Chapter 20 Electrostatics and Coulomb s Law 20.1 Introduction electrostatics Separation of Electric Charge by Rubbing

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1 I wish to give an account of some investigations which have led to the conclusion that the caies of negative electicity ae bodies, which I have called copuscles, having a mass vey much smalle than that of the atom of any known element, and ae of the same chaacte fom whateve souce the negative electicity may be deived. J.J.Thomson 20.1 Intoduction In chapte 1 we attempted to explain the wold in tems of the fewest numbe of fundamental quantities. Up to now only length, mass, and time have concened us. Now we will conside the fouth fundamental quantity, electic chage. Ou knowledge of electic chage is not new. The ealiest known expeiments on electostatics wee pefomed by Thales of Miletus (ca B.C.) aound 600 B.C., when he found that ambe, when ubbed with fu, attacted light objects. Today, we say that the ambe possesses an electical chage. (The wod electic is deived fom the Geek wod elekton, meaning ambe.) The study of electic chages at est unde the action of electic foces is called electostatics Sepaation of Electic Chage by Rubbing To help undestand some of the chaacteistics of electic chage, let us pefom the following expeiment. We attach a pith ball, a spongy mateial like cok, by a piece of sting to the ceiling, and it hangs feely, as shown in figue 20.1(a). We move an ambe od towad the pith ball, eventually touching it [figue 20.1(b)]. We obseve that nothing happens to the pith ball duing the touching pocess. Now we ub the ambe od with a piece of fu and then cause the od to touch the pith ball, figue 20.1(c). This time the pith ball flies away, as seen in figue 20.1(d). How can this behavio be explained? When the od was not ubbed and was touched to the pith ball, nothing happened to the pith ball. But when the od was ubbed with fu and touched the pith ball, the pith ball was epelled fom the od. We must conclude that something happened to the od duing the ubbing pocess. Let us assume that ubbing the od caused an amount of something, which we will call electic chage, to be deposited on the od. The type of chage deposited on the od could be abitaily called an A chage, o a B chage, o a blue chage, o a ed chage, but we will, again abitaily, call it negative chage. When the negatively chaged od touches the pith ball, some of this negative chage on the od flows onto the pith ball, leaving the pith ball negatively chaged. The pith ball then flies away fom the od, indicating that like chages epel each othe. It is because like chages epel each othe that the oiginal chages on the od flowed to the pith ball. 20-1

2 Sting Sting Sting Pith ball Ambe od Pith ball Rubbed Ambe od Pith ball (a) (b) (c) Glass od ubbed with silk (d) Repulsion (e) (f) Repulsion Attaction (g) Figue 20.1 Sepaation of chage by ubbing. The thought that should now occu to us is, Does ubbing any mateial geneate electic chage? Let us epeat the expeiment, only this time with a glass od. Fist, we touch the unubbed glass od to a new pith ball. As befoe nothing happens to the pith ball. Then we ub the glass od with silk and touch it to the pith ball, figue 20.1(e). We see that the pith ball flies away, as in figue 20.1(f). Electic chage has again been geneated by ubbing the od. But is this the same chage that was geneated when the ambe od was ubbed? At this point we cannot answe the question. To be completely geneal, let us assume that it is a diffeent electic chage, one that we will abitaily call positive chage. The positive chage appeas to flow fom the glass od to the pith ball duing contact, leaving the pith ball also positively chaged. The pith ball then flies away fom the glass od, again indicating that like chages epel each othe. Is this eally a diffeent chage, that is, a positive one? The expeimental esults ae the same fo both ods. Both pith balls ae epelled fom the od. 20-2

3 Let us now hang two sepaate pith balls fom the ceiling, and chage the one on the left negatively by touching it with the ubbed ambe od. The pith ball to the ight we chage positively by touching it with the glass od ubbed with silk. If the chages on both balls ae eally the same, then the balls should epel each othe, just as the balls wee epelled fom the ods in figues 20.1(d) and 20.1(f). We obseve, howeve, that the two pith balls ae attacted to each othe, as shown in figue 20.1(g). Theefoe ubbing the glass od with silk does indeed poduce a diffeent chage, a positive one, than that poduced by ubbing an ambe od with fu, the negative chage. Also note that the unlike chages attact each othe. This expeiment can be continued by ubbing diffeent ods with vaious mateials, but only these two types of chages, negative and positive, ae eve found. We conclude that thee ae two and only two types of electic chages in natue -- negative chage and positive chage. As a esult of these expeiments, the fundamental pinciple of electostatics can be stated: Like electic chages epel each othe, wheeas unlike electic chages attact each othe. To give a moe moden desciption of electostatics we need to discuss atomic stuctue Atomic Stuctue In an attempt to find simplicity in natue, the Geek philosophes Leucippus and Democitus suggested in the fifth centuy B.C. that matte is composed of vey small paticles called atoms. The wod atom comes fom a Geek wod that means that which is indivisible. Howeve, it was not until the ealy nineteenth centuy that John Dalton ( ), an English chemist, poposed that to evey known chemical element thee coesponds an atom of matte. Evey mateial in the wold is just some combination of these indivisible atoms. Howeve, in 1897 J. J. Thompson ( ), an English physicist, discoveed the electon, a negatively chaged paticle having a mass m e = kg. Although it had been known that thee was such a thing as negative electical chage, it was not known what the caie of that negative chage was. This newly discoveed electon, howeve, was the basic o elementay paticle caying the smallest amount of negative chage. All othe negative chage that occu in electostatic expeiments ae multiples of the electonic chage. The finding of the electon, howeve, pesented a athe difficult poblem. Whee did it come fom? The only place it could come fom is the inteio of the indivisible atom, which could not then be indivisible. The indivisible atom must have some stuctue. Because the atom is geneally neutal, thee must be some positive chage within the atom to neutalize the negative electon. In the ealy 1900 s Enest Ruthefod ( ), a Bitish physicist, bombaded atoms with alpha paticles (positively chaged paticles) and by obseving the effects of the collision, developed the nuclea model of the atom. His model of the atom consisted of a small, dense, positively chaged nucleus with negative electons obiting about it, somewhat in the manne of the planets obiting about the sun in the sola 20-3

4 system. Ruthefod found this positive paticle of the nucleus and named it the poton in The poton has a positive chage equal in magnitude to the chage on the electon. The mass of the poton, m p = kg, is about 1836 times geate than the electon mass. In 1920 Ruthefod suggested that thee is pobably anothe paticle within the nucleus, a neutal one, to which he gave the name the neuton. The neuton was discoveed some twelve yeas late in 1932 by the English physicist, James Chadwick. In tems of these paticles, o building blocks, the diffeent atoms ae fomed. The diffeence between one chemical element and anothe is in the numbe of potons, and electons within it. As seen in figue 20.2(a), the chemical element hydogen contains a nucleus which consists of one positive poton about which obits the lighte, negative electon. Sometimes the electon is to the ight and p+ e e 2p+ 2n e (a) Hydogen (b) Helium e e 3p+ 3n e e 4p+ 4n e e (c) Lithium e (d) Beyllium Figue 20.2 Atomic stuctue. sometimes to the left of the nucleus. Sometimes it is above and sometimes below the nucleus. By symmety, the electon s mean position coincides with the position of the positive nucleus. Theefoe, the atom as a whole acts as though it wee electically neutal. The next chemical element helium is fomed by the addition of anothe poton to the nucleus, and anothe electon to the obit, figue 20.2(b). Two neutons ae also found in the nucleus of helium. The next chemical element, lithium is fomed by the addition of anothe poton, anothe electon, and anothe 20-4

5 neuton, figue 20.2(c). In this way all of the chemical elements ae fomed, although in the highe elements thee ae usually moe neutons than potons. Because each element contains the same numbe of electons and potons, each element is electically neutal. Although this model of the atom is quite useful, it is not completely coect. An electon moving in a cicle is an acceleated chage, and it has been found that wheneve a chage is acceleated, it adiates enegy. Theefoe the adiating electon should lose enegy and spial into the nucleus and the atom should cease to exist. The wold, which is made up of atoms, should also cease to exist. Since the wold continues to exist, the above model of the atom can not be completely coect. Also since like chages epel each othe, the potons in the nucleus should also epel each othe and the nucleus should blow itself apat. Hence, the whole wold should blow itself apat. But it does not. Theefoe, thee must be some othe foce within the nucleus holding the potons togethe. This new foce is called the stong nuclea foce. Because the electon is the smallest unit of chage eve found, the fundamental unit of chage, the Coulomb, named afte the Fench physicist Chales A de Coulomb ( ), is defined in tems of a cetain numbe of these electonic chages. That is, 1 Coulomb of chage is equal to electonic chages, and the chage on one electon is Coulomb. The unit fo electical chage, the Coulomb, will be abbeviated as a capital C, in keeping with the SI convention that units named afte a peson ae symbolized by capital lettes. Because electic chage only comes in multiples of the electonic chage, it is said that electic chage is quantized. Also, the total net chage of any system is constant, a esult known as the law of consevation of electic chage. Although no electic chages have eve been found caying factional potions of the electonic chage, the latest hypothesis in elementay paticle physics is that potons and neutons ae made up of moe elementay paticles called quaks. (Muay Gell-Mann, 1964) It is pesently poposed that thee ae six quaks: the (1) up (u), (2) down (d), (3) stange (s), (4) cham (c), (5) bottom (b), and (6) top (t) quak. The chages on these quaks ae factional as shown in table Table 20.1 Quaks and thei chages. QUARK u d s c b t CHARGE (Faction of Electon Chage) 2/3 1/3 1/3 2/3 1/3 2/3 20-5

6 A poton is assumed to be made up of two up quaks and one down quak, as shown in figue The neuton is assumed to be made up of one up quak u u d chage 2/3 +2/3-1/3 = 1 Figue 20.3 The quak configuation of a poton. and two down quaks as shown in figue Of couse, individual quaks have not yet been found, and indeed most theoies of paticle physics pedict that an isolated single quak cannot exist. Thee is, howeve, stong indiect evidence fom scatteing expeiments that quaks do indeed exist. The quak model has also pedicted the existence of othe elementay paticles which have been found, indicating that the quak hypothesis is on vey good expeimental gound. Of couse, if the existence of quaks ae definitely confimed, the next question that would then have to be asked is, What ae quaks made of? u d d chage 2/3-1/3-1/3 = 0 Figue 20.4 The quak configuation of a neuton. In tems of the atomic concept of electic chage, the geneation of chage by ubbing is explained as follows. When an ambe od is ubbed with fu, the ubbing causes electons to be stipped fom the fu atoms and deposited on the ambe od, making the ambe od negative. The fu, which now has a deficiency of electons, becomes positive. Hence, chage has not been ceated, it has meely been sepaated. When the pith ball is touched by the negative ambe od, electons flow fom the od to the pith ball, leaving the pith ball negatively chaged. When the glass od is ubbed with silk, electons ae stipped fom the atoms in the glass od and ae deposited on the silk cloth making the silk cloth negative. The glass od, having a deficiency of electons, becomes positive. When the glass od 20-6

7 touches the neutal pith ball, electons fom the oiginally neutal pith ball flow onto the glass od, neutalizing some of its positive chage. The deficiency of electons on the pith ball causes the pith ball to become positive. This coectly explains the pevious expeiment whee, at that time, we incoectly assumed that positive chages flowed fom the glass od to the pith ball. In all these cases of chaging bodies by ubbing, the chage that moves fom one body to anothe is always the negative electonic chage. Recall that the electon is 1/1836 lighte than the poton and hence easie to move. Also the heavie potons ae tightly bound into the nucleus and ae not as easy to detach fom the atom as the weakly bound electons. We use the lette q to epesent the electic chage on a body and, in geneal, the net chage q on a body consisting of both potons and electons is given by q = (N p N e )e whee N p is equal to the numbe of potons, N e is equal to the numbe of electons, and e is equal to the basic unit of chage, namely C Measuement of Electic Chage The Electoscope One of the taditional means fo measuing electic chage is the electoscope, as seen in figue 20.5(a). An electoscope is a device made of two stips of thin gold (a) (b) Figue 20.5 An electoscope. leaf o aluminum foil fastened to the end of a metallic od. The metallic od is housed in an insulated enclosue. The top of the od is connected to a metallic ball at the top of the electoscope. If a negatively chaged od is touched to the metal ball on the top of the electoscope, electons move fom the negative od to the ball of the electoscope, down though the metallic connecting od to the aluminum leaves, some moving to the ight leaf and some to the left leaf. The negative chage on the aluminum leaves epel each othe, theeby sepaating the leaves, as shown in figue 20-7

8 20.5(b). The amount of sepaation of the leaves becomes a measue of the quantity of chage. If instead of the negative od, a positive od wee touched to the electoscope, the leaves would sepaate in the same way, only this time positive chage would be on each leaf. Electons on the electoscope would flow to the positive od leaving a deficiency of negative chages on the leaves of the electoscope. Hence, the leaves would have positive chages on them. Conductos and Insulatos Two electoscopes ae set up, as shown in figue When the negatively chaged od is touched to the metal bulb between the two electoscopes, the electoscope to Negatively chaged od Rubbe band Coppe wie Metal ball Glass od Figue 20.6 A conducto and an insulato. the ight shows a chage but the one to the left does not. The eason fo this phenomenon can be explained by the connecting mateial between the metal bulb and the electoscope. The coppe wie is called a conducto because it allows the chage deposited on the bulb to be conducted to the electoscope. The ubbe band is called an insulato because it does not pemit the chage to each the electoscope. (The ubbe band insulates the electoscope fom the chaged bulb.) In geneal, most substances fall into eithe one of these categoies. Mateials that pemit the fee flow of electic chage though them ae called conductos. Mateials that do not pemit the fee flow of electic chage though them ae called insulatos o dielectics. Most metals ae good conductos of electic chage, wheeas most nonmetals ae insulatos. (Thee ae a few mateials called semiconductos, whose chaacteistics lie between those of conductos and those of insulatos.) An inteesting chaacteistic of all conductos is that wheneve an electic chage is placed on a conducting body, that chage will edistibute itself until all of the chage is on the outside of the body. Fo example, if electic chages ae placed on a solid metallic sphee, the chages exet foces of epulsion on one anothe and 20-8

9 the chages ty to move as fa apat as they can. The geatest sepaation they can achieve is when they ae on the outside of the sphee. Chaging by Induction Most bodies ae electically neutal, that is, they contain the same numbe of positive chages as negative chages. If a negatively chaged od is bought into the vicinity of an unchaged conducting sphee the negative od epels the electons on the sphee, as shown in figues 20.7(a) and 20.7(b). The sphee still contains equal (a) Negative od fa away fom metallic sphee (b) Negative od bought close to, but not touching, metal sphee (c) Gounding the metal sphee. (d) The sphee is chaged positively. Figue 20.7 Chaging by induction. numbes of positive and negative chages but they ae now edistibuted so that the ight side of the sphee has a negative distibution of chage wheeas the left side has a positive one. (If the od is emoved, the chage would edistibute itself to its initial neutal configuation.) If you wee to touch you finge to the sphee you would povide a path fo the electons to escape to the gound, as shown in figue 20.7(c). When you emove you finge fom the sphee and emove the negative od, the sphee is chaged positively. A positive chage has been induced on the sphee without having touched it with a positive chage Coulomb s Law As has just been seen, electic chages exet foces on each othe. But what ae the magnitudes of these foces? Chales Augustin de Coulomb ( ) invented a tosion balance in 1777 in which a quantity of foce could be measued by the amount of twist it poduced in a thin wie. In 1785 he used this tosion balance to measue the foce between electical chages. If vey small spheical chages, q 1 and q 2, called point chages, ae sepaated by a distance between thei 20-9

10 centes, figue 20.8, then Coulomb found that the foce between the chages could be stated as: The foce between the point chages q 1 and q 2 is diectly popotional q q 1 2 Figue 20.8 Coulomb s Law. to the poduct of thei chages and invesely popotional to the squae of the distance sepaating them. The diection of this foce lies along the line sepaating the chages. This esult is known as Coulomb s law. Coulomb s law can be stated mathematically as F = k q 1q 2 (20.1) 2 whee k is a constant depending upon the units employed and on the medium in which the chages ae located. Fo a vacuum k = N m 2 /C 2 Fo ai the value of k is so close to the value of k fo a vacuum that the same value will be used fo both. To simplify the solution of poblems in this text the value of k used will be ounded off to the value k = N m 2 /C 2 To simplify moe advanced theoies of electomagnetism, Coulomb s law is also witten in the fom F = 1 q 1 q 2 (20.2) 4 o 2 whee k = 1 (20.3) 4 o and ε o = C 2 /N m 2 and is called the pemittivity of fee space. If the chages ae placed in a medium othe than ai o vacuum, then thee will be a diffeent value fo the pemittivity ε fo that medium and hence a diffeent value of k. In this text only the simple fom, equation 20.1, of Coulomb s law will be used. If the chages ae much lage than point chages, Coulomb s law can still be used if the distance sepaating the chages is quite lage compaed to the size of the electic chage. Unde these cicumstances the chages appoximate point chages. Let us conside some examples of the use of Coulomb s law. Notice the similaity between the fom of Coulomb s law of electostatics and Newton s law of univesal gavitation: 20-10

11 F = Gm 1m 2 2 (6.37) Let us compae the electic and gavitational foces between an electon and a poton in a hydogen atom. The mass of the poton is m p = kg, wheeas the mass of the electon is m e = kg. The adius of the lowest enegy obit fo the electon is = m. The chage on the electon and poton is C. The gavitational foce between the electon and the poton is found fom Newton s law of univesal gavitation, equation 6.37, as F g = Gm pm e 2 = (6.67 % Nm 2 /kg 2 )(1.67 % kg)(9.11 % kg) (5.29 % m) 2 = N The electic foce between the electon and the poton is found fom Coulomb s law of electostatics, equation 20.1, as F e = kq pq e 2 = (9.00 % 109 Nm 2 /C 2 )(1.60 % C) 2 (5.29 % m) 2 = N Although both foces seem quite small, let us compae the elative magnitude of these foces by taking the atio of the electic foce to the gavitational foce, that is, o F e F g = 8.23 % 10 8 N 3.63 % N F e = ( )F g = 2.27 % 1039 That is, fo the electon-poton system discussed hee, the electic foce is times geate than the gavitational foce. Because magnitudes ae sometimes had to visualize in scientific notation fo the beginning student, we can also wite this numbe as F e = 2,270,000,000,000,000,000,000,000,000,000,000,000,000 F g Theefoe, on the atomic level the gavitational foce is an extemely weak foce, wheeas the electic foce is vey lage. Hence, in the solution of electostatics poblems the gavitational foce can be ignoed when compaed to the electic foce. On an atomic level, the gavitational foce is indeed vey weak. By contast, howeve, when extemely lage masses ae involved, such as in a lage dying sta, the gavitational foce is so geat, that electons ae diven ight into the nuclei of 20-11

12 atoms, conveting the nuclea potons into neutons, and ceating a neuton sta. The density of the matte in such a neuton sta is so geat that one tablespoon of that matte would weigh 10 billion tons if it wee located in the gavitational field at the suface of the eath. Let us now conside the electostatic foce between the two potons in a helium nucleus. The potons ae sepaated by a distance of about m. The foce between the two potons is F pp = kq pq p 2 = (9.00 % 109 Nm 2 /C 2 )(1.60 % C) 2 (2.40 % m) 2 = 40.0 N If we compae this epulsive foce between the two potons in the helium nucleus to the attactive electostatic foce between the electon and the poton in the hydogen atom, we see that thei atio is o F pp F ep = 40.0 N 8.23 % 10 8 N F pp = F ep = 4.86 % 108 The foce between the two potons in the helium nucleus is an enomous foce and hence the helium nucleus should blow itself apat. The fact that it does not is an indication of the existence of anothe foce, the stong nuclea foce, that holds the potons togethe within the nucleus. Fom the calculation, the nuclea foce holding the nucleus togethe is at least 10 8 times geate than the electic foce holding the atom togethe, that is, F N = 10 8 F A Let us conside some examples of the use of Coulomb s law. Example 20.1 Coulomb s law fo two point chages. A point chage, q 1 = 2.00 µc is placed m fom anothe point chage q 2 = 5.00 µc. Calculate the magnitude and diection of the foce on each chage. F 12 F 21 +q 1 -q 2 Diagam fo example

13 Solution The foce acting on chage q 1, is a foce of attaction caused by the negative chage of q 2. This foce will be called F 12 (foce on chage 1 caused by chage 2). Since this foce is in the positive x-diection it can be witten as F 12 = if 12 The magnitude of the foce F 12 is found by equation 20.1 as and the foce is F 12 = kq 1q 2 2 F 12 = 9.00 % 10 9 Nm 2 C 2 (2.00 % 10 6 C)(5.00 % 10 6 C) (0.500 m) 2 F 12 = N F 12 = (0.360 N)i The foce acting on chage q 2, F 21, is a foce of attaction caused by chage q 1. Since this foce points in the negative x-diection, it can be witten as F 21 = F 21 i The magnitude of the foce F 21 is found fom Coulomb s law and is given by and the foce is F 21 = kq 2q 1 2 F 21 = 9.00 % 10 9 Nm 2 C 2 (5.00 % 10 6 C)(2.00 % 10 6 C) (0.500 m) 2 F 21 = N F 21 = (0.360 N)i Note that the magnitudes of the foces on q 1 and q 2 ae identical, but thei diections ae opposite. This could have been deduced immediately fom Newton s thid law, fo if chage 1 exets a foce on chage 2, then chage 2 must exet an equal but opposite foce on chage 1. To go to this Inteactive Example click on this sentence

14 Example 20.2 The effect of touching two chaged sphees. Two identical metal sphees ae placed m apat. A chage q 1 of 9.00 µc is placed on one sphee while a chage q 2 of 3.00 µc is placed upon the othe. (a) What is the foce on each of the sphees? (b) If the two sphees ae bought togethe and touched and then etuned to thei oiginal positions, what will be the foce on each sphee? F 12 F F F 12 +q 1 -q 2 +q 1 +q 2 (a) Diagam fo example 20.2 Solution (b) a. The magnitude of the attactive foce on sphee 1 is found fom Coulomb s law as F 12 = kq 1q 2 2 F 12 = 9.00 % 10 9 Nm 2 C 2 (9.00 % 10 6 C)(3.00 % 10 6 C) (0.200 m) 2 F 12 = 6.08 N and the foce is found as F 12 = (6.08 N)i The foce on sphee 2 is equal and opposite to this. b. When the sphees ae touched togethe, the 3.00 µc of chage q 2, neutalizes µc of chage q 1, leaving a net chage of C C = C This chage of C is then equally distibuted between sphee one and sphee two, giving q 1 = q 2 = C. Since the chages on both sphees ae now positive, the foce on each sphee is now epulsive. When the sphees ae emoved to the oiginal distance the magnitude of the foce on each sphee is now F = kq 2 q 1 2 = 9.00 % 10 9 Nm 2 (3.00 % 10 6 C)(3.00 % 10 6 C) C 2 (0.200 m)

15 F = 2.03 N To go to this Inteactive Example click on this sentence Multiple Discete Chages If thee ae thee o moe chages pesent, then the foce on any one chage is found by the vecto addition of the foces associated with the othe chages. That is, the esultant foce on any one chage is given by F = F 1 + F 2 + F 3 + F (20.4) Example 20.3 Multiple chages in a line. Thee chages ae placed on the line as shown. The sepaation of the chages ae 12, = m and 23 = m. If q 1 = 1.00 µc, q 2 = 2.00 µc, and q 3 = 3.00 µc, find: (a) the esultant foce on chage q 1, (b) the esultant foce on chage q 2, and (c) the esultant foce on chage q q 1 q 2 q 3 Diagam fo example Solution The esultant foce on each chage is equal to the vecto sum of all the foces acting on that chage. a. As can be seen in diagam 20.3(a), the foce on chage 1 is F 1 = F 12 + F 13 whee F 12 is the foce on chage 1 caused by chage 2, and F 13 is the foce on chage 1 caused by chage 3. Because q 2 is negative, F 12 is a foce of attaction to the ight and is given by F 12 = if

16 F 13 F 12 q 1 q 2 q 3 Diagam fo example 20.3(a). Since q 3 is positive, F 13 is a foce of epulsion to the left and is given by The total foce on chage 1 is theefoe F 13 = if 13 whee while Theefoe, F 1 = if 12 if 13 F 12 = kq 1q F 12 = 9.00 % 10 9 Nm 2 (1.00 % 10 6 C)(2.00 % 10 6 C) C 2 (0.500 m) 2 F 12 = N F 13 = kq 1q F 13 = 9.00 % 10 9 Nm 2 (1.00 % 10 6 C)(3.00 % 10 6 C) C 2 (1.00 m) 2 F 13 = N F 1 = if 12 if 13 = ( N)i ( N)i F 1 = ( N)i That is, the esultant foce on chage 1 is a foce of N to the ight. b. The esultant foce on chage 2 is o as can be seen fom diagam 20.3b F 2 = F 21 + F 23 Fom Newton s thid law and F 2 = if 23 if 21 F 21 = F 12 = ( N)i 20-16

17 F 21 F 23 q 1 q 2 q 3 Diagam fo example 20.3(b). F 23 = kq 2q F 23 = 9.00 % 10 9 Nm 2 (2.00 % 10 6 C)(3.00 % 10 6 C) C 2 (0.500 m) 2 F 23 = N Theefoe, the net foce on chage 2 is to the ight. F 2 = if 23 if 21 = (0.216 N)i ( N)i F 2 = (0.144 N)i c. The esultant foce on chage 3 is F 3 = F 31 + F 32 and as can be seen in diagam 20.3(c) this can be witten as F 3 = if 31 if F 32 F 31 q 1 q 2 q 3 Diagam fo example 20.3(c). But by Newton s thid law and F 31 = F 13 = ( N)i 20-17

18 F 32 = F 23 = ( N)i The net foce on chage 3 is theefoe to the left. F 3 = ( N)i ( N)i F 3 = (0.189 N)i To go to this Inteactive Example click on this sentence. Example 20.4 Multiple chages not on a line. Find the esultant foce on chage q 2 in the diagam if q 1 = 13.0 µc, q 2 = 4.00 µc, q 3 = 5.00 µc. 13 = m, 23 = m, y F 2 F 23 F 21y q 2 F 21 θ F 21x x θ 13 q 1 q 3 Diagam fo example 20.4 Solution The esultant foce on chage q 2 is found as F 2 = F 21 + F

19 whee F 21 is the foce on chage 2 caused by chage 1, and F 23 is the foce on chage 2 caused by chage 3. The distance 12 is found fom the diagam and the Pythagoean theoem as 12 = = (0.500 m) 2 + (0.800 m) 2 12 = m The magnitude F 21 is found fom Coulomb s law as F 21 = kq 2q F 21 = 9.00 % 10 9 Nm 2 (4.00 % 10 6 C)(13.00 % 10 6 C) C 2 (0.940 m) 2 F 21 = N while the magnitude of F 23 is F 23 = kq 2q F 23 = 9.00 % 10 9 Nm 2 (4.00 % 10 6 C)(5.00 % 10 6 C) C 2 (0.800 m) 2 F 23 = N The addition of the two vectos is an example of the addition of vectos discussed in chapte 3. The esultant vecto is given by F 2 = if 2x + jf 2y The magnitude of the esultant vecto is found, as befoe, as The foce F 21 is given by F 2 = (F 2x ) 2 + (F 2y ) 2 F 21 = if 21x + jf 21y The vecto F 21 has an x-component given by F 21x = F 21 cosθ The angle θ is found fom the geomety of the diagam as Theefoe, the x-component of F 21 is = tan = F 21x = F 21 cosθ = (0.530 N)cos = N 20-19

20 Similaly, the y-component of F 21 is F 21y = F 21 sinθ = (0.530 N)sin = N The vecto F 23 is in the y diection and is given by F 23 = jf 23 and has no x-component. Theefoe, the x-component of the esultant vecto is F 2x = F 21x = N and as can be seen fom the diagam the y-component of the esultant vecto is The esultant foce on chage 2 is F 2y = F 23 + F 21y = N N = N F 2 = if 2x + jf 2y = (0.281 N)i + (0.730 N)j The magnitude of the esultant foce on chage 2 is theefoe F 2 = (F 2x ) 2 + (F 2y ) 2 = (0.281 N) 2 + (0.730 N) 2 F 2 = N The angle φ that F 2 makes with the x-axis is found fom $ = tan 1 F 2y F 2x = tan φ = To go to this Inteactive Example click on this sentence. Example 20.5 Moe multiple chages. Find the esultant foce on chage q 2 in the diagam if q 1 = 3.00 µc, q 2 = 5.00 µc, q 3 = 4.00 µc, 12 = m and 23 = Solution The esultant foce on chage 2 is found fom the vecto sum 20-20

21 q q 1 12 F 21 q θ 2 F 23 F 2 Diagam fo example 20.5 F 2 = F 21 + F 23 Because F 21 is pependicula to F 23, the magnitude of the esultant foce is whee and F 2 = (F 21 ) 2 + (F 23 ) 2 F 21 = kq 2q F 21 = 9.00 % 10 9 Nm 2 (5.00 % 10 6 C)(3.00 % 10 6 C) C 2 (0.500 m) 2 F 21 = N F 23 = kq 2q F 23 = 9.00 % 10 9 Nm 2 (5.00 % 10 6 C)(4.00 % 10 6 C) C 2 (0.500 m) 2 F 23 = N As can be seen fom the diagam, the esultant foce on chage 2 is F 2 = if 21 jf 23 F 3 = (0.540 N)i (0.720)j The magnitude of the esultant foce on chage 2 is theefoe F 2 = (F 21 ) F 23 = (0.540 N) 2 + (0.720 N) 2 F 2 = N 20-21

22 The diection of the esultant foce is detemined by = tan 1 F 23 F 21 = tan θ = To go to this Inteactive Example click on this sentence. Example 20.6 Chaged pith balls. Two equally chaged pith balls ae sepaated by m as shown in figue 20.9(a). Find the chage on each ball and the tension in the sting if the mass of each ball is kg, and the length l of the sting is m. l T T y (a) θ (b) T x θ w F e Figue 20.9 Chaged pith balls. Solution Let us conside the foces acting on the ball at the ight. The foces acting ae the tension T in the sting, the weight w of the ball, and the electic foce of epulsion on the ball F e as shown in figue 20.9(b). Since the ball is in equilibium the fist condition of equilibium is applied as Σ F y = 0, Σ F x = 0 T y w = 0, F e T x = 0 T sin θ = w, F e = T cos θ F e = T cos θ = 1 w T sin θ tan θ 20-22

23 F e = w tan θ kq 2 = mg 2 tan θ Solving fo the chage q on each ball we get q = 2 mg k tan The angle θ, found fom the geomety of the figue 20.9(a), is θ = cos 1 /2 l = cos cm 25.0 cm = Theefoe, the chage on each ball is 2 mg q = k tan (0.100 m) 2 (5.00 % 10 = 3 kg)(9.80 m/s 2 ) (9.00 % 10 9 Nm 2 /C 2 ) tan The tension in the sting is = C T sin θ = w T = w sin θ = mg sin θ = ( kg)(9.80 m/s 2 ) sin = N To go to this Inteactive Example click on this sentence. In section 20.5, and in paticula in equation 20.1, we wote Coulomb s law in tems of its magnitude only. In this section we have been using the unit vectos i and j to descibe the diection of the foce vecto. Coulomb s law can be witten in a moe geneal fom as follows. Let q 1 be the pimay chage that we ae inteested in. We will intoduce a unit vecto o that points eveywhee adially away fom the chage q 1 as shown in figue 20.10(a)

24 F 21 o +q 2 o F 12 F 21 +q 2 F 12 +q 1 -q 1 (a) Foce on + chage (b) Foce on chage Figue Coulomb s law in vecto fom. foce If a chage + q 2 is bought into the vicinity of chage q 1, it will expeience the F 21 = k q 2q 1 2 o (20.5) whee F 21 is the foce on chage q 2 caused by chage q 1. If q 1 and q 2 ae of like sign then the foce on the seconday chage q 2 is in the same diection as the unit vecto o, and the foce is one of epulsion as expected and is shown in figue 20.10(a). If the pimay chage is negative, that is, q 1, then the chages ae of opposite sign and the foce on chage q 2 is in the opposite diection of the unit vecto o, and the foce is one of attaction as seen in figue 20.10(b). By Newton s thid law, the foce on chage q 1 is equal and opposite to the foce on chage q 2 as expected. That is, F 12 = F Foces Caused by a Continuous Distibution of Chage As we have seen in the last section, when thee ae multiple discete chages in a egion, then the foce on any one of those chages is found by the vecto sum of the foces associated with the othe chages. That is, the esultant foce on any one chage was given by equation 20.4 as F = F 1 + F 2 + F 3 + F 4 +. A shothand notation fo equation 20.4 can be witten as N F= i=1 F i (20.6) 20-24

25 whee, again, Σ, the Geek lette sigma, means the sum of and the sum goes fom i = 1 to i = N. Besides the foces caused by a discete distibution of chage, the foce on a single chage q o caused by a continuous distibution of chage can be handled in a simila way. The unifom distibution of chage can be boken up into a lage numbe of infinitesimal elements of chage, dq, and each element of chage will poduce an element of foce df on the discete chage q o. Coulomb s Law can then be witten as df= k q odq (20.7) 2 o whee is the distance fom the element of chage dq to the single discete chage q o. o is a unit vecto that points fom the element of chage dq, and points towad the discete chage q o. The total foce F on the discete chage q o, caused by the foces fom the entie distibution of all the dq s is again a sum, but since the elements of chage dq, ae infinitesimal the sum becomes the integal of all the elements of foce df. That is, the foce is found as F= df= k q odq 2 o (20.8) Remembe that q o is a constant in this integation. As an example of the foce caused by a continuous distibution, let us find the foce acting on a chage q o located at the oigin of a semicicula ing of chage as shown in figue The ing has a adius and caies a continuous distibution of chage q. A small element of this chage dq is shown in the ight quadant of the diagam. This element of chage dq exets the foce df on the discete chage q o, given by equation 20.7, and is shown in the diagam. The total foce F is found fom equation 20.8, which is a vecto integation. df df cosθ df sinθ Chapte 20 Electostatics and Coulomb s Law dq df y θ θ q θ o θ dq df x df sinθ df cosθ Figue Foce on a point chage caused by a continuous distibution of chage. The vecto integation can be simplified into a scala integation by noting that the element of chage dq has a mio image of chage of the same magnitude on the othe half of the semicicula ing as shown in the second quadant of the diagam. Each chage poduces an element of foce df. This element of foce can be boken up into two components, one, df cosθ, which lies along the x-axis, while the df

26 second component, df sinθ, lies along the negative y-axis. When we add up (integate) the effect of all the chages dq we see that half the components df cosθ, ae in the positive x-diection, while the othe half ae in the negative x-diection. Thus, the sum of all the x-components of the foces will be zeo, and we need only conside the y-components. Both y-components, df sinθ, ae in the negative y-diection, and the total foce can be witten as the scala integation F = 2dF sin = 2k q odq 2 sin (20.9) Notice that, in geneal, k and q o, ae constants and, fo this paticula poblem,, the adius of the ing is also a constant. Hence, they can all be taken outside the integal sign. The integal is now given as F = 2kq o 2 sin dq (20.10) Notice that the integation is ove dq. Rathe than integating ove a chage, it is easie to integate ove a geometical figue. We define a linea chage density λ as the amount of chage pe unit length, that is, = q s (20.11) whee q is the total chage on the semicicula ing and s is the total length of the semicicula ing and also epesents the ac of the semicicula ing. Solving equation fo q gives q = λs and upon diffeentiating fo the element of chage, dq, we get Replacing equation into equation we get dq = λ ds (20.12) F = 2kq o 2 sin ds (20.13) Now the sinθ and ds ae not independent and the elation between them must be stated befoe the integation can begin. The angle θ is elated to the adius and the ac s of a cicle by s = θ Diffeentiating this equation we get ds = dθ (20.14) 20-26

27 whee ds is an element of an ac of the semicicula ing, is the adius of the ing, and dθ is the small angle subtended by the ac ds. Replacing equation into equation gives F = 2kq o 2 sin d Assuming λ is a constant, it can be taken out of the integal to yield F = 2kq o 0 /2 sin d (20.15) Notice that the integation is now ove the angle θ, and the integation is fom θ = 0 to θ = π/2. Integating equation gives F = 2kq o /2 ( cos ) 0 F = 2kq o cos 2 cos 0 F = 2kq o (0 1) F = 2kq o (20.16) Equation gives the foce acting on the discete chage q o in tems of the linea chage density λ. It can be expessed in tems of the total chage q on the semicicula ing by witing equation as = q s = q (20.17) Notice that the ac s of the semicicula ing is equal to half the cicumfeence of a cicle, namely, π. Substituting equation into equation we get F = 2kq o q F = 2kq oq 2 (20.18) Equation gives the foce on a discete chage q o by a continuous distibution of chage q along the suface of a semicicula ing of adius. The foce on a discete chage q o caused by any othe continuous distibution of chage can be found in the same way. The Language of Physics Electostatics The study of electic chages at est unde the action of electic foces (p. )

28 The Fundamental Pinciple of Electostatics Like electic chages epel each othe while unlike electic chages attact each othe (p. ). Quaks Elementay paticles of matte. Thee ae six quaks. They ae: up, down, stange, cham, bottom, and top. The poton and neuton ae made of quaks, but the electon is not (p. ). Conductos Mateials that pemit the fee flow of electic chage though them (p. ). Insulatos o Dielectics Mateials that do not pemit the fee flow of electic chage though them (p. ). Coulomb s law The foce between point chages q 1 and q 2 is diectly popotional to the poduct of thei chages and invesely popotional to the squae of the distance sepaating them. The diection of the foce lies along the line sepaating the chages (p. ). Linea chage density λ is defined as the amount of chage pe unit length (p. ). Summay of Impotant Equations Coulomb s law F = k q 1q 2 (20.1) 2 Foce caused by multiple discete chages F = F 1 + F 2 + F 3 + F 4 + (20.4) Coulomb s law in vecto fom F 21 = k q 2q 1 (20.5) Coulomb s law in diffeential fom df= k q odq (20.7) Foced caused by a continuous chage distibution (20.8) 2 2 o o F= df= k q odq Linea chage density = q s (20.11) Linea element of chage dq = λ ds (20.12) 2 o 20-28

29 Questions fo Chapte Descibe the pocess of chaging by induction. 2. What did Ben Fanklin have to do with the classification of electical chage? 3. The diffeence between one chemical element and anothe is the numbe of potons that each contains. Is it possible fo one chemical element to have the same numbe of neutons as anothe chemical element? 4. Discuss the planetay model of the atom and state its good points and its limitations. 5. Is it possible fo a quak to exist if it has not been seen? If it cannot be isolated does it make any sense to descibe paticles as though they wee made up of quaks? 6. Descibe the pocess of lightning in the atmosphee. How do the eath and the clouds pick up electic chage? If ai is an insulato, how can a lightning bolt eve each the gound? 7. Can you think of a way that you could use electostatics to measue the humidity of the atmosphee? 8. How could you paint a metallic object with a minimum of paint using the pinciples of electostatics? 9. Descibe the phenomenon known as St. Elmo s Fie in tems of electostatics. 10. Why do clothes taken fom a clothes dye sometimes cling to the body? Poblems fo Chapte 20 Section 20.5 Coulomb s Law. 1. A point chage of 4.00 µc is placed 25.0 cm fom anothe point chage of 5.00 µc. Calculate the magnitude and diection of the foce on each chage. 2. If the foce of epulsion between two potons is equal to the weight of the poton, how fa apat ae the potons? 3. What equal positive chages would have to be placed on the eath and the moon to neutalize the gavitational foce between them? 4. What is the velocity of an electon in the hydogen atom if the centipetal foce is supplied by the coulomb foce between the electon and poton? The adius of the electon obit is m. 5. Two identical metal sphees ae placed 15.0 cm apat. A chage of 6.00 µc is placed on one sphee while a chage of 2.00 µc is placed upon the othe. What is the foce on each sphee? If the two sphees ae bought togethe and touched and then sepaated to thei oiginal sepaation, what will be the foce on each sphee? 20-29

30 6. Two identical metal sphees, caying diffeent opposite chages, attact each othe with a foce of N when they ae 5.00 cm apat. The sphees ae then touched togethe and then emoved to the oiginal sepaation whee now a foce of epulsion of N is obseved. What is the chage on each sphee afte touching and befoe touching? Section 20.6 Multiple Chages. 7. Thee chages q 1 = 2.00 µc, q 2 = 5.00 µc, and q 3 = 8.00 µc ae placed along the x-axis at 0.00, 45.0 cm, and 72.4 cm espectively. Find the foce on each chage. 8. Two chages q 1 = 4.00 µc and q 2 = 8.00 µc ae sepaated by the distance = 50.0 cm as shown. Whee should a thid chage be placed on the line between them such that the esultant foce on it will be zeo? Does it matte if the thid chage is positive o negative? q 1 q 2 Diagam fo Poblem Repeat poblem 8 but now let chage q 1 be negative, and find any position on the line, eithe to the left of q 1, between q 1 and q 2, o to the ight of q 2, whee a thid chage can be placed that expeiences a zeo esultant foce. 10. Thee chages of 2.00 µc, 4.00 µc, and 6.00 µc ae placed at the vetices of an equilateal tiangle of length 10.0 cm on a side. Find the esultant foce on each chage. 11. If q 1 = 5.00 µc = q 2 = q 3 = q 4 ae located on the cones of a squae of length 20.0 cm, find the esultant foce on q 3. q 1 q 1 q 2 q 3 l q 3 q 4 q 2 Diagam fo poblem 11. Diagam fo poblem Chages of 2.54 µc, 7.86 µc, 5.34 µc, and 3.78 µc ae placed on the cones of a squae of side 23.5 cm. Find the esultant foce on the fist chage. 13. Find the foce on chage q 3 = 1.00 µc, if q 1 = µc and q 2 = 5.00 µc. The distance sepaating chages q 1 and q 2 is 5.00 cm, and l = 1.00 m

31 14. Find the esultant foce on chage q 3 in the diagam if q 1 = 2.00 µc, q 2 = 7.00 µc, q 3 = 5.00 µc, 12 = m, 23 = m. q 3 23 θ q 1 q 12 2 Diagam fo poblem 14. Section 20.7 Foce Caused by a Continuous Distibution of Chage. 15. A thin nonconducting od of length l caies a unifom chage pe unit length λ. The od lies on the + x-axis with one end at the point x o and the othe end at x = x o + l. A point chage q o lies on the x-axis at the oigin. Find the foce acting on chage q o. 16. This is simila to poblem 15 except the point chage q o is located on the y-axis at the point y o. (a) Find the x-component of the foce on the point chage caused by the line of chage. (b) Find the y-component of the foce on the point chage caused by the line of chage. Additional Poblems. 17. Find the foce on chage q 5 = 5.00 µc, located at the cente of a squae 25.0 cm on a side if q 1 = q 2 = 3.00 µc and q 3 = q 4 = 6.00 µc. q 1 q 2 q l = 25 cm q 3 q 4 q 1 q 2 Diagam fo poblem 17. Diagam fo poblem Two small, equally chaged, sphees of mass gm ae suspended fom the same point by a silk fibe 50.0 cm long. The epulsion between them keeps them 15.0 cm apat. What is the chage on each sphee? 19. Two pith balls of 10.0 gm mass ae hung fom ends of a sting 25.0 cm long, as shown. When the balls ae chaged with equal amounts of chage, the theads sepaate to an angle of What is the chage on each ball? 20-31

32 20. Two 10.0 gm pith balls ae hung fom the ends of two 25.0 cm long stings as shown. When an equal and opposite chage is placed on each ball, thei sepaation is educed fom 10.0 cm to 8.00 cm. Find the tension in each sting and the chage on each ball. q 1 q 2 8 cm 10 cm Diagam fo poblem A chage of 15.0 µc is on a metallic sphee 10.0 cm adius. It is then touched to a sphee of 5.00 cm adius, until the suface chage density is the same on both sphees. What is the chage on each sphee afte they ae sepaated? (The suface chage density σ is the chage pe unit aea and is given by σ = q/a.) 22. Two small sphees caying chages q 1 = 7.00 µc and q 2 = 5.00 µc ae sepaated by 20.0 cm. If q 2 wee fee to move what would its initial acceleation be? Sphee 2 has the mass m 2 = 15.0 gm. 23. Whee should a fouth chage, q 4 = 3.00 µc, be placed to give a net foce of zeo on chage q 3? Chage q 1 = 2.00 µc, q 2 = 4.00 µc, and q 3 = 2.00 µc. q 1 1 m +q q o 1 m q 3 2a x θ -q q 2 +q q o Diagam fo poblem 23. Diagam fo poblem Chage q 1 = 3.00 µc, is located at the coodinates (0,2) and chage q 2 = 6.00 µc, is located at the coodinates (1,0) of a Catesian coodinate system. Find the coodinates of a thid chage that will expeience a zeo net foce. 25. The configuation of a positive chage q sepaated by a distance 2a fom a negative chage q is called an electic dipole. Show that the foce exeted by an electic dipole on a point chage q o, located as shown in the diagam vaies as 1/

33 while the foce between a point chage q and the point chage q o vaies as 1/ 2. Which foce is the weake? 26. A plastic od 50.0 cm long, has a chage + q 1 = 2.00 µc at each end. The od is then hung fom a sting and placed so that each chage is only 5.00 cm fom negative chages q 2 = 10.0 µc as shown in the diagam. Find the toque acting on the sting. 27. A chage of 5.00 µc is unifomly distibuted ove a coppe ing 2.00 cm in adius. What foce will this ing exet on a point chage of 8.00 µc that is placed 3.00 m away fom the ing. Indicate what assumptions you make to solve this poblem. 28. A chage of 2.50 µc is placed at the cente of a hollow sphee of chage of 8.00 µc. What is the esultant foce on the chage placed at the cente of the sphee? Indicate what assumptions you make to solve this poblem. q 2 q 1 q 1 q 2 Diagam fo poblem A thin nonconducting od of length l caies a nonunifom chage density λ = Ax 2. The od lies on the x-axis with one end at the oigin and the othe end at x = l. A point chage q o lies on the x-axis at the point x o. Find the foce acting on chage q o. 30. Find the foce acting on a point chage q o located at the oigin of a semicicula ing of chage as shown in figue The ing has a adius and caies a nonunifom continuous distibution of chage density λ = A sinθ. Inteactive Tutoials 31. Coulomb s law. Two chages q 1 = C and q 2 = C ae sepaated by a distance = 1.00 m. Calculate the electostatic foce F of epulsion acting on chage 1 as the distance of sepaation is inceased fom = 1 to = 10 m. Show how the foce F vaies with the distance. 32. Coulomb s law. Two electons of chage q 1 = q 2 = e = C ae positioned at the coodinates (0,1) and (0, 1) (in metes) of a Catesian coodinate system. Calculate the net foce F 3 on an electon q 3 as it is moved fom x = 0 to x = 10.0 m along the x-axis

34 33. Coulomb s law and multiple chages. Two chages q 1 = C and q 2 = C lie on the x-axis and ae sepaated by the distance 12 = m. A thid chage q 3 = C is located a distance 23 = m fom chage q 2, and the line between chage 2 and 3 makes an angle φ = with espect to the x-axis. Find the esultant foce on (a) chage 3, (b) chage 2, and (c) chage Coulomb s law and a continuous chage distibution. A od of chage of length L = m lies on the x-axis. One end of the od lies at the oigin and the othe end is on the positive x-axis. A chage q = C is unifomly distibuted ove the od. Find the foce exeted on a point chage q = C that lies on the x-axis at distance x 0 = m fom the oigin of the coodinate system. To go to these Inteactive Tutoials click on this sentence. To go to anothe chapte, etun to the table of contents by clicking on this sentence