Ouestions '

ber. A gamma ruy entered the chamber from the bottom and at one point trans-formed into an electron and a positron. Because those new particles were chargedand moving, each left a trail of tiny bubbles. (The trails were curved because a mag-netic field had been set up in the chamber.) The gammaray,being electrically neutral,left no trail. Still, you can tell exactly where it underwent pair production -at the tipof the curved V, which is where the trails of the electron and positron begin.

Electric Charge The strength of a particle's electrical in-teraction with objects around it depends on its electric charge,which can be either positive or negative. Charges with thesame sign repel each other, and charges with opposite signs

attract each other. An object with equal amounts of the twokinds of charge is electrically neutral, whereas one with animbalance is electrically charged.

Conductors are materials in which a signiflcant numberof charged particles (electrons in metals) are free to move.The charged particles in nonconductors, or insulators, are notfree to move.

The Coulomb and Ampere The SI unit of charge is thecoulomb (C). It is deflned in terms of the unit of current, theampere (A), as the charge passing a particular point in 1 sec-

ond when there is a current of 1 ampere at that point:

1C : (1 A)(1 r).

This is based on the relation between current i and the ratedqldt at which charge passes a point:

(or nearly at rest) and separated by a distance r:

I I o.,ll a"lF- ^ + (Coulomb'slaw).4 7Tt11 f '

(2r-4)

Coulomb's Law Coulomb's law describes the electrostaticforce between small (point) electric charges q1 and ez at rest

Here 80 : 8.85 x 10 -tz gz75 ' mz is the permittivity constant,and Il4rres - k: 8.99 x 10r Y. m2lc2.

The force of attraction or repulsion between pointcharges at rest acts along the line joining the two charges. Ifmore than two charges are present, Eq. 2I-4 holds for eachpair of charges. The net force on each charge is then found,using the superposition principle, as the vector sum of theforces exerted on the charge by all the others.

The two shell theorems for electrostatics are

A shell of uniform charge attracts or repels a charged

';:::::'*n':,:#;':,":!::*:,::"'if atttheshett'scharge

If a charged particle is located inside a shell of uniformcharge, there is no net electrostatic force on the particle

from the shell.

The Elementary Charge Electric charge is quantized:any charge can be written as ne,whete n is a positive or nega-tive integer and e rs a constant of nature called the elementarycharge (: 1 .602 x 10 -tq C). Electric charge is conserved: thenet charge of any isolated system cannot change.

.dqi - d, (electric current). (2r-3)

1 Figure 2l-I3 shows four situations in which charged particlesare flxed in place on an axis. In which situations is there a point tothe left of the particles where an electron will be in equilibrium?

cl

-5q

3 Figure 2l-I5 shows four situations in which flve chargedparticles are evenly spaced along an axis. The charge valuesare indicated except for the central particle, which has thesame charge in all four situations. Rank the situations accord-itrg to the magnitude of the net electrostatic force on thecentral particle , greatest first.

+e +e

Flffi" Al"tS Question 3.

4 Figure 2I-16 shows three pairs of identical spheres that areto be touched together and then separated. The initial chargeson them are indicated. Rank the pairs according to (a) the mag-

+q -q +3q

(b)(")

-qo

-5q

FlG" 2t-13 Question 1.

e Figure 2I-L4 shows twocharged particles on an axis. Thecharges are free to move.However, a third charged parti-

+3q +q(1)

(2)

(3)

(4)

(d,)(')

o

-5q -q

Flffi, en-14 QuestionZ.

cle can be placed at a certain point such that all three particlesare then in equilibrium. (a) Is that point to the left of the firsttwo particles, to their right, or between them? (b) Should thethird particle be positively or negatively charged? (c) Is theequilibrium stable or unstable?

ffihmpter RR I Electric Fields

duces a torque on an electric dipole moment F to align 1 with E. gecause theoven's E otiillates, the water molecules continuously nip-nop in a frustrated at-tempt to align with E.

Energy is transferred from the electric field to the thermal energy of the water(and thus of the food) where three water molecules happened to have bonded to-gether to form a group.The flip-flop breaks some of the bonds. When the mole-cules reform the bonds, energy is transferred to the random motion of the groupand then to the surrounding molecules. Soon, the thermal energy of the water isenough to cook the food. Sometimes the heating is surprisitrg. If you heat a jellydonut, for example, the jelly (which holds a lot of water) heats far more than thedonut material (which holds much less water). Although the exterior of the donutmay not be hot, biting into the jelly can burn you. If water molecules were notelectric dipoles, we would not have microwave ovens.

A neuftal water molecule (FI rO) in its vapor state has anelectric dipole moment of magnitude 6.2 x 10-30 C.m.(u) How far apart are the molecule's centers of positiveand negative charge?

A molecule's dipole moment depends onor negativethe magnitude q of the molecule's positive

charge and the charge separation d.

eafeufafdons; There are 10 electrons and 10 protons ina neutral water molecule; so the magnitude of its dipolemoment is

p - qd : (10e) (d),

in whi ch d is the separation we are seeking and e is theelementary charge. Thus,

p 6.2 x 10-30 C.mt\e (10)(1.60 x 10-1e C)

This distance is not only small, but it is also actuallysmaller than the radius of a hydrogen atom.

(b) If the molecule is placed in an electric fleld of 1.5 x104 N/C, what maximum torque can the fleld exert on it?(Such a fleld can easily be set up in the laboratory.)

dipole is maximum whenis 90".

90" in Eq. 22-33 yields

r - pE sin g

(.) FIow much work must an external agent do to rotatethis molecule by 180" in this fleld, starting from its fullyaligned position, for which 0 : 0?

The work done by an external agent (bVmeans of a torque applied to the molecule) is equal tothe change in the molecule's potential energy due to thechange in orientation.

Ca faffom; From Eq. 22-40,we flnd

W- (Jtro" Uo

- ,pE - (2)(6.2 x 10-30 C.-)(1.5 x 104 N/C)L

' .9 x I0-2s J. (Answer)I

d

Electric Field One way to explain the electrostatic forcebetween two charges is to assume that each charge sets up anelectric fleld in the space around it. The electrostatic force act-ing on any one charge is then due to the electric fleld set up atits location by the other charge

Definition of Electric Field The electric fietd E at anypoint is defined in terms of the electrostatic force F thatwould be exerted on a positive test charge qsplaced there:

(22-t)

visualizing the direction and magnitude of electric flelds. Theelectric field vector at any point is tangent to a field line throughthat point. The density of fleld lines in any region is proportionalto the magnitude of the electric field in that region. Field iinesoriginate on positive charges and terminate on negative charges.

Field Due to a Point Charge The magnitude of theelectric field E set up by a point rhurg. q at a distan ce r fromthe charge is

__) FE-_qo

(22-3)

The direction of E ir away from the point charge if the chargeis positive and toward it if the charge is negative.Electric Field Lines Electric fietd lines provide a means for

Field Due to an Electric Dipole An electric dipote con-sists of two particles with charges of equal magnitude q bfiopposite sign, separated by a small distance d. Their electricdipole moment n has magnitude qd and points from thenegative charge to the positive charge. The magnitude of theelectric field set up by the dipole at a distant point on thedipole axis (which runs through both charges) is

Force on a Point Charge in an Electric Field When a

point charge q is qfaced in an external electric field E, th"electrostatic force F that acts on the point charge is

F-qE. (22-28)

Force F has the same direction as E it q is positive and theopposite direction if q is negative.

Dipole in an Electric Field When an electricdipole moment p' is placed in an electric fleld E,exerts a torqu e i on the dipole:

i--i*8.

Questiorrs ,,

dipole ofthe fleld

(22-34)

(22-e)

where z is the distance between the point and the center of thedipole.

Field Due to a Continuous Charge DistributionThe electric fleld due to a continuous charge distribution isfound by treating charge elements as point charges and thensumming, via integration, the electric field vectors producedby all the charge elements.

The dipole has a potential ener gy U associated with its orien-tation in the field:

(Ji--7.8. (22-38)

This potential energy is defined to be zero when F is perpe!_dicular to E;it is least ((I: -pE) when P'is aligned with Eand greatest (U - pE) when l is directed opposite E.

T Figure 22-2I shows three affangements of electric fieldlines. In each arrangement, a proton is released from rest atpoint A and is then accelerated through point B by the electricfield. Points A and B have equal separations in the threearrangements. Rank the arrangements according to the linearmomentum of the proton at point B,greatest first.

(") (b)

Ffrffi, *P-*1 Question 1.

tr Figure 22-22 shows four situations in which four chargedparticles are evenly spaced to the left and right of a centralpoint. The charge values are indicated. Rank the situationsaccording to the magnitude of the net electric field at thecentral point, greatest first.

(1)+e -e -e +e

(2)+e +e -e -e

(3)-e +e +e +e

(4)-e -e +e -e

la-l-- d,+d+d'-)FEffi" 2?-ZA Question2.

axis (other than at an infinite distance) is there a point atwhich their net electric fleld is zero: between the charges, to theirleft, or to their right? (b) Is there a point of zero net electric fleldanywhere off the axis (other than at aninflnite distance)?

4 Figure 22-24 shows two square arrays of charged particles.The squares, which are centered on point P, are misaligned.The particles are separated by either d or dlz along theperimeters of the squares. What are the magnitude and direc-tion of the net electric fleld at P?

(c)

-2q

o

-5q+2rl

+2q -Zq

+3q

-2q

-2q

F[G" 2Z"A& Question 4.

+6q

S In Fig. 22-25, two particlesof charge - q are arranged sym-metrically about the y axis; eachproduces an electric fleld atpoint P on that axis. (a) Are themagnitudes of the fields at Pequal? (b) Is each electric fielddirected toward or away fromthe charge producing it? (c) Is

FflG. AA-e5 Question 5.

the magnitude of the net electric fleld at P equal to the sum ofthe magnitudes ,E of the two field vectors (is it equal to 2E)?(d) Do the x components of those two field vectors add or can-cel? (e) Do their y components add or cancel? (f ) Is the direc-tion of the net field at P that of the canceling components or

+6q

+3q

-q

S Figure 22-23 shows twocharged particles flxed in placeon an axis. (u) Where on the

+q -3q

Ff,G. 2&-fr3 Question 3.

#ha,pter 2S I Gauss' Law

ffiffiKP#$ruT 4 Thefigure shows two large, parallel,nonconducting sheets with identi-cal (positive) uniform surfacecharge densities, and a sphere witha uniform (positive) volumecharge density. Rank the four

o4c3

o1

ol

a-_-]-- d-+- d---'.l--- a-*- a

numbered points according to the rnagnitude of the net electric field there, greatest first.

Gauss' Law Gauss' law and Coulomb's law are differentways of describing the relation between charge and electricfield in static situations. Gauss' law is

eoO : e"n" (Gauss' law), (23-6)

in which e"n is the net charge inside an imaginary closed sur-face (a Gaussian surface) and Q is the net flu* of the electricfield through the surface:

- ^t = .--? (electric flux through ao-? E'dA Gaussiansurface). Q3-4)

Coulomb's law can be derived from Gauss' law.

Applications of Gauss' Law Using Gauss' law and, insome cases, symmetry arguments, we can derive several im-portant results in electrostatic situations. Among these are:

L. An excess charge on an isolated conductor ts located en-tirely on the outer surface of the conductor.

2. The external electric field near the surface of a charged con-ductor is perpendicular to the surface and has magnitude

E,- (line of charge), (23-12)2Tesr

where r is the perpendicular distance from the line ofcharge to the point.

4. The electric field due to an infinite nonconducting sheet

with uniform surface charge density o is perpendicular tothe plane of the sheet and has magnitude

aE -

- (conductingsurface).

(̂j0

aE - ^ (sheet of charge).

2"0

5. The electric fleld outside a spherical shell of charge with radiusR and total charge q is directed radially and has magnitude

E- | q-

4nes 12 (spherical shell, for r > R). (23-15)

Here r is the distance from the center of the shell to the pointat which E is measured. (Th" charge behaves, for externalpoints, as if it were all located at the center of the sphere.) Thefield inside a uniform spherical shell of charge is exactly zero:

E : 0 (spherical shell, for r < R). (23-16)

6. The electric field inside a uniform sphere of charge is di-rected radially and has magnitude

(23-r3)

(23-20)

(23-rr)

Within the conductor, E : 0.

3. The electric field at any point due to an infinite line ofcharge with uniform linear charge density ,1, is perpendicu-lar to the line of charge and has magnitude

$ Figure 23-20 shows, in cross

section, a central metal ball,two spherical metal shells, andthree spherical Gaussiansurfaces of radii R,2R, and 3R,all with the same center. Theuniform charges on the threeobjects are; ball, Q; smallershell, 3Q;larger shell, 5Q. Rankthe Gaussian surfaces accord-ing to the magnitude of the

on the surfaces, greatest flrst, andindicate whether the magnitudesare uniform or variable alongeach surface.

S A surface has the area vec-tor A- Qi + gi) ttrt. What is

the flux of a uniform electricfield through it if the fleld is(q) E - +i Nrc and (b) E -4K N/C?

4 Figure 23-22 shows, in cross

section, three solid cylinders, each

Shell

Gaussiansurface

F$ffi" AS-40 Question 1.

electric field at any point on the surface , greatest first.

A Figure 23-2I shows, in cross section, two Gaussian spheres

and two Gaussian cubes that are centered on a positively chargedparticle. (u) Rank the net flux through the four Gaussian sur-faces, greatest first. (b) Rank the magnitudes of the electric fields

F!ffi. A3-tr1 QuestionZ.

of length L and uniform charge Q. Concentric with each cylinderis a cylindrical Gaussian surface, with all three surfaces havingthe same radius. Rank the Gaussian surfaces according to theelectric field at any point on the surface,greatest first.

Review & Summal.y

you may have so much excess charge that the potential difference between yourbody and your surroundings is 5 kV or more. If you touch a computer keyboardwhile charged like this, the excess charge on your body can flow through the com-puter's circuit chips, overloading and ruining them.

There are countless examples in which contact between a person and someother type of material leaves the person so highly charged that the person mightdischarge with a spark. Children sliding down a plastic slide on a dry day havebeen measured to have a potential of about 60 kV. If such a child reaches for anyconducting object (such as another person), the child probably will discharge tothe object with a very painful spark.

-Such a spark discharge would be disastrous in a hospital operatitrg roomwhere a flammable gas (such as an anesthetic gas) is present. To drain the chargethey collect as they move around, a surgical team wears conducting shoes andstands on a conducting floor. Spark discharges have also caused a number of flresat self-serve gasoline stations when a customer has slid back into a car seat towait for the car's tank to be filled. Contact with the car seat can leave the cus-tomer with so much charge that a spark might jump between flngers and pumpnozzle when the customer goes back to remove the nozzle from the ear. Thespark can ignite the gasoline vapor that surrounds the nozzle.

If an isolated conductor is placed in an external electric field, as in Fig. 24-20, allpoints of the conductor still come to a single potential regardless of whether theconductor has an excess charge.The free conduction electrons distribute themselveson the surface in such a way that the electric fleld they produc e at interior pointscancels the external electric field that would otherwise be there. Furthermore, theelectron distribution causes the net electric field at all points on the surface to beperpendicular to the surface. If the conductor in Fig. 24-20 could be somehow re-moved, leaving the surface charges frozen in place, the pattern of the electric fieldwould remain absolutely unchanged, for both exterior and interior points.

Ffffi. tr4-ffiffi An uncharged conduc-tor is suspended in an external elec-tric field. The free electrons in theconductor distribute themselves onthe surface as shown, so as to reducethe net electric fleld inside the con-ductor to zero and make the net fieldat the surface perpendicular to thesurface.

Electric Potential Energy The change LU in the electricpotential ener gy U of a point charge as the charge moves froman initial point i to a final point f tn an electric field is

L(J: Uf - (Jr - -W,

W^V-q

The SI unit of potential is the volt: I volt - 1 joule percoulomb.

Pot*ntial and potential difference can also be written interms of the electric potential ener gy U of a particle of chargeq rn an electric field:

v _ , , (24_5)

q

(24-r)

(24-B)

(24-6)

where W'is the work done by the electrostatic force (due tothe external electric field) on the point charge during themove from i to f.If the potential energy is defined to be zeroat inflnity, the electric potential energy U of the point chargeat a particular point is

(Jt - -w*. (24-2)

Here W* is the work done by the electrostatic force onthe point charge as the charge moves from inflnity to the par-ticular point.

Electric Potential Difference and Electric PotentialWe define the potential difference LV between two points iand f in an electric field as

where q is the charge of a pafircle on which workthe fleld. The potential at a point is

(24-7)

is done by

Equipotential Surfaces The points on an equipotentialsurface all have the same electric potential. The work doneon a test charge in moving it from one such surface to anotheris independent of the locations of the initial and final pointson these surfaces and of the path that joins the points.The electric fleld E is always directed perpendicularly tocorresponding equipotential surfaces.

Finding V from i The electric potential difference be-tween two points i and/is

Lv: v-v-+-!qqLu

LV: V- V: -+q

v-v- di, (24-rB)

C'hapter 2,& I Electric Potential

where the integral is taken over any path connecting the points. Ifwe choose [ : 0, we have, for the potential at a particular point,

(24-re)

Potential Due to Point Charges The electric potentialdue to a single point charge at a distance r from that pointcharge is

V- Iq4rres r

where I/ has the same sign as q.The potential duetion of point charges is

v_2u:*t+Potential Due to j"'El."tr3;;from an electric dipole with dipole momentqd,the electric potential of the dipole is

Calculating F from V The componenttion is the negative of the rat e at which thewith distance in that direction:

AVDLs-ds

of E in any direc-potential changes

(24-40)

The x, !,and e components of E may be found from

E,: -#i E,: -# Q4-4r)(24-26)

to a collec-

(24-27)

At a distance rmagnitude p -

When E is uniform, Eq. 24-40 reduces to

AV(24-42)E_ As)

where s is perpendicular to the equipotential surfaces. Theelectric field is zero parallel to an equipotential surface.

Electric Potential Energy of a System of PointCharges The electric potential energy of a system of pointcharges is equal to the work needed to assemble the systemwith the charges initially at rest and infinitely distant fromeach other. For two charges at separation r,

u-w- r QQz4rres r

(24-43)

Potential of a Charged Conductor An excess chargeplaced on a conductor will, in the equilibrium state, be locatedentirely on the outer surface of the conductor. The charge willdistribute itself so that the entire conductor, including interiorpoints, is at a uniform potential.

1 Figure 24-2I shows four pairs of charged particles. Foreach pair, let V - 0 at infinity and considet Vn"tat points on thex axis. For which pairs is there a point at which Vn"t: 0 (a) be-tween thegarticles and (b) to the right of the particles? (c) At such

a point is En", due to the particles equal to zero? (d) For each patr,are there off-axis points (other than at infinity) whereVn"l: 0?

point P at the center of the squareif the electric potential is zero atinfinity?

& Figure 24-24 shows three sets ofcross sections of equipotential sur-

faces; all three cover the same size

region of space. (u) Rank thearrangements according to themagnitude of the electric field pre-sent in the region, greatest first. (b)In which is the electric field di-rected down the page?

-4q+6q-2q +3q

-!q -?q +:l

d

+5q o -5oPI

-q -2q +4q

Flffi. #4-23 Question 3.

(1) (2)

-6q -2q(3) (4)

Ffffi" &e-A1 Questions I and7.

2 Figure 24-22 shows four arrangements of charged particles, allthe same distance from the origin. Rank the situations accordingto the net electric potential at the origin, most positive first. Takethe potential to be zero at infinity.

20v *--140v -*-10v40

60 ---120 **-3080

100 ---100 **-50(1)

+l2q +q

(a) (b)

(3)(2)

FlG" 24-44 Question 4.

Ffrffi. 2.4-22 QuestionZ.

3 In Fig. 24-23, eight particles form a square, with distance dbetween adjacent particles. What is the electric potential at

5 Figure 24-25 shows three pathsalong which we can move the posi-tively charged sphere A closer topositively charged sphere B,whichis held fixed in place. (a) Wouldsphere Abe moved to a higher orlower electric potential? Is the

(c)

FIG" A4-e5 Question 5.

Ouestions

Capacitor; Capacitance A capacitor consists of two iso-lated conductors (the plates) with charges + q and - q. Itscapacitance C is defined from

q-CV, (2s-r)

where I/ is the potential difference between the plates. The SI unitof capacitance is the farad (1. farad : 1 F - 1 coulomb per volt).

Determining Capacitance We generally determine thecapacitance of a particular capacitor configuration by(1) assuming a charge q to have been placed on the plates,(2) finding the electric field E d.r" to this charge, (3) evaluat-ing the potential difference V,, and (4) calculating C fromEq. 25-L Some specific results are the following:

A parallel-plate capacitor with flat parallel plates of areaA and spacing dhas capacttance

(2s-e)

A cylindrical capacitor (two long coaxial cylinders) oflength L and radii a and b has capacitance

and (n capacitors in series). (25-20)

Equivalent capacrtances can be used to calculate the cap aci-tances of more complicated series-parallel combinations.

Potential Energy and Energy Density The electricpotential energy U of a charged capacitor,

1 (1%-3,1

A spherical capacitor with concentric spherical plates ofradii a and b has capacitance

abC - 4rres

D-A

If we let b -+ oo and a - R in Eq. 25-t7, wecapacitance of an isolated sphere of radius R:

(25-27,25-22)

is equal to the work required to charge the capacitor. Thisenergy can be associated with the capuiitor', eleitric fleld EBy extension we can associate stored energy with any electricfield. In vacuuffi, the energy density tt) or potential energy perunit volume, within an electric field of magnitude E is given by

u - L"oE' (2s-2s)

Capacitance with a Dielectric If the space between theplates of a capacitor is completely filled with a dielectric material,the capacitance C is increased by a factor rc, called the dielectricconstant, which is characteristic of the material. In a region that iscompletely filled by a dielectric, all electrostatic equations con-taining s6 rnust be modified by replacing es with Kro.

The effects of adding a dielectric can be understood phys-ically in terms of the action of an electric field on the perma-nent or induced electric dipoles in the dielectric slab. Theresult is the formation of induced charges on the surfaces ofthe dielectric, which results in a weakening of the fleld withinthe dielectric for a given amount of free charge on the plates.

Gauss' Law with a Dielectric When a dielectric is pre-sent, Gauss'law may be generaltzedto

,1 eoAL:

d

(2s-r7)

obtain the

(2s-18)

fr q2u - b -lcv!,

(2s-r4)

C - 4rresR.

Capacitors in Parallel and in Series The equivalentcapacitances C"n of combinations of individual capacitors con-nected in parallel and in series can be found from

c"q,: C j (n capacitors in parallel) (25-19)

(2s-36)

Here q is the free charge;any induced surface charge is accountedfor by including the dielectric constant xinside the integral.j :1'

1 Figure 25-19 shows plotsof charge versus potential dif-ference for three parallel-plate capacitors that have theplate areas and separationsgiven in the table. Which plotgoes with which capacitor? F$ffi" A$-19 Question L.

uncharged capacitors of capacr-tance C. When the switch is

closed and the circuit reachesequilibrium, what are (u) thepotential difference across eachcapacitor and (b) the charge onthe left plate of each capacitor? Fnffi" AS-AS Question2.

Capacitor Area Separation

d

d

2d

2 Figure 25-20 shows an open switch, o battery of potentialdifference V, d current-measuring meter A, and three identical

(c) During charging, what net charge passes through the meter?

S For each circuit in Fig. zs-ZI,are the capacitors connected inseries, in parallel, or in neither mode?

*J-Fl-(a) (b)

T

2-tJ

A2A

A

Flfi. 2S-At Question 3.