Chapter 25. Electric Potential

Transcription

1 Chapte 25 Electic Potential CHPTER OUTLINE 25.1 Potential Diffeence and Electic Potential 25.2 Potential Diffeences in a Unifom Electic Field 25.3 Electic Potential and Potential Enegy Due to Point Chages 25.4 Obtaining the Value of the Electic Field fom the Electic Potential 25.5 Electic Potential Due to Continuous Chage Distibutions 25.6 Electic Potential Due to a Chaged Conducto 25.7 The Millikan Oil-Dop Epeiment 25.8 pplications of Electostatics Pocesses occuing duing thundestoms cause lage diffeences in electic potential between a thundecloud and the gound. The esult of this potential diffeence is an electical dischage that we call lightning, such as this display ove Tucson, izona. ( Keith Kent/ Photo Reseaches, Inc.) 762

2 The concept of potential enegy was intoduced in Chapte 8 in connection with such consevative foces as the gavitational foce and the elastic foce eeted by a sping. By using the law of consevation of enegy, we wee able to avoid woking diectly with foces when solving vaious poblems in mechanics. The concept of potential enegy is also of geat value in the study of electicity. Because the electostatic foce is consevative, electostatic phenomena can be conveniently descibed in tems of an electic potential enegy. This idea enables us to define a scala quantity known as electic potential. Because the electic potential at any point in an electic field is a scala quantity, we can use it to descibe electostatic phenomena moe simply than if we wee to ely only on the electic field and electic foces. The concept of electic potential is of geat pactical value in the opeation of electic cicuits and devices we will study in late chaptes Potential Diffeence and Electic Potential When a test chage q 0 is placed in an electic field E ceated by some souce chage distibution, the electic foce acting on the test chage is q 0 E. The foce q 0 E is consevative because the foce between chages descibed by Coulomb s law is consevative. When the test chage is moved in the field by some etenal agent, the wok done by the field on the chage is equal to the negative of the wok done by the etenal agent causing the displacement. This is analogous to the situation of lifting an object with mass in a gavitational field the wok done by the etenal agent is mgh and the wok done by the gavitational foce is mgh. When analyzing electic and magnetic fields, it is common pactice to use the notation ds to epesent an infinitesimal displacement vecto that is oiented tangent to a path though space. This path may be staight o cuved, and an integal pefomed along this path is called eithe a path integal o a line integal (the two tems ae synonymous). Fo an infinitesimal displacement ds of a chage, the wok done by the electic field on the chage is F ds q 0 E ds. s this amount of wok is done by the field, the potential enegy of the chage field system is changed by an amount du q 0 E ds. Fo a finite displacement of the chage fom point to point B, the change in potential enegy of the system U U B U is U q 0 B Ed s (25.1) The integation is pefomed along the path that q 0 follows as it moves fom to B. Because the foce q 0 E is consevative, this line integal does not depend on the path taken fom to B. Fo a given position of the test chage in the field, the chage field system has a potential enegy U elative to the configuation of the system that is defined as U 0. Dividing the potential enegy by the test chage gives a physical quantity that depends only on the souce chage distibution. The potential enegy pe unit chage U/q 0 is Change in electic potential enegy of a system 763

3 764 CHPTER 25 Electic Potential PITFLL PREVENTION 25.1 Potential and Potential Enegy The potential is chaacteistic of the field only, independent of a chaged test paticle that may be placed in the field. Potential enegy is chaacteistic of the chage field system due to an inteaction between the field and a chaged paticle placed in the field. independent of the value of q 0 and has a value at evey point in an electic field. This quantity U/q 0 is called the electic potential (o simply the potential) V. Thus, the electic potential at any point in an electic field is V U q 0 (25.2) The fact that potential enegy is a scala quantity means that electic potential also is a scala quantity. s descibed by Equation 25.1, if the test chage is moved between two positions and B in an electic field, the chage field system epeiences a change in potential enegy. The potential diffeence V V B V between two points and B in an electic field is defined as the change in potential enegy of the system when a test chage is moved between the points divided by the test chage q 0 : Potential diffeence between two points V U q 0 B Ed s (25.3) PITFLL PREVENTION 25.2 Voltage vaiety of phases ae used to descibe the potential diffeence between two points, the most common being voltage, aising fom the unit fo potential. voltage applied to a device, such as a television, o acoss a device is the same as the potential diffeence acoss the device. If we say that the voltage applied to a lightbulb is 120 volts, we mean that the potential diffeence between the two electical contacts on the lightbulb is 120 volts. The electon volt Just as with potential enegy, only diffeences in electic potential ae meaningful. To avoid having to wok with potential diffeences, howeve, we often take the value of the electic potential to be zeo at some convenient point in an electic field. Potential diffeence should not be confused with diffeence in potential enegy. The potential diffeence between and B depends only on the souce chage distibution (conside points and B without the pesence of the test chage), while the diffeence in potential enegy eists only if a test chage is moved between the points. Electic potential is a scala chaacteistic of an electic field, independent of any chages that may be placed in the field. If an etenal agent moves a test chage fom to B without changing the kinetic enegy of the test chage, the agent pefoms wok which changes the potential enegy of the system: W U. The test chage q 0 is used as a mental device to define the electic potential. Imagine an abitay chage q located in an electic field. Fom Equation 25.3, the wok done by an etenal agent in moving a chage q though an electic field at constant velocity is (25.4) Because electic potential is a measue of potential enegy pe unit chage, the SI unit of both electic potential and potential diffeence is joules pe coulomb, which is defined as a volt (V): That is, 1 J of wok must be done to move a 1-C chage though a potential diffeence of 1 V. Equation 25.3 shows that potential diffeence also has units of electic field times distance. Fom this, it follows that the SI unit of electic field (N/C) can also be epessed in volts pe mete: 1 W q V 1 V 1 N C 1 Theefoe, we can intepet the electic field as a measue of the ate of change with position of the electic potential. unit of enegy commonly used in atomic and nuclea physics is the electon volt (ev), which is defined as the enegy a chage field system gains o loses when a chage of magnitude e (that is, an electon o a poton) is moved though a potential diffeence of 1 V. Because 1 V 1 J/C and because the fundamental chage is C, the electon volt is elated to the joule as follows: J C V m 1 e V CV J (25.5)

4 SECTION 25.2 Potential Diffeences in a Unifom Electic Field 765 Fo instance, an electon in the beam of a typical television pictue tube may have a speed of m/s. This coesponds to a kinetic enegy of J, which is equivalent to ev. Such an electon has to be acceleated fom est though a potential diffeence of 2.6 kv to each this speed. Quick Quiz 25.1 In Figue 25.1, two points and B ae located within a egion in which thee is an electic field. The potential diffeence V V B V is (a) positive (b) negative (c) zeo. Quick Quiz 25.2 In Figue 25.1, a negative chage is placed at and then moved to B. The change in potential enegy of the chage field system fo this pocess is (a) positive (b) negative (c) zeo. PITFLL PREVENTION 25.3 The Electon Volt The electon volt is a unit of enegy, NOT of potential. The enegy of any system may be epessed in ev, but this unit is most convenient fo descibing the emission and absoption of visible light fom atoms. Enegies of nuclea pocesses ae often epessed in MeV. E 25.2 Potential Diffeences in a Unifom Electic Field B Equations 25.1 and 25.3 hold in all electic fields, whethe unifom o vaying, but they can be simplified fo a unifom field. Fist, conside a unifom electic field diected along the negative y ais, as shown in Figue 25.2a. Let us calculate the potential diffeence between two points and B sepaated by a distance s d, whee s is paallel to the field lines. Equation 25.3 gives V B V V B Ed s B (E cos 0)d s B E ds Because E is constant, we can emove it fom the integal sign; this gives V E B d s E d (25.6) The negative sign indicates that the electic potential at point B is lowe than at point ; that is, V B V. Electic field lines always point in the diection of deceasing electic potential, as shown in Figue 25.2a. Figue 25.1 (Quick Quiz 25.1) Two points in an electic field. Potential diffeence between two points in a unifom electic field d d B q B m E g (a) Figue 25.2 (a) When the electic field E is diected downwad, point B is at a lowe electic potential than point. When a positive test chage moves fom point to point B, the chage field system loses electic potential enegy. (b) When an object of mass m moves downwad in the diection of the gavitational field g, the object field system loses gavitational potential enegy. (b)

5 766 CHPTER 25 Electic Potential Now suppose that a test chage q 0 moves fom to B. We can calculate the change in the potential enegy of the chage field system fom Equations 25.3 and 25.6: θ Figue 25.3 unifom electic field diected along the positive ais. Point B is at a lowe electic potential than point. Points B and C ae at the same electic potential. s d Change in potential enegy when a chaged paticle is moved in a unifom electic field B C E (25.7) Fom this esult, we see that if q 0 is positive, then U is negative. We conclude that a system consisting of a positive chage and an electic field loses electic potential enegy when the chage moves in the diection of the field. This means that an electic field does wok on a positive chage when the chage moves in the diection of the electic field. (This is analogous to the wok done by the gavitational field on a falling object, as shown in Figue 25.2b.) If a positive test chage is eleased fom est in this electic field, it epeiences an electic foce q 0 E in the diection of E (downwad in Fig. 25.2a). Theefoe, it acceleates downwad, gaining kinetic enegy. s the chaged paticle gains kinetic enegy, the chage field system loses an equal amount of potential enegy. This should not be supising it is simply consevation of enegy in an isolated system as intoduced in Chapte 8. If q 0 is negative, then U in Equation 25.7 is positive and the situation is evesed: system consisting of a negative chage and an electic field gains electic potential enegy when the chage moves in the diection of the field. If a negative chage is eleased fom est in an electic field, it acceleates in a diection opposite the diection of the field. In ode fo the negative chage to move in the diection of the field, an etenal agent must apply a foce and do positive wok on the chage. Now conside the moe geneal case of a chaged paticle that moves between and B in a unifom electic field such that the vecto s is not paallel to the field lines, as shown in Figue In this case, Equation 25.3 gives V B Ed s E B d s Es U q 0 V q 0 E d (25.8) whee again we ae able to emove E fom the integal because it is constant. The change in potential enegy of the chage field system is U q 0 V q 0 Es (25.9) Finally, we conclude fom Equation 25.8 that all points in a plane pependicula to a unifom electic field ae at the same electic potential. We can see this in Figue 25.3, whee the potential diffeence V B V is equal to the potential diffeence V C V. (Pove this to youself by woking out the dot poduct E s fo s :B, whee the angle between E and s is abitay as shown in Figue 25.3, and the dot poduct fo s :C, whee 0.) Theefoe, V B V C. The name equipotential suface is given to any suface consisting of a continuous distibution of points having the same electic potential. The equipotential sufaces of a unifom electic field consist of a family of paallel planes that ae all pependicula to the field. Equipotential sufaces fo fields with othe symmeties ae descibed in late sections. 9 V 8 V 7 V C B E D Quick Quiz 25.3 The labeled points in Figue 25.4 ae on a seies of equipotential sufaces associated with an electic field. Rank (fom geatest to least) the wok done by the electic field on a positively chaged paticle that moves fom to B; fom B to C; fom C to D; fom D to E. 6 V Figue 25.4 (Quick Quiz 25.3) Fou equipotential sufaces. Quick Quiz 25.4 Fo the equipotential sufaces in Figue 25.4, what is the appoimate diection of the electic field? (a) Out of the page (b) Into the page (c) Towad the ight edge of the page (d) Towad the left edge of the page (e) Towad the top of the page (f) Towad the bottom of the page.

6 SECTION 25.2 Potential Diffeences in a Unifom Electic Field 767 Eample 25.1 The Electic Field Between Two Paallel Plates of Opposite Chage battey poduces a specified potential diffeence V between conductos attached to the battey teminals. 12-V battey is connected between two paallel plates, as shown in Figue The sepaation between the plates is d 0.30 cm, and we assume the electic field between the plates to be B d V = 12 V Figue 25.5 (Eample 25.1) 12-V battey connected to two paallel plates. The electic field between the plates has a magnitude given by the potential diffeence V divided by the plate sepaation d. unifom. (This assumption is easonable if the plate sepaation is small elative to the plate dimensions and if we do not conside locations nea the plate edges.) Find the magnitude of the electic field between the plates. Solution The electic field is diected fom the positive plate () to the negative one (B), and the positive plate is at a highe electic potential than the negative plate is. The potential diffeence between the plates must equal the potential diffeence between the battey teminals. We can undestand this by noting that all points on a conducto in equilibium ae at the same electic potential 1 ; no potential diffeence eists between a teminal and any potion of the plate to which it is connected. Theefoe, the magnitude of the electic field between the plates is, fom Equation 25.6, E V B V d 12 V m V/m The configuation of plates in Figue 25.5 is called a paallel-plate capacito, and is eamined in geate detail in Chapte 26. Eample 25.2 Motion of a Poton in a Unifom Electic Field Inteactive poton is eleased fom est in a unifom electic field that has a magnitude of V/m (Fig. 25.6). The poton undegoes a displacement of 0.50 m in the diection of E. () Find the change in electic potential between points and B. Solution Because the positively chaged poton moves in the diection of the field, we epect it to move to a position of lowe electic potential. Fom Equation 25.6, we have V Ed ( V/m)(0.50 m) V (B) Find the change in potential enegy of the poton field system fo this displacement. Solution Using Equation 25.3, U q 0 V e V ( C)( V) J The negative sign means the potential enegy of the system deceases as the poton moves in the diection of the electic field. s the poton acceleates in the diection of the field, it gains kinetic enegy and at the same time the system loses electic potential enegy. (C) Find the speed of the poton afte completing the 0.50 m displacement in the electic field. Solution The chage field system is isolated, so the mechanical enegy of the system is conseved: K U 0 ( 1 2 mv 2 0) e V 0 v = 0 Figue 25.6 (Eample 25.2) poton acceleates fom to B in the diection of the electic field. v (2e V ) m 2( C)( V) kg E d m/s What If? What if the situation is eactly the same as that shown in Figue 25.6, but no poton is pesent? Could both pats () and (B) of this eample still be answeed? v B 1 The electic field vanishes within a conducto in electostatic equilibium; thus, the path integal between any two points in the conducto must be zeo. moe complete discussion of this point is given in Section 25.6.

7 768 CHPTER 25 Electic Potential nswe Pat () of the eample would emain eactly the same because the potential diffeence between points and B is established by the souce chages in the paallel plates. The potential diffeence does not depend on the pesence of the poton, which plays the ole of a test chage. Pat (B) of the eample would be meaningless if the poton is not pesent. change in potential enegy is elated to a change in the chage field system. In the absence of the poton, the system of the electic field alone does not change. t the Inteactive Woked Eample link at you can pedict and obseve the speed of the poton as it aives at the negative plate fo andom values of the electic field. d θ ds ˆ q B B 25.3 Electic Potential and Potential Enegy Due to Point Chages In Section 23.4 we discussed the fact that an isolated positive point chage q poduces an electic field that is diected adially outwad fom the chage. To find the electic potential at a point located a distance fom the chage, we begin with the geneal epession fo potential diffeence: V B V B Ed s whee and B ae the two abitay points shown in Figue t any point in space, the electic field due to the point chage is E k e q ˆ/ 2 (Eq. 23.9), whee ˆ is a unit vecto diected fom the chage towad the point. The quantity E ds can be epessed as Figue 25.7 The potential diffeence between points and B due to a point chage q depends only on the initial and final adial coodinates and B. The two dashed cicles epesent intesections of spheical equipotential sufaces with the page. PITFLL PREVENTION 25.4 Simila Equation Waning Do not confuse Equation fo the electic potential of a point chage with Equation 23.9 fo the electic field of a point chage. Potential is popotional to 1/, while the field is popotional to 1/ 2. The effect of a chage on the space suounding it can be descibed in two ways. The chage sets up a vecto electic field E, which is elated to the foce epeienced by a test chage placed in the field. It also sets up a scala potential V, which is elated to the potential enegy of the two-chage system when a test chage is placed in the field. Ed s k e Because the magnitude of ˆ is 1, the dot poduct ˆ ds ds cos, whee is the angle between ˆ and ds. Futhemoe, ds cos is the pojection of ds onto ; thus, ds cos d. That is, any displacement ds along the path fom point to point B poduces a change d in the magnitude of, the position vecto of the point elative to the chage ceating the field. Making these substitutions, we find that E ds (k e q/ 2 )d; hence, the epession fo the potential diffeence becomes V B V k e q B V B V k e q 1 B 1 (25.10) This equation shows us that the integal of E ds is independent of the path between points and B. Multiplying by a chage q 0 that moves between points and B, we see that the integal of q 0 E ds is also independent of path. This latte integal is the wok done by the electic foce, which tells us that the electic foce is consevative (see Section 8.3). We define a field that is elated to a consevative foce as a consevative field. Thus, Equation tells us that the electic field of a fied point chage is consevative. Futhemoe, Equation epesses the impotant esult that the potential diffeence between any two points and B in a field ceated by a point chage depends only on the adial coodinates and B. It is customay to choose the efeence of electic potential fo a point chage to be V 0 at. With this efeence choice, the electic potential ceated by a point chage at any distance fom the chage is V k e q 2 ˆd s q d 2 k e q B (25.11)

8 SECTION 25.3 Electic Potential and Potential Enegy Due to Point Chages Electic potential (V) 1 0 y Figue 25.8 The electic potential in the plane aound a single positive chage is plotted on the vetical ais. (The electic potential function fo a negative chage would look like a hole instead of a hill.) The ed line shows the 1/ natue of the electic potential, as given by Equation Figue 25.8 shows a plot of the electic potential on the vetical ais fo a positive chage located in the y plane. Conside the following analogy to gavitational potential: imagine tying to oll a mable towad the top of a hill shaped like the suface in Figue Pushing the mable up the hill is analogous to pushing one positively chaged object towad anothe positively chaged object. Similaly, the electic potential gaph of the egion suounding a negative chage is analogous to a hole with espect to any appoaching positively chaged objects. chaged object must be infinitely distant fom anothe chage befoe the suface in Figue 25.8 is flat and has an electic potential of zeo. We obtain the electic potential esulting fom two o moe point chages by applying the supeposition pinciple. That is, the total electic potential at some point P due to seveal point chages is the sum of the potentials due to the individual chages. Fo a goup of point chages, we can wite the total electic potential at P in the fom V k e i q i i (25.12) whee the potential is again taken to be zeo at infinity and i is the distance fom the point P to the chage q i. Note that the sum in Equation is an algebaic sum of scalas athe than a vecto sum (which we use to calculate the electic field of a goup of chages). Thus, it is often much easie to evaluate V than to evaluate E. The electic potential aound a dipole is illustated in Figue Notice the steep slope of the potential between the chages, epesenting a egion of stong electic field. Electic potential due to seveal point chages 2 Electic potential (V) Figue 25.9 The electic potential in the plane containing a dipole.

9 770 CHPTER 25 Electic Potential q 1 12 q 2 We now conside the potential enegy of a system of two chaged paticles. If V 2 is the electic potential at a point P due to chage q 2, then the wok an etenal agent must do to bing a second chage q 1 fom infinity to P without acceleation is q 1 V 2. This wok epesents a tansfe of enegy into the system and the enegy appeas in the system as potential enegy U when the paticles ae sepaated by a distance 12 (Fig a). Theefoe, we can epess the potential enegy of the system as 2 (a) U k e q 1 q 2 12 (25.13) P V = k e q (b) ctive Figue (a) If two point chages ae sepaated by a distance 12, the potential enegy of the pai of chages is given by k e q 1 q 2 / 12. (b) If chage q 1 is emoved, a potential k e q 2 / 12 eists at point P due to chage q 2. t the ctive Figues link at you can move chage q 1 o point P and see the esult on the electic potential enegy of the system fo pat (a) and the electic potential due to chage q 2 fo pat (b). q 2 q 2 Note that if the chages ae of the same sign, U is positive. This is consistent with the fact that positive wok must be done by an etenal agent on the system to bing the two chages nea one anothe (because chages of the same sign epel). If the chages ae of opposite sign, U is negative; this means that negative wok is done by an etenal agent against the attactive foce between the chages of opposite sign as they ae bought nea each othe a foce must be applied opposite to the displacement to pevent q 1 fom acceleating towad q 2. In Figue 25.10b, we have emoved the chage q 1. t the position that this chage peviously occupied, point P, we can use Equations 25.2 and to define a potential due to chage q 2 as V U/q 1 k e q 2 / 12. This epession is consistent with Equation If the system consists of moe than two chaged paticles, we can obtain the total potential enegy by calculating U fo evey pai of chages and summing the tems algebaically. s an eample, the total potential enegy of the system of thee chages shown in Figue is U k e q 1 q 2 12 q 1 q 3 13 q 2 q 3 23 (25.14) Physically, we can intepet this as follows: imagine that q 1 is fied at the position shown in Figue but that q 2 and q 3 ae at infinity. The wok an etenal agent must do to bing q 2 fom infinity to its position nea q 1 is k e q 1 q 2 / 12, which is the fist tem in Equation The last two tems epesent the wok equied to bing q 3 fom infinity to its position nea q 1 and q 2. (The esult is independent of the ode in which the chages ae tanspoted.) q 1 13 Figue Thee point chages ae fied at the positions shown. The potential enegy of this system of chages is given by Equation PITFLL PREVENTION 25.5 Which Wok? Thee is a diffeence between wok done by one membe of a system on anothe membe and wok done on a system by an etenal agent. In the pesent discussion, we ae consideing the goup of chages to be the system and an etenal agent is doing wok on the system to move the chages fom an infinite sepaation to a small sepaation. q 3 Quick Quiz 25.5 spheical balloon contains a positively chaged object at its cente. s the balloon is inflated to a geate volume while the chaged object emains at the cente, does the electic potential at the suface of the balloon (a) incease, (b) decease, o (c) emain the same? Does the electic flu though the suface of the balloon (d) incease, (e) decease, o (f) emain the same? Quick Quiz 25.6 In Figue 25.10a, take q 1 to be a negative souce chage and q 2 to be the test chage. If q 2 is initially positive and is changed to a chage of the same magnitude but negative, the potential at the position of q 2 due to q 1 (a) inceases (b) deceases (c) emains the same. Quick Quiz 25.7 Conside the situation in Quick Quiz 25.6 again. When q 2 is changed fom positive to negative, the potential enegy of the two-chage system (a) inceases (b) deceases (c) emains the same. 2 The epession fo the electic potential enegy of a system made up of two point chages, Equation 25.13, is of the same fom as the equation fo the gavitational potential enegy of a system made up of two point masses, Gm 1 m 2 / (see Chapte 13). The similaity is not supising in view of the fact that both epessions ae deived fom an invese-squae foce law.

10 SECTION 25.3 Electic Potential and Potential Enegy Due to Point Chages 771 Eample 25.3 The Electic Potential Due to Two Point Chages Inteactive chage q C is located at the oigin, and a chage q C is located at (0, 3.00) m, as shown in Figue 25.12a. () Find the total electic potential due to these chages at the point P, whose coodinates ae (4.00, 0) m. Solution Fo two chages, the sum in Equation gives V P k e q 1 1 (B) Find the change in potential enegy of the system of two chages plus a chage q C as the latte chage moves fom infinity to point P (Fig b). Solution When the chage q 3 is at infinity, let us define U i 0 fo the system, and when the chage is at P, U f q 3 V P ; theefoe, q 2 2 V P ( Nm 2 /C 2 ) C 4.00 m V U q 3V P 0 ( C)( V) J C 5.00 m Theefoe, because the potential enegy of the system has deceased, positive wok would have to be done by an etenal agent to emove the chage fom point P back to infinity. What If? You ae woking though this eample with a classmate and she says, Wait a minute! In pat (B), we ignoed the potential enegy associated with the pai of chages q 1 and q 2! How would you espond? nswe Given the statement of the poblem, it is not necessay to include this potential enegy, because pat (B) asks fo the change in potential enegy of the system as q 3 is bought in fom infinity. Because the configuation of chages q 1 and q 2 does not change in the pocess, thee is no U associated with these chages. Howeve, if pat (B) had asked to find the change in potential enegy when all thee chages stat out infinitely fa apat and ae then bought to the positions in Figue 25.12b, we would need to calculate the change as follows, using Equation 25.14: U k e q 1q 2 12 q 1q 3 13 ( Nm 2 /C 2 ) ( C)( C) 3.00 m ( C)( C) 4.00 m ( C)( C) 5.00 m J q 2q µ C y 6.00 µ C y 3.00 m 3.00 m 2.00 µ C 4.00 m P 2.00 µ C 4.00 m 3.00 µ C (a) (b) Figue (Eample 25.3) (a) The electic potential at P due to the two chages q 1 and q 2 is the algebaic sum of the potentials due to the individual chages. (b) thid chage q C is bought fom infinity to a position nea the othe chages. Eploe the value of the electic potential at point P and the electic potential enegy of the system in Figue 25.12b at the Inteactive Woked Eample link at

11 772 CHPTER 25 Electic Potential 25.4 Obtaining the Value of the Electic Field fom the Electic Potential The electic field E and the electic potential V ae elated as shown in Equation We now show how to calculate the value of the electic field if the electic potential is known in a cetain egion. Fom Equation 25.3 we can epess the potential diffeence dv between two points a distance ds apat as dv Ed s (25.15) If the electic field has only one component E, then E ds E d. Theefoe, Equation becomes dv E d, o E dv d (25.16) That is, the component of the electic field is equal to the negative of the deivative of the electic potential with espect to. Simila statements can be made about the y and z components. Equation is the mathematical statement of the fact that the electic field is a measue of the ate of change with position of the electic potential, as mentioned in Section Epeimentally, electic potential and position can be measued easily with a voltmete (see Section 28.5) and a mete stick. Consequently, an electic field can be detemined by measuing the electic potential at seveal positions in the field and making a gaph of the esults. ccoding to Equation 25.16, the slope of a gaph of V vesus at a given point povides the magnitude of the electic field at that point. When a test chage undegoes a displacement ds along an equipotential suface, then dv 0 because the potential is constant along an equipotential suface. Fom Equation 25.15, we see that dv Eds 0; thus, E must be pependicula to the displacement along the equipotential suface. This shows that the equipotential sufaces must always be pependicula to the electic field lines passing though them. s mentioned at the end of Section 25.2, the equipotential sufaces fo a unifom electic field consist of a family of planes pependicula to the field lines. Figue 25.13a shows some epesentative equipotential sufaces fo this situation. E (a) (b) (c) Figue Equipotential sufaces (the dashed blue lines ae intesections of these sufaces with the page) and electic field lines (ed-bown lines) fo (a) a unifom electic field poduced by an infinite sheet of chage, (b) a point chage, and (c) an electic dipole. In all cases, the equipotential sufaces ae pependicula to the electic field lines at evey point.

12 SECTION 25.4 Obtaining the Value of the Electic Field fom the Electic Potential 773 If the chage distibution ceating an electic field has spheical symmety such that the volume chage density depends only on the adial distance, then the electic field is adial. In this case, E ds E d, and we can epess dv in the fom dv E d. Theefoe, E dv d (25.17) Fo eample, the electic potential of a point chage is V k e q/. Because V is a function of only, the potential function has spheical symmety. pplying Equation 25.17, we find that the electic field due to the point chage is E k e q/ 2, a familia esult. Note that the potential changes only in the adial diection, not in any diection pependicula to. Thus, V (like E ) is a function only of. gain, this is consistent with the idea that equipotential sufaces ae pependicula to field lines. In this case the equipotential sufaces ae a family of sphees concentic with the spheically symmetic chage distibution (Fig b). The equipotential sufaces fo an electic dipole ae sketched in Figue 25.13c. In geneal, the electic potential is a function of all thee spatial coodinates. If V() is given in tems of the Catesian coodinates, the electic field components E, E y, and E z can eadily be found fom V(, y, z) as the patial deivatives 3 E V Fo eample, if V 3 2 y y 2 yz, then E y V y E z V z V (3 2 y y 2 y z) (3 2 y) 3y d d ( 2 ) 6 y (25.18) Finding the electic field fom the potential Quick Quiz 25.8 In a cetain egion of space, the electic potential is zeo eveywhee along the ais. Fom this we can conclude that the component of the electic field in this egion is (a) zeo (b) in the diection (c) in the diection. Quick Quiz 25.9 In a cetain egion of space, the electic field is zeo. Fom this we can conclude that the electic potential in this egion is (a) zeo (b) constant (c) positive (d) negative. Eample 25.4 The Electic Potential Due to a Dipole n electic dipole consists of two chages of equal magnitude and opposite sign sepaated by a distance 2a, as shown in Figue The dipole is along the ais and is centeed at the oigin. () Calculate the electic potential at point P. q a y a q P Solution Fo point P in Figue 25.14, V k e q i q k e i a q a 2k e qa 2 a 2 Figue (Eample 25.4) n electic dipole located on the ais. 3 In vecto notation, E is often witten in Catesian coodinate systems as whee is called the gadient opeato. E V î ĵ y kˆ z V

13 774 CHPTER 25 Electic Potential (B) Calculate V and E at a point fa fom the dipole. Solution If point P is fa fom the dipole, such that a, then a 2 can be neglected in the tem 2 a 2 and V becomes and using Equation 25.16, E dv d d d 2 k e q a k e q a 2 2 (a 2 2 ) 2 V 2k e qa 2 ( a) Using Equation and this esult, we can calculate the magnitude of the electic field at a point fa fom the dipole: E dv d 4 k e qa 3 ( a) (C) Calculate V and E if point P is located anywhee between the two chages. Solution Using Equation 25.12, q i q V k e k e i a q a 2 k e q a 2 2 We can check these esults by consideing the situation at the cente of the dipole, whee 0, V 0, and E 2k e q/a 2. What If? What if point P in Figue happens to be located to the left of the negative chage? Would the answe to pat () be the same? nswe The potential should be negative because a point to the left of the dipole is close to the negative chage than to the positive chage. If we edo the calculation in pat () with P on the left side of q, we have q i q V k e k e i a q a Thus, the potential has the same value but is negative fo points on the left of the dipole. 2 k e qa 2 a 2 dq 25.5 Electic Potential Due to Continuous Chage Distibutions P Figue The electic potential at the point P due to a continuous chage distibution can be calculated by dividing the chage distibution into elements of chage dq and summing the electic potential contibutions ove all elements. Electic potential due to a continuous chage distibution We can calculate the electic potential due to a continuous chage distibution in two ways. If the chage distibution is known, we can stat with Equation fo the electic potential of a point chage. We then conside the potential due to a small chage element dq, teating this element as a point chage (Fig ). The electic potential dv at some point P due to the chage element dq is dv k e (25.19) whee is the distance fom the chage element to point P. To obtain the total potential at point P, we integate Equation to include contibutions fom all elements of the chage distibution. Because each element is, in geneal, a diffeent distance fom point P and because k e is constant, we can epess V as dq V k e (25.20) dq In effect, we have eplaced the sum in Equation with an integal. Note that this epession fo V uses a paticula efeence: the electic potential is taken to be zeo when point P is infinitely fa fom the chage distibution. If the electic field is aleady known fom othe consideations, such as Gauss s law, we can calculate the electic potential due to a continuous chage distibution using Equation If the chage distibution has sufficient symmety, we fist evaluate E at any point using Gauss s law and then substitute the value obtained into Equation 25.3 to detemine the potential diffeence V between any two points. We then choose the electic potential V to be zeo at some convenient point. PROBLEM-SOLVING HINTS

14 SECTION 25.5 Electic Potential Due to Continuous Chage Distibutions 775 PROBLEM-SOLVING HINTS Calculating Electic Potential Remembe that electic potential is a scala quantity, so vecto components do not eist. Theefoe, when using the supeposition pinciple to evaluate the electic potential at a point due to a system of point chages, simply take the algebaic sum of the potentials due to the vaious chages. Howeve, you must keep tack of signs. The potential is positive fo positive chages and negative fo negative chages. Just as with gavitational potential enegy in mechanics, only changes in electic potential ae significant; hence, the point whee you choose the potential to be zeo is abitay. When dealing with point chages o a chage distibution of finite size, we usually define V 0 to be at a point infinitely fa fom the chages. You can evaluate the electic potential at some point P due to a continuous distibution of chage by dividing the chage distibution into infinitesimal elements of chage dq located at a distance fom P. Then, teat one chage element as a point chage, such that the potential at P due to the element is dv k e dq/. Obtain the total potential at P by integating dv ove the entie chage distibution. In pefoming the integation fo most poblems, you must epess dq and in tems of a single vaiable. To simplify the integation, conside the geomety involved in the poblem caefully. Study Eamples 25.5 though 25.7 below fo guidance. nothe method that you can use to obtain the electic potential due to a finite continuous chage distibution is to stat with the definition of potential diffeence given by Equation If you know o can easily obtain E (fom Gauss s law), then you can evaluate the line integal of Eds. This method is demonstated in Eample Eample 25.5 Electic Potential Due to a Unifomly Chaged Ring () Find an epession fo the electic potential at a point P located on the pependicula cental ais of a unifomly chaged ing of adius a and total chage Q. Solution Figue 25.16, in which the ing is oiented so that its plane is pependicula to the ais and its cente is at the oigin, helps us to conceptualize this poblem. Because the ing consists of a continuous distibution of chage athe a dq 2 a 2 Figue (Eample 25.5) unifomly chaged ing of adius a lies in a plane pependicula to the ais. ll elements dq of the ing ae the same distance fom a point P lying on the ais. P than a set of discete chages, we categoize this poblem as one in which we need to use the integation technique epesented by Equation To analyze the poblem, we take point P to be at a distance fom the cente of the ing, as shown in Figue The chage element dq is at a distance 2 a 2 fom point P. Hence, we can epess V as V k e dq dq k e 2 a 2 Because each element dq is at the same distance fom point P, we can bing 2 a 2 in font of the integal sign, and V educes to k k e Q V e dq (25.21) 2 a 2 2 a 2 The only vaiable in this epession fo V is. This is not supising because ou calculation is valid only fo points along the ais, whee y and z ae both zeo. (B) Find an epession fo the magnitude of the electic field at point P.

15 776 CHPTER 25 Electic Potential Solution Fom symmety, we see that along the ais E can have only an component. Theefoe, we can use Equation 25.16: E k e Q ( 2 a 2 ) 3/2 (25.22) E d V d k e Q d d ( 2 a 2 ) 1/2 k e Q( 1 2 )( 2 a 2 ) 3/2 (2) To finalize this poblem, we see that this esult fo the electic field agees with that obtained by diect integation (see Eample 23.8). Note that E 0 at 0 (the cente of the ing). Could you have guessed this? Eample 25.6 Electic Potential Due to a Unifomly Chaged Disk unifomly chaged disk has adius a and suface chage density. Find V 2k e [( 2 a 2 ) 1/2 ] (25.23) () the electic potential and (B) the magnitude of the electic field along the pependicula cental ais of the disk. Solution () gain, we choose the point P to be at a distance fom the cente of the disk and take the plane of the disk to be pependicula to the ais. We can simplify the poblem by dividing the disk into a seies of chaged ings of infinitesimal width d. The electic potential due to each ing is given by Equation Conside one such ing of adius and width d, as indicated in Figue The suface aea of the ing is d 2 d. Fom the definition of suface chage density (see Section 23.5), we know that the chage on the ing is dq d 2 d. Hence, the potential at the point P due to this ing is whee is a constant and is a vaiable. To find the total electic potential at P, we sum ove all ings making up the disk. That is, we integate dv fom 0 to a: V k e a 0 dv k e dq 2 k e 2 d d 2 k 2 e a ( 2 2 ) 1/2 2 d 0 This integal is of the common fom u n du and has the value u n1 1 /(n 1), whee n and u 2 2. This gives 2 (B) s in Eample 25.5, we can find the electic field at any aial point using Equation 25.16: E dv d 2k e 1 2 a 2 (25.24) The calculation of V and E fo an abitay point off the ais is moe difficult to pefom, and we do not teat this situation in this tet. a d 2 2 d = 2πdπ Figue (Eample 25.6) unifomly chaged disk of adius a lies in a plane pependicula to the ais. The calculation of the electic potential at any point P on the ais is simplified by dividing the disk into many ings of adius and width d, with aea 2 d. P Eample 25.7 Electic Potential Due to a Finite Line of Chage od of length located along the ais has a total chage Q and a unifom linea chage density Q /. Find the electic potential at a point P located on the y ais a distance a fom the oigin (Fig ). P y Solution The length element d has a chage dq d. Because this element is a distance 2 a 2 fom point P, we can epess the potential at point P due to this element as dv k e dq k e d 2 a 2 To obtain the total potential at P, we integate this epession ove the limits 0 to. Noting that k e and ae a O Figue (Eample 25.7) unifom line chage of length located along the ais. To calculate the electic potential at P, the line chage is divided into segments each of length d and each caying a chage dq d. dq d

16 SECTION 25.5 Electic Potential Due to Continuous Chage Distibutions 777 constants, we find that V k e This integal has the following value (see ppendi B): Evaluating V, we find 0 d V k e Q d 2 a 2 k e Q 0 2 a 2 ln ( 2 a 2 ) ln 2 a 2 a (25.25) What If? What if we wee asked to find the electic field at point P? Would this be a simple calculation? d 2 a 2 nswe Calculating the electic field by means of Equation would be a little messy. Thee is no symmety to appeal to, and the integation ove the line of chage would epesent a vecto addition of electic fields at point P. Using Equation 25.18, we could find E y by eplacing a with y in Equation and pefoming the diffeentiation with espect to y. Because the chaged od in Figue lies entiely to the ight of 0, the electic field at point P would have an component to the left if the od is chaged positively. We cannot use Equation to find the component of the field, howeve, because we evaluated the potential due to the od at a specific value of ( 0) athe than a geneal value of. We would need to find the potential as a function of both and y to be able to find the and y components of the electic field using Equation Eample 25.8 Electic Potential Due to a Unifomly Chaged Sphee n insulating solid sphee of adius R has a unifom positive volume chage density and total chage Q. () Find the electic potential at a point outside the sphee, that is, fo R. Take the potential to be zeo at. Solution In Eample 24.5, we found that the magnitude of the electic field outside a unifomly chaged sphee of adius R is E k e whee the field is diected adially outwad when Q is positive. This is the same as the field due to a point chage, which we studied in Section In this case, to obtain the electic potential at an eteio point, such as B in Figue 25.19, we use Equation 25.10, choosing point as : V B V k e Q V B 0 k e Q V B (fo R) Because the potential must be continuous at R, we can use this epession to obtain the potential at the suface of the sphee. That is, the potential at a point such as C shown in Figue is V C k e Q 2 (fo R ) k e Q R 1 B 1 1 B 0 Q (fo R) (B) Find the potential at a point inside the sphee, that is, fo R. Solution In Eample 24.5 we found that the electic field inside an insulating unifomly chaged sphee is E k e Q R 3 (fo R ) We can use this esult and Equation 25.3 to evaluate the potential diffeence V D V C at some inteio point D: V D V C 2 3 R V 0 V 0 Q E d k e Q R 3 d k e Q R 2R 3 (R 2 2 ) V R Figue (Eample 25.8) unifomly chaged insulating sphee of adius R and total chage Q. The electic potentials at points B and C ae equivalent to those poduced by a point chage Q located at the cente of the sphee, but this is not tue fo point D. V 0 = 3k eq 2R V D = k e Q 2R 3 2 R 2 V B = k eq R Figue (Eample 25.8) plot of electic potential V vesus distance fom the cente of a unifomly chaged insulating sphee of adius R. The cuve fo V D inside the sphee is paabolic and joins smoothly with the cuve fo V B outside the sphee, which is a hypebola. The potential has a maimum value V 0 at the cente of the sphee. We could make this gaph thee dimensional (simila to Figues 25.8 and 25.9) by evolving it aound the vetical ais. D ( C B (

17 778 CHPTER 25 Electic Potential Substituting V C k e Q /R into this epession and solving fo V D, we obtain V D k e Q 2R R (fo R) (25.26) t R, this epession gives a esult that agees with that fo the potential at the suface, that is, V C. plot of V vesus fo this chage distibution is given in Figue Electic Potential Due to a Chaged Conducto In Section 24.4 we found that when a solid conducto in equilibium caies a net chage, the chage esides on the oute suface of the conducto. Futhemoe, we showed that the electic field just outside the conducto is pependicula to the suface and that the field inside is zeo. We now show that evey point on the suface of a chaged conducto in equilibium is at the same electic potential. Conside two points and B on the suface of a chaged conducto, as shown in Figue long a suface path connecting these points, E is always pependicula to the displacement ds; theefoe E ds 0. Using this esult and Equation 25.3, we conclude that the potential diffeence between and B is necessaily zeo: V B V B Eds 0 (a) R This esult applies to any two points on the suface. Theefoe, V is constant eveywhee on the suface of a chaged conducto in equilibium. That is, the suface of any chaged conducto in electostatic equilibium is an equipotential suface. Futhemoe, because the electic field is zeo inside the conducto, we conclude that the electic potential is constant eveywhee inside the conducto and equal to its value at the suface. (b) k e Q R V E k e Q k e Q 2 Because this is tue, no wok is equied to move a test chage fom the inteio of a chaged conducto to its suface. Conside a solid metal conducting sphee of adius R and total positive chage Q, as shown in Figue 25.22a. The electic field outside the sphee is k e Q / 2 and points adially outwad. Fom Eample 25.8, we know that the electic potential at the inteio and suface of the sphee must be k e Q /R elative to infinity. The potential outside the sphee is k e Q /. Figue 25.22b is a plot of the electic potential as a function of, and Figue 25.22c shows how the electic field vaies with. (c) Figue (a) The ecess chage on a conducting sphee of adius R is unifomly distibuted on its suface. (b) Electic potential vesus distance fom the cente of the chaged conducting sphee. (c) Electic field magnitude vesus distance fom the cente of the chaged conducting sphee. R B E Figue n abitaily shaped conducto caying a positive chage. When the conducto is in electostatic equilibium, all of the chage esides at the suface, E 0 inside the conducto, and the diection of E just outside the conducto is pependicula to the suface. The electic potential is constant inside the conducto and is equal to the potential at the suface. Note fom the spacing of the positive signs that the suface chage density is nonunifom.

18 SECTION 25.6 Electic Potential Due to a Chaged Conducto 779 When a net chage is placed on a spheical conducto, the suface chage density is unifom, as indicated in Figue 25.22a. Howeve, if the conducto is nonspheical, as in Figue 25.21, the suface chage density is high whee the adius of cuvatue is small (as noted in Section 24.4), and it is low whee the adius of cuvatue is lage. Because the electic field just outside the conducto is popotional to the suface chage density, we see that the electic field is lage nea conve points having small adii of cuvatue and eaches vey high values at shap points. This is demonstated in Figue 25.23, in which small pieces of thead suspended in oil show the electic field lines. Notice that the density of field lines is highest at the shap tip of the left-hand conducto and at the highly cuved ends of the ight-hand conducto. In Eample 25.9, the elationship between electic field and adius of cuvatue is eploed mathematically. Figue shows the electic field lines aound two spheical conductos: one caying a net chage Q, and a lage one caying zeo net chage. In this case, the suface chage density is not unifom on eithe conducto. The sphee having zeo net chage has negative chages induced on its side that faces the chaged sphee and positive chages induced on its side opposite the chaged sphee. The boken blue cuves in the figue epesent the coss sections of the equipotential sufaces fo this chage configuation. s usual, the field lines ae pependicula to the conducting sufaces at all points, and the equipotential sufaces ae pependicula to the field lines eveywhee. PITFLL PREVENTION 25.6 Potential May Not Be Zeo The electic potential inside the conducto is not necessaily zeo in Figue 25.22, even though the electic field is zeo. Fom Equation 25.15, we see that a zeo value of the field esults in no change in the potential fom one point to anothe inside the conducto. Thus, the potential eveywhee inside the conducto, including the suface, has the same value, which may o may not be zeo, depending on whee the zeo of potential is defined. Quick Quiz Conside stating at the cente of the left-hand sphee (sphee 1, of adius a) in Figue and moving to the fa ight of the diagam, passing though the cente of the ight-hand sphee (sphee 2, of adius c) along the way. The centes of the sphees ae a distance b apat. Daw a gaph of the electic potential as a function of position elative to the cente of the left-hand sphee. Q Q = 0 Coutesy of Haold M. Waage, Pinceton Univesity Figue Electic field patten of a chaged conducting plate placed nea an oppositely chaged pointed conducto. Small pieces of thead suspended in oil align with the electic field lines. The field suounding the pointed conducto is most intense nea the pointed end and at othe places whee the adius of cuvatue is small. Figue The electic field lines (in ed-bown) aound two spheical conductos. The smalle sphee has a net chage Q, and the lage one has zeo net chage. The boken blue cuves ae intesections of equipotential sufaces with the page.