ELECTRICITY& MAGNETISM

NOTES

PHYSICS B2B

BAKERSFIELD COLLEGE

Rick Darke (Instructor)

CHARGE

Electric charge is a fundamental prop-erty associated with 2 of the 3 subatomicparticles making up most matter. 'Whatcharge is' is unknown. 'How charge be-haves' is well-known.

e- electrons (negative charge)

p protons (positive charge)

n neutrons (no net charge)

e-

pp

nn

e-

helium-4 atom (4He)

CHARGE

Charge is a scalar quantity that can be positive or negative. The standardunit of charge (S.I.) is the coulomb (C). Traditionally the symbols q and Qare used to denote charge. The smallest common unit of charge is thecharge on an electron or proton:

e- qelectron = -1.6 x 10-19 C = -e

p qproton = +1.6 x 10-19 C = +e

n qneutron = 0 C

The symbol e is used to represent the physical quantity fundamental charge(1.6 x 10-19 C).

QUARKS

In 1963 Murray Gell-Mann and Georg Zweig proposed that hadrons (pro-tons, neutrons, and other particles) are composed of 2 or 3 quarks. In 1979Sheldon Glashow and Abdus Salam won the Nobel Prize in physics for workleading to the theory of how quarks interact with each other.

Gell-Mann Zweig Glashow Salam

CHARGE OF QUARKS

proton neutron

u u

u d d d

u = up quark d = down quarkqu = +(2/3)e qd = -(1/3)e

CHARGE MOBILITY

Most materials can be classified as conductors, insulators, or semiconduc-tors, depending upon how easy it is for charge to move through the material.

conductor - a sub-stance that allowscharge to move throughit with relative ease.examples: aluminum,gold, copper (shown)

insulator - a substancethrough which chargemove very reluctantly, ornot at all.examples: polystyrene,rubber, glass (shown)

semiconductor - a sub-stance with electricalproperties intermediateto those of conductorsand insulators.examples: germanium,silicon (shown)

CONDUCTIVITIES OF MATERIALS

class examples conductivity (ΩΩΩΩΩ.m)-1

silver (Ag) 6.3 x 107

copper (Cu) 5.9 x 107conductors

aluminum (Al) 3.6 x 107

carbon (C) 2.9 x 104

silicon (Si) .016semi-

germanium (Ge) 2.17conductors

GaAs 1.0 x 10-6

glass ~10-12

nylon 6,6 ~10-12insulatorsquartz ~10-18

note: Most conductors become more conductive as their temperature in-creases, and most semiconductors become less conductive as their tem-perature increases.

LAW OF CHARGES

Charges of the same sign are mutually repulsive, and charges of oppositesign are mutually attractive.

q1 F21 F12 q2+ _

r

F21 q1 q2 F12+ +

r

F21 q1 q2 F12_ _

r

note: The notation F21 is taken to mean "the force due to charge q2 andacting on charge q1".

COULOMB'S LAW: QUALITATIVE

The electric force between two point charges q1 and q2 is: (1) directedalong the line joining the charges; (2) directly proportional to the product ofthe magnitudes of the charges; (3) inversely proportional to the distancebetween the charges squared; and (4) attractive if the charges have oppo-site signs and repulsive if they have like signs (Law of Charges).

F21 F12q1 q2r

note: Newton's Third Law insures that F12 = - F21 and that the two forces liealong the same line of action.

COULOMB'S LAW: QUANTITATIVE

The electric force due to point charge q1 acting on point charge q2 is givenby:

F12q1 q2r

F12 = keq1q2n12/r2

note: n12 is a unit vector in the direction of r2 - r1 and ke is the Coulombconstant 9.0 x 109 Nm2/C2.

USING COULOMB'S LAW

Effective use of Coulomb's Law to compute the electric force (magnitudeand direction) due to point charge q1 acting on point charge q2:

F12q1 q2r

(1) Compute the magnitude of the force using:

F12 = ke|q1||q2|/r2

(2) Use the Law of Charges to determine the direction of the force.

(3) Express the combined magnitude and direction as the answer.

PRACTICE PROBLEM: 1-D COULOMB'S LAW

Find the electric force on the -2.0 μC charge below due to the +4.0 μCcharge to the left of it.

+4.0 μC -2.0 μC3.0 cm

PRACTICE PROBLEM: 2-D COULOMB'S LAW

Find the electric force on the +1.0 μC charge located in the diagram belowdue to all other charges.

y

-3.0 μC

3.0 cm

+1.0 μC +5.0 μC

4.0 cm x

H20 MOLECULE CHARGE DISTRIBUTION

+ + + +

+ +

104oH

H

O

_ _ _ _

_

_

PRACTICE PROBLEM: 2-D ELECTRIC FIELD

Find the electric field at point P on the y-axis in the diagram below due tothe two charges.

y

P

6.0 cm

+1.0 μC -5.0 μC

8.0 cm x

PRACTICE PROBLEM: 1-D ELECTRIC FIELD

Find all positions along the x-axis (other than infinity) where the electric fieldwould be zero..

+4.0 mC -2.0 mC

xx = 0.0 m x = +1.0 m

ELECTRIC MONOPOLE FIELD

Electric field lines in the vicinity of a singlepoint charge (electric monopole). Thecharge could be positive or negative, aseither would show the same radial pat-tern. The dark lines in the photo are smallpieces of thread suspended in oil, whichalign with the electric field produced bythe single charged conductor at the cen-ter.

ELECTRIC DIPOLE FIELD

Electric field lines in the vicinity of a pointcharge distribution where the charges havethe same magnitude, but opposite sign(electric dipole). The dark lines in thephoto are small pieces of thread sus-pended in oil, which align with the electricfield produced by the two charges.

ELECTRIC FIELD

Electric field lines in the vicinity of a pointcharge distribution where the charges havethe same magnitude and sign. The darklines in the photo are small pieces ofthread suspended in oil, which align withthe electric field produced by the twocharges.

ELECTRIC FIELD: CHARGED CONDUCTORS

Electric field pattern in the vicinity of twooppositely charged conductors. Note thatall field lines are perpendicular to the con-ductors at their surfaces. The dark linesin the photo are small pieces of threadsuspended in oil, which align with the elec-tric field produced by the two chargedconductors.

CHARGE ON AN ISOLATED CONDUCTOR

Any excess charge on an isolated con-ductor resides entirely on its surface.

The electric field is 0 everywhere in-side the conductor.

The electric field just outside acharged conductor is perpendicular to itssurface (at the surface).

The surface charge density is highestat sharp points (small radius of curvatureof surface) on the surface.

PRACTICE PROBLEM: KINEMATICS

A proton is released from rest at x = 0.0 cm in a region of uniform electricfield. The magnitude of the electric field is 1500 N/C, and it is directed in thepositive x-direction. Calculate the following quantities: (a) the force acting onthe proton in the electric field; (b) the velocity of the proton when it reachesthe position x = 5.0 cm; (c) the work done on the proton by the electric fieldfor this displacement; (d) the change in the electric potential energy asso-ciated with the proton for this displacement. The mass of a proton is 1.67x 10-27 kg.

USE OF ke and εεεεεo

The Coulomb constant ke (8.99 x 109 N.m2/C2) has been used in someformulae in physics, while the permitivity of free space εεεεεo (8.85 x 10-12 C2/N.m2) has been used in others. The difference in usage is historical prefer-ence, where the two constants are related by:

εo = 1/(4πke)

examples:

ke is usually used in the relations: F12 = ke|q1||q2|/r2

and V = keq/r

εεεεεo is usually used in the relations: C = kAεo/d

and E = σ/εo

PRACTICE PROBLEM: CHARGED PLATES

Two identical parallel rectangular plates with surface area .05 m2 are placeda distance 1.0 cm apart. Charges of +5.0 nC and -5.0 nC are uniformlydistributed over the surfaces of the plates. Calculate: (a) the magnitude ofthe surface charge density on the plates; and (b) the magnitude of theelectric field that this charge distribution creates between the plates.

PARALLEL-PLATE CAPACITOR

A parallel-plate capacitor consists of two par-allel conducting plates with identical surfacearea (A) separated by a distance (d). Theplates are charged equally and oppositely (+qand -q) by applying an electric potential differ-ence (V) across the them. This creates uni-form surface charge densities (+σ and -σ) onthe two plates, producing an electric field (E)between the plates that is uniform betweenthe plates and practically zero outside. Thecapacitor stores energy (U) in the process.

useful relations:

surface charge density: σ = q/Aelectric field: E = σ/εo = V/dcapacitance of capacitor: C = q/V = Aεo/denergy stored in capacitor: U = (1/2)CV2 = (1/2)(1/C)q2

E

+q

-q

A

+ _

V

++

+++

+++

+++

++++

+++

+++

+

PRACTICE PROBLEM: CAPACITOR

A parallel-plate capacitor has an area of 2.0 x 10-4 m2 and a plate separationof 1.0 mm. The capacitor is connected to a 3.0-volt battery. Find the follow-ing quantities: (a) the capacitance of the capacitor; (b) the charge on theplates of the capacitor; (c) the charge density on the plates of the capacitor;(d) the magnitude of the electric field between the plates of the capacitor;and (e) the energy stored in the capacitor.

PRACTICE PROBLEM: CAPACITANCE

A 24-volt source is connected to the parallel combination of capacitorsshown below. Find the following quantities: (a) the charge on each capacitor;(b) the voltage drop across each capacitor; (c) the energy stored in eachcapacitor; and (d) the equivalent capacitance of the combination.

24 V+

.003 F .006 F .012 F_

PRACTICE PROBLEM: CAPACITANCE

A 9-volt source is connected to the parallel combination of capacitors shownbelow. Find the following quantities: (a) the charge on each capacitor; (b) thevoltage drop across each capacitor; (c) the energy stored in each capacitor;and (d) the equivalent capacitance of the combination.

.08 F9 V

+.08 F

_

.12 F

PRACTICE PROBLEM: CAPACITANCE

A 12-volt source is connected to the parallel combination of capacitorsshown below. Find the following quantities: (a) the charge on each capacitor;(b) the voltage drop across each capacitor; (c) the energy stored in eachcapacitor; and (d) the equivalent capacitance of the combination.

.009 F

12 V+

.006 F .012 F_

PRACTICE PROBLEM: CAPACITANCE

Calculate the equivalent capacitance for the combination of capacitors shownbelow with respect to terminals a-b.

4 mF

1 mF

a6 mF

b3 mF

2 mF

8 mF

TABLE OF DIELECTRIC MATERIALS

material κκκκκ (@ 20°C) Emax (V/m)

vacuum 1.0000 ---------dry air 1.0006 3 x 106

teflon 2.1 60 x 106

polystyrene 2.6 24 x 106

paper 3.7 16 x 106

pyrex 5.6 14 x 106

neoprene 6.7 12 x 106

mica 7 150 x 106

water 80 ---------SrTiO3 233 15 x 106

EXERCISE IN CAPACITOR ENERGETICS

ACTION 1: Start with a 2.0-F air-spaced capacitor with uncharged plates(no voltage dropped across them). Use a variable voltage source to drop 1.0volt across the plates of the capacitor.

ACTION 1

C = 2.0 FV = 0.0 Vq = 0.0 CU = 0.0 J

C =V = 1.0 Vq =U =

+ _

V

+ _

V

EXERCISE IN CAPACITOR ENERGETICS

ACTION 2: Insert a slab of dielectric material with a dielectric constant of3.0 between the plates of the capacitor. Do not disconnect the voltagesource or change the voltage drop across the plates.

ACTION 2

C = 2.0 FV = 1.0 Vq = 2.0 CU = 1.0 J

C =V =q =U =

+ _

V

+ _

V

κ =3.0

EXERCISE IN CAPACITOR ENERGETICS

ACTION 3: Use the voltage source to increase the voltage drop across theplates of the capacitor to 5.0 V. Do not change anything else.

ACTION 3

C = 6.0 FV = 1.0 Vq = 6.0 CU = 3.0 J

C =V = 5.0 Vq =U =

+ _

V

+ _

V

κ =3.0

κ =3.0

EXERCISE IN CAPACITOR ENERGETICS

ACTION 4: Disconnect the leads of the voltage source from the plates ofthe capacitor and remove the source. Do not change anything else.

ACTION 4

C = 6.0 FV = 5.0 Vq = 30 CU = 75 J

C =V =q =U =

+ _

V

κ =3.0

κ =3.0

EXERCISE IN CAPACITOR ENERGETICS

ACTION 5: Remove the slab of dielectric material from between the platesof the capacitor. Do not change anything else.

ACTION 5

C = 6.0 FV = 5.0 Vq = 30 CU = 75 J

C =V =q =U =

κ =3.0

RESISTIVITIES OF MATERIALS

material ρρρρρo @ 20 °C (ΩΩΩΩΩ.m) ααααα (°C-1)

silver 1.6 x 10-8 .0038copper 1.7 x 10-8 .0039aluminum 2.8 x 10-8 .0039tungsten 5.6 x 10-8 .0045nichrome 1.5 x 10-6 .0004carbon 3.5 x 10-5 -.0005silicon 640 -.0750glass ~1012

quartz ~1018

note: To obtain the resistivity (ρ) of a material in the table above at aspecific celsius temperature (T), use:

ρ = ρo(1 + α(T - To), where To = 20 °C

CYLINDRICAL RESISTORS

A

R = IVE

R = ρL/AP = IV = I2R = V2/RE = V/L L

+ _

VL = length of cylindrical resistor (m)A = cross-sectional area of resistor (m2)ρ = resistivity of conducting material (Ω.m)T = temperature of resistive material (°C)R = resistance of resistor (Ω)V = voltage dropped across resistor (V)I = current through resistor (A)E = uniform electric field inside resistor (V/m)P = power dissipated by resistor (W)

PRACTICE PROBLEM: RESISTANCE

A nichrome wire with a diameter of 0.50mm and length of 5.0 meters is used asa heating element. In use the elementhas a temperature of 1500 °C when 120volts is dropped across the ends. Find:(a) the resistivity of nichrome at this tem-perature; (b) the resistance of the ele-ment; (c) the current through the element;and (d) the power dissipated by the el-ement.

PRACTICE PROBLEM: RESISTANCE

Suppose that a platinum thermistor isused to interpret the temperature of theenvironment in which it is immersed. Ithas a resistance of 50.0 Ω at 20°C. Whenit is immersed in a vessel containing melt-ing indium the resistance of the thermistoris found to increase to 130.0 W. Use thisinformation to find the melting point ofindium.

PRACTICE PROBLEM: RESISTANCE

An immersion heater whose resistance inoperation is 24 ohms is placed in a mugwith 400 grams of water initially at a tem-perature of 20 °C. If the heater is con-nected to a 120-volt source, how longwill it be until the water comes to a boil?Assume the mug is a good thermal insu-lator and does not absorb heat duringthe process.

KIRCHHOFF'S CIRCUIT LAWS

Kirchhoff's Current Law (KCL): The sum of all currents directed into ajunction is equal to the sum of all currents directed out of the junction: Σ Iin= Σ Iout

15 ΩJ

18 Ω K

20 Ω

2.0 A 1.0 A 0.6 A 60 V

+1.0 0.4

_ A 30 Ω A 30 Ω

note: At junction J, Σ Iin = 2.0 A and Σ Iout = 1.0 A + 1.0 A = 2.0 A, andat junction K, Σ Iin = 1.0 A and Σ Iout = 0.4 A + 0.6 A = 1.0 A

KIRCHHOFF'S CIRCUIT LAWS

Kirchhoff's Voltage Law (KVL): The sum of all voltage gains around anyloop in a circuit is equal to zero: Σ ΔVloop = 0

15 ΩJ

18 Ω K

20 Ω

2.0 A 1.0 A 0.6 A 60 V

+1.0 0.4

_ A 30 Ω A 30 Ω

note: Around the outermost mesh, Σ ΔVloop = 60 V - (15 Ω)(2.0 A) -(18 Ω)(1.0 A) - (20 Ω)(0.6 A) = 0 V