WIXLIB

Chapter 19Electric Charges, Forces, and FieldsOutline19-1Electric Charge19-2Insulators and Conductors19-3Coulomb’s Law (and net vector force)19-4The Electric Field19-5Electric Field Lines19-6Shield and Charging by Induction19-7Electric Flux and Gauss’s Law

19-1 Electric ChargeObjectives: What is electric charge How it is created Electric charge propertiesThe effect of electric charge have been known fora long time.

Figure 19-1Charging an Amber Rod(a). Uncharged amberexert no force on papers.(b). Rod amber is rubbedagainst a piece of fur.(c). Amber becomecharged and then attractsthe papers.

Figure 19-2Likes Repel, Opposites AttractAmber (-1) glass ( )Amber (-) Amber (-): Repel: Attract

Conclusions:There are two type of charges (in amber andglass), which is illustrated in Fig. 19-2 Positive charge ( )Negative charge (-)Properties The “like” or same charges repel. The opposite or different charges attract.

Object with zero net charge are said to beelectrically neutral.A good example of neutral object is the atom,as shown in Fig. 19-3.

Figure 19-3The Structure of an Atom:proton and neutron.

All electrons have the same electric charge.Electron is negative, and is defined with amagnitude e, giving by:Magnitude of an Electron’s Charge, ee 1.60 10 19 C(19 1)SI unit: coulomb, CThe electron has a mass:M e 9.11 10 31 kg(19 2)

In contrast, the charge on proton (oneconstituent of the nucleus) is positive ( e).The mass of proton ism p 1.673 10 27 kg(19 3)The other constituent of the nucleus is theneutron, which has zero charge. Its mass ismn 1.675 10 27kg(19 4)

Charge Separation /TransferFigure 19-4Charge Transfer(a) Initially, an amber rod and apiece of fur is electrically neutral:no more negative/positivecharge!(b) Charges is transferred fromone to the other.(c) At the end, the fur and theamber have charges of equalmagnitude, but opposite sign.

Conservation of Electric ChargeThe total electric charge of the universe isconstant:No physical process can increase ordecrease the total amount of electriccharge.

Some conceptsDue to the movement of electrons, chargeis transferred from one object to another.Positive Ion: The atom that loses anelectron is said to be a positive ion;Negative Ion: The atom that receives anextra electron is said to be a negative ion;

19-2 Insulators and Conductors Insulator: Charges cannot move about freely on it.Examples: amber, glass, wood, rubber. Conductor: Charges can move freely on it.Examples: different metals and metal wires. Semiconductor: Materials have properties of electricconduction between insulator and conductor.And semiconductors have wide application in electronics!

Figure 19-6Charging a Conductor(a). An uncharged metal sphereis touched by a charged rod, andsome charge is transferred to it.(b). Charges move freely on aconductor, and spread outuniformly because of the samecharge repel.

19-3 Coulomb’s LawCoulomb’s law for the magnitude of the electrostatic forcebetween two point chargesq1 q2F k,2rwith k 8.99 109 N .m 2 / C 2SI unit: newton, N(19 5)

Figure 19-7Forces Between Point Charges

Note: the mass of proton is 2000 times that of the electron !

Note: the mass of proton is 2000 times of the electron !

An Example:Calculate the magnitude of electric force between a electronand a proton at a distance of r 5.29 x 10 -11 m.According to Coulomb’s lawq1 q2F kr2 (8.99 109 N .m 2 / C 2 ) 8.22 10 8 N 1.60 10 9 C 1.60 10 9 C(5.29 10 11 m) 2

An Example 19-9:The attraction electrostatic force between the point charges 8.44c10-6 C and Q has a magnitude of 0.975 N when theseparation between the charges is 1.31 m. Find the sign andmagnitude of the charge Q.

Negative charge andThe magnitude of the charge Q isQ 2(1.31 m ) ( 0.975 N )2r F 52.2110C 922 6kq(8.99 10 N m / C )(8.44 10 C )

Example 19-1: The Bohr OrbitIn the Borh’s Hydrogen model, the electron is imagined tomove in a circular orbit about a stationary proton. The forceresponsible for the electron circular motion is the electric forcebetween the electron and the proton.Given that the radius of the electron’s orbit is 5.29x10-11m, andits mass is me 9.11x10-31km. Find the electron’s speed.

Solution1. Set the coulomb force between the electron andproton equal to the centripetal force required for theelectro’s circular orbit:q1 q 2 m e a cpk2re2v2k 2 merr2. Solve for the speed of the electron, v:v ekm er3. Substitute numerical data:v (1 . 6 0 1 0 1 9 C )8 . 9 9 1 0 9 N .m 2 / C 2(9 .1 1 1 0 31 k g )(5 .2 9 1 0 11 m ) 2 .1 9 1 0 6 m / s

SuperpositionThe electric force, like all forces, is avector. Hence, a charge experiencesforces due to two or more charges isthe vector sum of all the forces.For examples, in Fig. 19-8, the total forceon charge 1 is the vector sum of the forcesdue to charges 2, 3 ,4

Vector sum/ Net force:JG JGJGJGF 1 F 12 F 13 F 14Figure 19-8Superposition of Forces

Example 19-15: Net ForceFor three charges, given q 12 uC and d 16 cm.(a) Find the direction and magnitude of the net electrostatic force exertedon the point charge q2 in the following figure. (b) How will your answerchange if the distance d were tripled?Note: 1 uC 10-6 C

Solution(a) Write Coulomb’s law using vector notation:GGGq1q3q1q2F2 F21 F23 k 2 xˆ k 2 xˆddSubstitute the charge magnitudes given in the figureGkkF2 2 [ q1q2 q1q3 ] xˆ 2 q ( 2.0q ) ( 2.0q )( 3.0q ) xˆdd 2kq4.0] xˆ [ 4.0]2 [d(8.99 10 N m92/ C2 )(12 10 6 C )( 0.16 m )22xˆ ( 200 N ) xˆThe net electrostatic force on q2 is 200 N 0.20 kN toward q3(b) If the distance d were tripled, the magnitude would be cut toa ninth and the direction would be unchanged.

Example 19-21: Superposition(a) Find the direction and magnitude of the net force exerted onthe point charge q3 in the figure. Let q 2.4 uC and d 33cm.(b) How would your answer change if the distance d weredouble?Picture the problem

Solution1. (a) FindGk q2 q3k ( 2.0q )( 3.0q )6.0kq 2F2 xˆ xˆ xˆ222dddG2. Find F4Gk q3 q4k ( 3.0q )( 4.0q )12kq 2F4 yˆ yˆ yˆ222dddG3. Find F1G k q1 q3k ( q )( 3.0q ) xˆyˆ ˆˆF1 cos 45 x sin 45 y ) 2 (222 2d2d()3.0 2kq 2 ( xˆ yˆ )24d()

4. Find the vector sum of the three forces:G 6.0kq 2 3.0 2kq 2 12kq 2 3.0 2kq 2 Fnet xˆ 2 224d4d 2 d d3.0kq 2 2.0 2 4 xˆ 4 2 4 yˆ 2 d( GFnet) ()3.0 ( 8.99 109 N m 2 / C 2 )( 2.4 10 6 C )( 0.33 m ) ( 2.3 N ) xˆ ( 5.2 N ) yˆ yˆ 22() (GF5. Find the direction of net from the x axis:θ tan 1Fnet, yFnet, x tan 15.2 N 66 2.3 NG6. Find the magnitude of FnetFnet Fnet2 , x Fnet2 , y ( 2.3 N ) ( 5.2 N ) 5.7 N2) 2.0 2 4 xˆ 4 2 4 yˆ 2

7. (b) If the distance d were doubled, the magnitudeof the force would be cut to one-fourth and the directionwould be unchanged.

SuperpositionFor case of more than two forces, the net force vector isJGJG JG JGF net F 1 F 2 F 3 .The X and Y components of the net force areFnet , x F1, x F2, x F3, x .Fnet , y F1, y F2, y F3, y .The magnitude and direction of the net force areFnet Fnet2 , x Fnet2 , yθ tan 1 ( Fnet , y / Fnet , x )

Summary1) There are two kinds of charge: positive ( ) and negative (-)charges.2) Like charges repel; different charges attract.3) Electric charge conservation: the total charges areconstant in the universe.4) Insulator, semiconductor, and conductor.5) Coulomb’s lawq1 q2F k,2r(19 5)6) Superposition (based on Coulomb’s law): net vector force.

Example 19-3: SuperpositionThree charges, each equal to 2.90 uC, are placed at three corners ofa square 0.500 m on a side, as shown in the diagram. Find themagnitude and direction of the net force on charge number 3Picture the problem

SolutionJG1. Find the magnitude of F 31F31 kq1 q 3(2r )2 ( 8 . 9 9 1 0 N .m9 0 .1 5 12( 2 .9 1 0 6 C ) 2/C ) 2 ( 0 .5 0 0 m ) 22( 2 .9 1 0 6 C ) 2/C ) ( 0 .5 0 0 m ) 22NJG2. Find the magnitude of F 32F32 kq2 q3(r )2 ( 8 . 9 9 1 0 N .m9 0 .3 0 2N2

JGJG3. Calculate the two orthogonal components of F 31 and F 32F3 1, x F3 1 c o s 4 5 .0 ( 0 .1 5 1 N ) ( 0 .7 0 7 ) 0 .1 0 7NF3 1, y F3 1 s i n 4 5 .0 ( 0 .1 5 1 N ) ( 0 .7 0 7 ) 0 .1 0 7NF 3 2 , x F 3 2 c o s 0 ( 0 . 3 0 2 N ) (1) 0 . 3 0 2F32 , y F32 s in 0 ( 0 .3 0 2 N )( 0 ) 0NNJG4. Calculate the net force F 3F3 , x F3 1, x F3 2 , x 0 .1 0 7 N 0 .3 0 2 N 0 .4 0 9F3 , y F3 1, y F3 2 , y 0 .1 0 7 N 0 N 0 .1 0 7F3 F 3 2, x F 3 2, y 5. Find the direction ofθ 3 ta n 1 (F3, yF3,xNN( 0 .4 0 9 ) 2 ( 0 .1 0 7 N ) 2 0 .4 2 3JGF3) ta n 1(0 .1 0 7 N) 1 4 .7 0 .4 0 9 NN

Figure 19-7 Forces Between Point Charges. Note: the mass of proton is 2000 times that of the electron ! Note: the mass of proton is 2000 times of the electron ! An Example: Calculate the magnitude of electric force between a electron and a proton at a distance of r 5.29 x 10 -11 m. 12 2 99 922 11 2 8