacitors E =

s/e

0 = Q/A

e

0 For uniform field E and given plate distance d V ab = E d = 1/

e

0 (Qd)/A

Units: [F] = [C 2 /Nm] = [C 2 /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 right Hand. Step 2: Find equivalent for C parallel left to right.

Step 3: Find equivalent for series.

Capacitor Networks 24.63

C1 = 6.9

m

F, C2= 4.6

m

F 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: Q C1 = Q C2 =

Energy Storage in Capacitors

V = Q/C W =

Energy Storage in Capacitors

24.24: plate C 920 pF, charge on each plate 2.55 m C 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 capacitor Need V ab for C, need E for V Take Gaussian surface as sphere and find enclosed charge ab : Note:  Cylindrical capacitor Find E E r Find V = V l/2pe ab 0 1/r = Find C Q = What is C dep. On?

Dielectrics

Dielectric constant K K= C/C 0 For constant charge: Q=C 0 V 0 =CV And V = V 0 /K Dielectrics are never perfect insulators: material leaks

Induced charge: Polarization

If V changes with K, E must decrease too: E = E 0 /K This can be visually understood considering that materials are made up of atoms and molecules:

Induced charge: Polarization – Molecular View

Dielectric breakdown

Change with dielectric: E 0

s-s

i

s

i =

s/e

0 =

s 

E = (

s-s

i )/

e

0 =

e

0 E 0 /K

 s

i

= s -s

and = E 0 /K /K (1-1/K) E =

s/e Empty space: K=1, e=e 0

RC Circuits Charging a capacitor

From now on instantaneous Quantities I and V in

small fonts

v ab v bc = i R = q/C Kirchhoff: As q increases towards Q f , i decreases  0 Separate variables: Integrate, take exp.:

Characteristic time constant

t

= RC !

Discharging a capacitor

Ex 26.37 - C= 455 [pF] charged with 65.5 [nC] per plate C then connected to ‘V’ with R= 1.28 [M

W

] a) i initial ?

b) What is the time constant of RC?

a) i = q/(RC) = [C/(

W

F] =! [A] b)

t

= RC =

Group Task: Find

t charged fully after 1[hr]? Y/N

Ex 26.82 C= 2.36 [

m

F] 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

Group Task

c) Power output source a) P R = b) P C = dU/dt = c) P

t

=

E

I = d) What is the answer to the questions after ‘a long time’?



all zero