chap163 - if r I.r 16.4 THE ELECTRlC FIELD 551 t-i I[an...

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Unformatted text preview: “if“ r... ‘_ I ._...r- 16.4 THE ELECTRlC FIELD 551 ‘ t-—--i-—- I'. [an lhl (e) t Figure 16.32 Electric field lines due to isolated point charges. (a) Field of negative point charge: (b) field of positive " point charge. These sketches show only field lines that lie in a two-dimensional plane. (c) A three-dimensional illustration of electric field lines due to a positive charge. The electric field is strong where the field lines are close together and weak . where they are far apart. Compare the lengths of the electric field vector arroWs in Fig. 16.24. -—---—u—-—-—-—-.n.-.n... .___.. -. W— I-r Figure 16.33 Electric field lines for a dipole. The electric field vectorfi at a point P is tangent to the field line through that point and is the sum of the Pia-J I'm: fields (11 and E.) due to each or the tWo point charges. (Text web- site tutorial: E-field of dipole) 'Csncséssslfisample 15-8.. _. Field Lines for 3 Thin Spherical Shell A thin metallic spherical shell of radius R carries a total shell is tiny. and you are looking at it from distant points; Charge Q, which is positive. The charge is spread out evenly (b) you are looking at the field inside the shell’s cavity. In over the shell‘s outside surface. Sketch the electric field (a). also sketch E field vectors at two different points out; lines in two dilTErent views of the situation: (a) the spherical side the shell. i'mifimu'a' on next page 592 CHAPTER 16 Electric Forces and Fields Conceptual Example 16.6 continge‘dl‘g" Strategy Since the charge on the shell is positive. field lines begin on the shell. A sphere is a highly symmetrical shape: standing at the center. it looks the same in any chosen direction. This symmetry helps in sketching the field lines. Solution in) A tiny spherical shell located far away can- not be distinguished from a point charge. The sphere looks like a point when seen from a great distance and the field lines look just like those emanating from a positive point charge (Fig. 16.34). The field lines show that the electric field is directed radially away from the center of the shell and that its magnitude decreases with increasing distance. as illustrated by the tool—Z vectors in Fig. 16.34. (b) Field lines begin on the positive charges on the shell sur- face. Some go outward. representing the electric field outside the shell. whereas others may perhaps go inward. represent- ing the field inside the shell. Any field lines inside must stun evenly spaced on the shell and point directly toward the center F: direction due to the symmetry of the sphere. But what would hap- pen to the field lines when they only end at the center if a negav / tive point charge is found there-— "'*'—-—-'_+Q'—-*-— but there is no point charge. If cross at the center. That can- not be right since the field must have a unique direction at every Figure 16.34 The inescapable conclusion: Field lines outside the shell are directed radially of the shell (Fig. 16.35): the lines cannot deviate from the radial reach the center? The lines can / \\\ the lines do not end. they would El point—field lines never cross. outward. there we no field finer inside the she” (Fig. l6.36). so E = () everywhere inside the shell. How does Gymnorchus r' navigate in muddy @ water? Foils-d \.;—_4_—:~.< -<-' —-- 1’ -<-! .""' —-:-+= i=0 74"“ 2i- /.i_\ ii. 2""\;.._ _ .4 ZR . _ "J. . _'+:.- . f f \ / 1 Figure 16.35 Figure 16.36 If there are field lines inside the Shell. they must start (In lines—and therefore no the shell and point radially electric field—inside the inward. Tlten what? shell. There can be no field Discussion We conclude that the electric field inside a spherical shell of charge is zero. This conclusion. which we reached using field lines and symmetry considerations. can also be proved using Coulomb’s law, the principle of super- position. and some calculus—a much more difficult method! The field line picture also shows that the eieetric field patient mirsitle a spherical .vheii is the some as if the change were all condensed into a point charge at the center of the sphere. ——--——_—..._.__.._____.,..____,____ Conceptual Practice Problem 16.8 Field Lines After a Negative Point Charge ls Inserted Suppose the spherical shell of evenly distributed posi- tive charge Q has a point charge —Q placed at its center. (:1) Sketch the field lines. [Hinc Since the charges are equal in magnitude. the number of lines starting on the shell is equal to the number ending on the point charge] (b) Defend your sketch using the principle ofsuperposition (total field = field due to shell + field due to point charge). Application of Electric Fields: Electrolocation Figure 15.37 The electric field generated by Gynnntrchns. The field is approximately that of a dipole. The head of the fish is positively charged and the tail is negatively charged. Long before scientists learned how to detect and measure electric fields. certain animals and fish evolted organs to produce and detect electric fields. Gynmurdms niloriens (see the Chapter Opener) has electrical organs running along the length of its body: these organs set up an electric field around the fish (Fig. 16.37). When a nearby object distorts the field lines. Gyinmrrelms detects the change through sensory receptors, mostly near the head. and responds accordingly. This extra sense enables the fish to detect prey or predators in muddy streams where eyes are less useful. Since Gynnmrchus relies primarily on electrolocation. where slight changes in the electric field are interpreted as the presence of nearby objects. it is important that it be able to create the same electric field over and over. For this reason. Gynnmrcims swims by undulating its long dorsal fin while holding its body rigid. Keeping the backbone straight keeps the negative and positive charge centers aligned and at a fixed distance apart. A swishing tail would cause variation in the electric field and that would make electrolocation much less accurate. I. 15.5 MOTION OF A POINT CHARGE IN A UNIFORM ELECTRIC FIELD The simplest example of how a charged object responds to an electric field is when the elec- g-ie field (due to other charges) is uniform—that is. has the same magnitude and direction at every point. The field due to a single point charge is not unifomt: it is radially directed and its magnitude follows the inverse square law. To create a uniform field requires a large num- ber of charges. The most common way to create a (nearly) uniform electric field is to put equal and opposite changes on two parallel metal plates (Fig. [6.38). If the charges are :I: Q and the plates have area A. the magnitude of the field equation box between the plates is r I Electric field between oppositely charged metal plates 1. _ i - -_ E _ £04 (16 6) The constant ea. called the permittivity of vacuum. is related to the Coulomb constant: 1. Permittivity of vacuum 1 6": —- = 8.85 x lo— h q (16-7) .The direction of the field is perpendicular to the plates. from the positively charged plate toward the negatively charged plate. Assuming the unifomt field E is knovm. a point charge (1 experiences an electric force F = git: (16-41)) If this is the only force acting on the point charge. then the net force is constant and therefore so is the acceleration: .. " F: a=%=‘{7 (16-8) With a constant acceleration, the motion can take one of two forms. If the initial veloc- ity of the point charge is zero or is parallel or antiparallel to the field, then the motion is along a straight line. If the point charge has an initial velocity component perpendicular to the field, then the trajectory is parabolic (just like a projectile in a uniform gravitational field ifother forces are negligible). All the tools developed in Chapters 2 and 3 to analyre motion with constant acceleration can be used here. The direction of the acceleration is either parallel to I“; (for a positive charge) or antipamllel to F1 (for a negative charge). fiHECKPomT 16.5 An electron moves in a region of uniform electric field in the +x-direction. The electric field is also in the +x-direction. Describe the subsequent motion of the electron. -—_- CONNECTION: If no forces act on a point charge other than the force due to a uniform electric field. then the acceleration is constant. All the principles vte learned for motion vt ith constant acceleration in a uniform gravitational field apply. I'lotvever. the accelera- tion does not hate the same magnitude and direction for all point charges in the same ficIdv—sec Eq. 1 lei-8). Figure 15.38 (a) Uniform electric field between two paralv lel metal plates with opposite charges +Q and -Q. The field has magnitude E = Q/(EnA) where A is the area of each plate. The direction of the field is per- pendicular to the plates. pointing away from the positive plate and tovt ard the negative plate. (b) Side view ofthe field lines. 594 CHAPTER 16 Electric Forces and Fields 'E-itamp'ie 16.9 Electron Beam A cathode ray tube (CRT) is used to accelerate electrons in some televisions. computer monitors, oscilloscopes. and x-ray tubes. Electrons from a heated filament pass through a hole in the cathode; they are then accelerated by an elec- tric field between the cathode and the anode (Fig. |6.39). Suppose an electron passes through the hole in the cathode at a velocity of 1.0 K 105 mls toward the anode. The electric field is uniform between the anode and cathode and has a magnitude of LO x 104 NC. ta) What is the acceleration of the electron? (b) If the anode and cathode are separated by 2.0 cm. what is the final velocity of the electron? Strategy Because the field is uniform. the acceleration of the electron is constant. Then we can apply Newton's second law and use any of the methods we previously developed for motion with constant acceleration. i' lt'ctron Given: initial speed vi = 1.0 x 10" mls: separation between plates :1 = 0.020 in: electric field magnitude E = 1.0 x 10‘ MC Look up: electron mass me = 9. l09 x 10"“ kg: electron charge q = —e = —l.602 x 10"” C Find: (a) acceleration; (b) final velocity Solution (a) First, check that gravity is negligible. The.- weight of the electron is ' Fg = mg = 9.109 x 10‘1‘I kg x 9.8 01st = 3.9 x 10‘5“»: The magnitude of the electric force is FE = eE: 1.602 x 10""C x 1.0 x 10‘ NIC: L6 x10‘”_1§t_' l \) l'lsitt‘s for horizontal tlL-llutitm ill] lit-uteri lilanltnt ' lsutiru- ul' t-lt t lrnnsi l’latts for _ \tt'lital titilt'ttinn i'Jt‘t‘lrttil lit-am Side view Figure 16.39 in a cathode ray tube [CRT]. electrons are accelerated to high speeds by an electric field bet“ een the cathode and anode. This CRT, used in an oscilloscope. also has two pairs of parallel plates that are used to deflect the electron beam horizontally {A} and vertically (B). Note that most of the deflection of the beam occurs after it has left the region between the plates. Between either set of plates. the force on an electron is constant so it mm es along a parabolic path. Once an electron leaves the plates. the electric field is essentially zero so it travels in a straight line path with constant velocity. c ourt'mtt'd rm new page 16.5 MDTION OF A POINT CHARGE IN A UNIFIIJRM ELECTRIC FIELD 595 Example 16.9 1:0ntir1u_e'd¢ Which is about l-‘l orders of magnitude larger. Gravity is iiiompletely negligible. While between the plates, the elec- iimn‘s acceleration is therefore a_£__eg= 1.602 x 10"9c x 1.0 x 10‘ NC 9.109 x 10‘31 kg - 1'11c _ me = l.‘l6 x 1015 mls2 .Totwo significant figures.a= 1.8 x 1015 mlsz. Since the charge on the electron is negative. the direction of the acceleration is p'pposile to the electric field, or to the right in the figure. ([9) The initial velocity of the electron is also to the right. We have a one-dimensional constant acceleration problem since "the initial velocity and the acceleration are in the same direc- ition. From Eq. (3—22). the final velocity is av." =\/v;l + lad = :10 x to5 me]2 + 2 x “1.76 x 10‘5 me: x 0.020 m = 8.4 x 10" this to the right Discussion The acceleration of the electrons seems large. This large value might cause some concem. but there is no law of physics against such large accelerations. Note that the final speed is less than the speed of light (3 x 103 mfs). the universe‘s ultimate speed limit. You may suspect that this problem can also be solved using energy methods. We could indeed find the work done by the electric force and use the work done to find the change in kinetic energy. The energy approach for electric fields is developed in Chapter 17. _________.._—-————--— Practice Problem 16.9 Slowing Some Protons If a beam of protons were projected horizontally to the right through the hole in the cathode (see Fig. 16.39) with an ini- tial speed of vi = 3.0 x 105 mls. with what speed would the protons reach the anode (if they do reach it)? --.;t\_pplication of Electric Field: Oscilloscope The electric field is used to speed '._up the electron beam in a CRT. In an oscilloscope—a device used to measure time- '_d'ependcnt quantities in circuits—it is also used to deflect the beam. An electric field is 'fipt used to deflect the electron beam in the CRT used in a TV or computer monitor: that 'iunction is performed by a magnetic field. rel-.1 I. ..- - - sighs!“ as. -..._= . . . . ._., $1.1: gut-s. Hem-Jr .Deflection of an Electronq Projected into a UniformE Field 'An electron is projected horizontally into the uniform elec- -tric field directed vertically downward between two par- allel plates (Fig. [6.40). The plates are 2.00 cm apart and an: of length 4.00 cm. The initial speed of the electron is '01: 8.00 x 10" mils. As it enters the region between the plates. the electron is midway between the two plates: as it / Ir— J.()0cm - —'j / .Figure 16.40 Ah electron deflected by an electric field. The trajectory is para- bPlic between the plates because the electric field exerts a constant ifglrce on the electron. After exiting the plates. it moves at constant .i'j'ilocity because the net force is zero. leaves. the electron just misses the upper plate. What is the magnitude of the electric field‘.’ Strategy Using the .t‘- and yvaxes in the figure, the elecv tric field is in the -y-dircction and the initial velocity of the electron is in the +.t'-direction. The electric force on the elec- tron is upward (in the +_v~direction) since it has a negative charge and is constant because the field is uniform. Thus. the acceleration of the electron is constant and directed upward. Since the acceleration is in the +_v-direction. the .v-component of the velocity is constant. The problem is similar to a projectile problem. but the constant acceleration is due to a uniform electric field instead of a uniform gravi‘ tational field. If the electron just misses the upper plate. its displacement is + l.00 cm in the ysdirection and +4.00 cm in the .r-direction. From '0. and Ar. we can find the time the electron spends between the plates. From Ay and the time. we can find (1.. From the acceleration we find the electric field using Newton's second law. SF = mi. comforted on new page 596 CHAPTER 16 Electric Forces and Fields Example 15-10Ffl'fifim We ignore the gravitational force on the electron because we assume it to be negligible. We can test this assumption later. Given: At- = 4.00 cm: Ay = 1.00 cm: a = 8.00 x I0“ mls Find: electric field strength. E Solution We start by finding the time the electron spends between the plates from Ar and or _ zr- —-——(—=5.0()x 10'“s -‘ 8.00 x 10' this Frotu the time spent between the plates and Ay. we find the component of the acceleration in the .v-direction. A_\' = go- (At)1 a. = —-, = -—~—T = 3.00 x to” ma: (my (5.00 x 10' s)‘ This acceleration is produced by the electric force acting on the electron since we assume that no other forces act. From Newton’s second law. Fr = :15, = Hie“)- Solving for E“ We have _ 9.109 x 10'“ kg x 8.00 x 10'4m/s2 ‘1 —l.602 x 10"“C = _4.55 x 103 NIC “1c“!- E,- _ a... 6 Application: Gel Electrophoresis Gel electrophoresis is a technique that usesfan . applied electric field to sort biological macromolecules (e.g.. proteins or nucleic acids) . based on size. The molecules to be sorted are chemically treated so they unfold into r'od- , like shapes and so they carry a net charge in solution. The molecules are deposited into -. ' a gel matrix and an electric field is applied (Fig. [6.41). Figure 15.4]. An apparatus used to perform gel electropho- resis. The molecules to be sorted are placed in wells in the gel. Then the power supply is turned on. subjecting the molecules to a large electric field and making them tnigrate through the gel. Since the field has no .r-compoucnt. its magnitude'ig 4.55 x 103 MC. Ii? . Discussion We have ignored the gravitational force Lontrl '5 the electron because we suspect that it is negligible in cofi'fFj -' parison with the electric force. This should be checked to'bfi I sure it is a valid assumption. F a: mag = 9.109 x 10'” kg x (9.80 leg downward); = 8.93 X 10"]. N downward I 171. = 4i: = —l.602 x 10“”c x {4.55 x :03 WC amnesia) . = 7.29 x 10'"1 N upward The electric force is stronger than the gravitational foreé; by a factor of approximately 10”. so the assumptionii, ' valid. ' Practice Problem 16.10 _Deflection of a Proton Projected into a Uniform E Field If the electron is replaced by a proton projected with fibe— same initial velocity. will the proton exit the region betwec't't'lil " the plates or will it hit one of the plates? 11' it does not strike t 5 one of the plates. by hour much is it deflected by the titnéiit _}_ leaves the region of electric field? I 15.6 CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM ”The electric force pulls the molecules toward one of the electrodes, depending on the sign of its charge. ' If no other forces acted. the molecules would move with constant acceleration, but a force due to the gel opposes their motion. This force is similar to viscous drag liege Section 9.]O)—it is proportional to the speed of the molecule. where the con- stant of proportionality depends on the size and shape of the molecule. Each mol- ecule reaches a terminal speed at which the electric and drag forces balance; smaller 'i‘nolecules move faster and large molecules move more slowly. so after a while. the molecules are sorted by size. The molecules can then be stained to make them visible (Fig. 16.42). Force and Torque on a Dipole in an Electric Field The electric force on a dipole in a uniform electric field is zero. because the forces on the two charges are equal in mag- nitude and opposite in direction (Fig. 16.43a). However. the torque on the dipole is not .egero unless H = 0 or 9 = [80“ . As shown in Problem 108. the magnitude of the torque lfor any angle 9 (as defined in Fig. 16.43a) is Torque on a dipole T = (1.9:! sin 9 (16-9) The direction of the torque tends to rotate the dipole toward the stable equi- librium position (Fig. 16.43b) and away from the unstable equilibrium position (Fig. 16.43c). If the electric field were nonuniform. the electric forces on the positive .and negative charge would not be equal; then the electric force on the dipole would be nonzero. 'i‘ 16.5 CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM In Section [6.1. we described how a piece of paper can be polarized by nearby charges. The polarization is the paper's response to an applied electric field. By applied we tncan a field due to charges outside the paper. The separation of charge in the paper produces ' an electric field of its own. The net electric field at any point-"whether inside or outside the paper—is the sum of the applied field and the field due to the separated charges in 'the paper. How much charge separation occurs depends on both the strength of the applied field and properties of the atoms and molecules that make up the paper. Some materials are tnore easily polarized than others. The most easily polarized materials are conduc- tors because they contain highly mobile charges that can move freely throug...
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