Chapter 2 Coulomb s Law


 Godfrey Warren
 4 years ago
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1 Chapte Coulomb s Law.1 lectic Chage Coulomb's Law...3 Animation.1: Van de Gaaff Geneato Pinciple of Supeposition...5 xample.1: Thee Chages lectic Field...7 Animation.: lectic Field of Point Chages lectic Field Lines Foce on a Chaged Paticle in an lectic Field lectic Dipole The lectic Field of a Dipole...1 Animation.3: lectic Dipole Dipole in lectic Field Potential neg of an lectic Dipole Chage Densit Volume Chage Densit Suface Chage Densit Line Chage Densit lectic Fields due to Continuous Chage Distibutions xample.: lectic Field on the Axis of a Rod xample.3: lectic Field on the Pependicula Bisecto xample.4: lectic Field on the Axis of a Ring...1 xample.5: lectic Field Due to a Unifoml Chaged Disk Summa PoblemSolving Stategies Solved Poblems Hdogen Atom Millikan OilDop xpeiment Chage Moving Pependiculal to an lectic Field lectic Field of a Dipole
2 .13.5 lectic Field of an Ac lectic Field Off the Axis of a Finite Rod Conceptual Questions Additional Poblems Thee Point Chages Thee Point Chages Fou Point Chages Semicicula Wie lectic Dipole Chaged Clindical Shell and Clinde Two Conducting Balls Toque on an lectic Dipole
3 Coulomb s Law.1 lectic Chage Thee ae two tpes of obseved electic chage, which we designate as positive and negative. The convention was deived fom Benjamin Fanklin s expeiments. He ubbed a glass od with silk and called the chages on the glass od positive. He ubbed sealing wax with fu and called the chage on the sealing wax negative. Like chages epel and opposite chages attact each othe. The unit of chage is called the Coulomb (C). The smallest unit of fee chage known in natue is the chage of an electon o poton, which has a magnitude of e = C (.1.1) Chage of an odina matte is quantized in integal multiples of e. An electon caies one unit of negative chage, e, while a poton caies one unit of positive chage, + e. In a closed sstem, the total amount of chage is conseved since chage can neithe be ceated no destoed. A chage can, howeve, be tansfeed fom one bod to anothe.. Coulomb's Law Conside a sstem of two point chages, and q, sepaated b a distance in vacuum. q1 The foce exeted b on q is given b Coulomb's law: q1 F qq 1 1 = k ˆ e (..1) whee ke is the Coulomb constant, and ˆ = / is a unit vecto diected fom q1 to q, as illustated in Figue..1(a). (a) (b) Figue..1 Coulomb inteaction between two chages Note that electic foce is a vecto which has both magnitude and diection. In SI units, the Coulomb constant is given b k e 3
4 k e = 1 = N m / C (..) whee 1 1 ε = = C N m (..3) 9 4 π ( N m C ) is known as the pemittivit of fee space. Similal, the foce on q1 due to q is given b F = F, as illustated in Figue..1(b). This is consistent with Newton's thid law. 1 1 As an example, conside a hdogen atom in which the poton (nucleus) and the electon 11 ae sepaated b a distance = m. The electostatic foce between the two 8 paticles is appoximatel F = k e / = 8. 1 N. On the othe hand, one ma show that the gavitational foce is onl neglected when dealing with electostatic foces! e e F g N. Thus, gavitational effect can be Animation.1: Van de Gaaff Geneato Conside Figue..(a) below. The figue illustates the epulsive foce tansmitted between two objects b thei electic fields. The sstem consists of a chaged metal sphee of a van de Gaaff geneato. This sphee is fixed in space and is not fee to move. The othe object is a small chaged sphee that is fee to move (we neglect the foce of gavit on this sphee). Accoding to Coulomb s law, these two like chages epel each anothe. That is, the small sphee expeiences a epulsive foce awa fom the van de Gaaff sphee. Figue.. (a) Two chages of the same sign that epel one anothe because of the stesses tansmitted b electic fields. We use both the gass seeds epesentation and the field lines epesentation of the electic field of the two chages. (b) Two chages of opposite sign that attact one anothe because of the stesses tansmitted b electic fields. The animation depicts the motion of the small sphee and the electic fields in this situation. Note that to epeat the motion of the small sphee in the animation, we have 4
5 the small sphee bounce off of a small squae fixed in space some distance fom the van de Gaaff geneato. Befoe we discuss this animation, conside Figue..(b), which shows one fame of a movie of the inteaction of two chages with opposite signs. Hee the chage on the small sphee is opposite to that on the van de Gaaff sphee. B Coulomb s law, the two objects now attact one anothe, and the small sphee feels a foce attacting it towad the van de Gaaff. To epeat the motion of the small sphee in the animation, we have that chage bounce off of a squae fixed in space nea the van de Gaaff. The point of these two animations is to undescoe the fact that the Coulomb foce between the two chages is not action at a distance. Rathe, the stess is tansmitted b diect contact fom the van de Gaaff to the immediatel suounding space, via the electic field of the chage on the van de Gaaff. That stess is then tansmitted fom one element of space to a neighboing element, in a continuous manne, until it is tansmitted to the egion of space contiguous to the small sphee, and thus ultimatel to the small sphee itself. Although the two sphees ae not in diect contact with one anothe, the ae in diect contact with a medium o mechanism that exists between them. The foce between the small sphee and the van de Gaaff is tansmitted (at a finite speed) b stesses induced in the intevening space b thei pesence. Michael Faada invented field theo; dawing lines of foce o field lines was his wa of epesenting the fields. He also used his dawings of the lines of foce to gain insight into the stesses that the fields tansmit. He was the fist to suggest that these fields, which exist continuousl in the space between chaged objects, tansmit the stesses that esult in foces between the objects..3 Pinciple of Supeposition Coulomb s law applies to an pai of point chages. When moe than two chages ae pesent, the net foce on an one chage is simpl the vecto sum of the foces exeted on it b the othe chages. Fo example, if thee chages ae pesent, the esultant foce expeienced b due to and q will be q3 q1 F3 = F13+ F 3 The supeposition pinciple is illustated in the example below. xample.1: Thee Chages Thee chages ae aanged as shown in Figue.3.1. Find the foce on the chage (.3.1) assuming that q 1 = 6. 1 C, q = q1 = 6. 1 C, q 3 = C and a =. 1 m. q 35
6 Figue.3.1 A sstem of thee chages Solution: Using the supeposition pinciple, the foce on q 3 is F = F + F = 1 qq ˆ qq + ˆ In this case the second tem will have a negative coefficient, since q is negative. The unit vectos ˆ 13 and ˆ 3 do not point in the same diections. In ode to compute this sum, we can expess each unit vecto in tems of its Catesian components and add the foces accoding to the pinciple of vecto addition. Fom the figue, we see that the unit vecto ˆ 13 which points fom q1 to q3 can be witten as ˆ ˆ ˆ ˆ 13 = cosθ i+ sin θ j = ( i +ˆj ) Similal, the unit vecto ˆ 3 = ˆi points fom q to q3. Theefoe, the total foce is 1 qq 1 3 qq 3 1 qq 1 3 ˆ ˆ ( q1) q3 ˆ F3 = ˆ ˆ = ( ) 13 3 i+ j + i ( a) a 1 qq 1 3 ˆ = 1 + ˆ a 4 i j 4 upon adding the components. The magnitude of the total foce is given b 6
7 1 1 qq 1 3 F3 = 1 + a (6. 1 C)(3. 1 C) (9. 1 N m / C ) (.74) 3. N = = (. 1 m) The angle that the foce makes with the positive x axis is F /4 φ = = = F 3, x 1+ /4 1 3, tan tan Note thee ae two solutions to this equation. The second solution φ = 8.7 is incoect because it would indicate that the foce has positive î and negative ˆj components. Fo a sstem of N chages, the net foce expeienced b the jth paticle would be whee F ij F N j = Fij i= 1 i j denotes the foce between paticles i and (.3.) j. The supeposition pinciple implies that the net foce between an two chages is independent of the pesence of othe chages. This is tue if the chages ae in fixed positions..4 lectic Field The electostatic foce, like the gavitational foce, is a foce that acts at a distance, even when the objects ae not in contact with one anothe. To justif such the notion we ationalize action at a distance b saing that one chage ceates a field which in tun acts on the othe chage. An electic chage q poduces an electic field evewhee. To quantif the stength of the field ceated b that chage, we can measue the foce a positive test chage q expeiences at some point. The electic field is defined as: F = lim q q e (.4.1) We take q to be infinitesimall small so that the field q geneates does not distub the souce chages. The analog between the electic field and the gavitational field g= lim F / m is depicted in Figue.4.1. m m 7
8 Figue.4.1 Analog between the gavitational field g and the electic field. Fom the field theo point of view, we sa that the chage q ceates an electic field which exets a foce F e = q on a test chage q. Using the definition of electic field given in q. (.4.1) and the Coulomb s law, the electic field at a distance fom a point chage q is given b 1 = q ˆ (.4.) Using the supeposition pinciple, the total electic field due to a goup of chages is equal to the vecto sum of the electic fields of individual chages: 1 q = = ˆ i i i i i (.4.3) Animation.: lectic Field of Point Chages Figue.4. shows one fame of animations of the electic field of a moving positive and negative point chage, assuming the speed of the chage is small compaed to the speed of light. Figue.4. The electic fields of (a) a moving positive chage, (b) a moving negative chage, when the speed of the chage is small compaed to the speed of light. 8
9 .5 lectic Field Lines lectic field lines povide a convenient gaphical epesentation of the electic field in space. The field lines fo a positive and a negative chages ae shown in Figue.5.1. (a) (b) Figue.5.1 Field lines fo (a) positive and (b) negative chages. Notice that the diection of field lines is adiall outwad fo a positive chage and adiall inwad fo a negative chage. Fo a pai of chages of equal magnitude but opposite sign (an electic dipole), the field lines ae shown in Figue.5.. Figue.5. Field lines fo an electic dipole. The patten of electic field lines can be obtained b consideing the following: (1) Smmet: Fo eve point above the line joining the two chages thee is an equivalent point below it. Theefoe, the patten must be smmetical about the line joining the two chages () Nea field: Ve close to a chage, the field due to that chage pedominates. Theefoe, the lines ae adial and spheicall smmetic. (3) Fa field: Fa fom the sstem of chages, the patten should look like that of a single point chage of value Q= Q. Thus, the lines should be adiall outwad, unless Q =. i i (4) Null point: This is a point at which =, and no field lines should pass though it. 9
10 The popeties of electic field lines ma be summaized as follows: The diection of the electic field vecto at a point is tangent to the field lines. The numbe of lines pe unit aea though a suface pependicula to the line is devised to be popotional to the magnitude of the electic field in a given egion. The field lines must begin on positive chages (o at infinit) and then teminate on negative chages (o at infinit). The numbe of lines that oiginate fom a positive chage o teminating on a negative chage must be popotional to the magnitude of the chage. No two field lines can coss each othe; othewise the field would be pointing in two diffeent diections at the same point..6 Foce on a Chaged Paticle in an lectic Field Conside a chage in Figue q moving between two paallel plates of opposite chages, as shown Figue.6.1 Chage moving in a constant electic field Let the electic field between the plates be = ˆ j, with >. (In Chapte 4, we shall show that the electic field in the egion between two infinitel lage plates of opposite chages is unifom.) The chage will expeience a downwad Coulomb foce F = q e (.6.1) Note the distinction between the chage q that is expeiencing a foce and the chages on the plates that ae the souces of the electic field. ven though the chage q is also a souce of an electic field, b Newton s thid law, the chage cannot exet a foce on itself. Theefoe, is the field that aises fom the souce chages onl. Accoding to Newton s second law, a net foce will cause the chage to acceleate with an acceleation 1
11 F q a = = = m m m e q ˆj (.6.) Suppose the paticle is at est ( v = ) when it is fist eleased fom the positive plate. The final speed v of the paticle as it stikes the negative plate is q v = a = (.6.3) m whee is the distance between the two plates. The kinetic eneg of the paticle when it stikes the plate is.7 lectic Dipole 1 K = mv = q (.6.4) An electic dipole consists of two equal but opposite chages, + q and q, sepaated b a distance a, as shown in Figue.7.1. Figue.7.1 lectic dipole The dipole moment vecto p which points fom q to + q (in the +  diection) is given b p = qa ˆj (.7.1) The magnitude of the electic dipole is p = qa, whee q >. Fo an oveall chageneutal sstem having N chages, the electic dipole vecto p is defined as i= N p qii i= 1 (.7.) 11
12 whee i is the position vecto of the chage qi. xamples of dipoles include HCL, CO, HO and othe pola molecules. In pinciple, an molecule in which the centes of the positive and negative chages do not coincide ma be appoximated as a dipole. In Chapte 5 we shall also show that b appling an extenal field, an electic dipole moment ma also be induced in an unpolaized molecule..7.1 The lectic Field of a Dipole What is the electic field due to the electic dipole? Refeing to Figue.7.1, we see that the xcomponent of the electic field stength at the point P is x q cosθ cosθ q x x = = x + a x + + a + 3/ 3/ 4 πε + ( ) ( ) (.7.3) whee ± = + a acos θ = x + ( a) (.7.4) Similal, the component is q sinθ sinθ q a + a = = x + a x + + a + 3/ 3/ 4 πε + ( ) ( ) (.7.5) In the pointdipole limit whee a, one ma veif that (see Solved Poblem.13.4) the above expessions educe to and x 3p = sinθ cosθ (.7.6) 3 p = 3cos θ 1) (.7.7) 3 ( whee sin θ = x / and cos θ = /. With3p cosθ = 3p and some algeba, the electic field ma be witten as 1 p 3( p ) () = (.7.8) Note that q. (.7.8) is valid also in thee dimensions whee = xˆi + ˆj + zkˆ. The equation indicates that the electic field 3 due to a dipole deceases with as 1/, 1
13 unlike the 1/ behavio fo a point chage. This is to be expected since the net chage of a dipole is zeo and theefoe must fall off moe apidl than 1/ at lage distance. The electic field lines due to a finite electic dipole and a point dipole ae shown in Figue.7.. Figue.7. lectic field lines fo (a) a finite dipole and (b) a point dipole. Animation.3: lectic Dipole Figue.7.3 shows an inteactive ShockWave simulation of how the dipole patten aises. At the obsevation point, we show the electic field due to each chage, which sum vectoiall to give the total field. To get a feel fo the total electic field, we also show a gass seeds epesentation of the electic field in this case. The obsevation point can be moved aound in space to see how the esultant field at vaious points aises fom the individual contibutions of the electic field of each chage. Figue.7.3 An inteactive ShockWave simulation of the electic field of an two equal and opposite chages..8 Dipole in lectic Field What happens when we place an electic dipole in a unifom field = ˆi, with the dipole moment vecto p making an angle with the xaxis? Fom Figue.8.1, we see that the unit vecto which points in the diection of p is cosθ ˆi + sinθ ˆ j. Thus, we have p= qa(cosθ ˆi+ sin θ ˆj ) (.8.1) 13
14 Figue.8.1 lectic dipole placed in a unifom field. As seen fom Figue.8.1 above, since each chage expeiences an equal but opposite foce due to the field, the net foce on the dipole is Fnet = F+ + F =. ven though the net foce vanishes, the field exets a toque a toque on the dipole. The toque about the midpoint O of the dipole is τ = F + F = ( acosθˆi+ asin θ ˆj ) ( F ˆi) + ( acosθˆi asin θ ˆj ) ( F ˆi ) = asin θ F ( kˆ) + asin θ F ( kˆ) = af sin θ ( kˆ ) + (.8.) whee we have used F+ = F = F. The diection of the toque is k ˆ, o into the page. The effect of the toque τ is to otate the dipole clockwise so that the dipole moment p becomes aligned with the electic field. With F = q, the magnitude of the toque can be ewitten as τ = aq ( )sin θ = ( aq ) sinθ = psinθ and the geneal expession fo toque becomes τ = p (.8.3) Thus, we see that the coss poduct of the dipole moment with the electic field is equal to the toque..8.1 Potential neg of an lectic Dipole The wok done b the electic field to otate the dipole b an angle dθ is dw = τ dθ = p sinθ dθ (.8.4) 14
15 The negative sign indicates that the toque opposes an incease inθ. Theefoe, the total amount of wok done b the electic field to otate the dipole fom an angle θ to θ is ( sin θ ) θ ( cosθ cos θ ) θ W = p d = p θ (.8.5) The esult shows that a positive wok is done b the field when cosθ > cosθ. The change in potential eneg U of the dipole is the negative of the wok done b the field: ( θ θ ) U = U U = W = p cos cos (.8.6) whee U = Pcosθ is the potential eneg at a efeence point. We shall choose ou efeence point to beθ = π so that the potential eneg is zeo thee, U =. Thus, in the pesence of an extenal field the electic dipole has a potential eneg U = pcosθ = p (.8.7) A sstem is at a stable equilibium when its potential eneg is a minimum. This takes place when the dipole p is aligned paallel to, making U a minimum with U min = p. On the othe hand, when p and ae antipaallel, U max =+p is a maximum and the sstem is unstable. If the dipole is placed in a nonunifom field, thee would be a net foce on the dipole in addition to the toque, and the esulting motion would be a combination of linea acceleation and otation. In Figue.8., suppose the electic field + at +q diffes fom the electic field at q. Figue.8. Foce on a dipole Assuming the dipole to be ve small, we expand the fields about x : d d + ( x+ a) ( x) + a, ( x a) ( x) a dx dx (.8.8) The foce on the dipole then becomes 15
16 F d ( ) ˆ d e = q + = qa = p dx i î (.8.9) dx An example of a net foce acting on a dipole is the attaction between small pieces of pape and a comb, which has been chaged b ubbing against hai. The pape has induced dipole moments (to be discussed in depth in Chapte 5) while the field on the comb is nonunifom due to its iegula shape (Figue.8.3). Figue.8.3 lectostatic attaction between a piece of pape and a comb.9 Chage Densit The electic field due to a small numbe of chaged paticles can eadil be computed using the supeposition pinciple. But what happens if we have a ve lage numbe of chages distibuted in some egion in space? Let s conside the sstem shown in Figue.9.1: Figue.9.1 lectic field due to a small chage element q i..9.1 Volume Chage Densit Suppose we wish to find the electic field at some point P. Let s conside a small volume element which contains an amount of chage q. The distances between chages within the volume element V i Vi i ae much smalle than compaed to, the distance between V i and P. In the limit whee Vi becomes infinitesimall small, we ma define a volume chage densit ρ( ) as qi dq ρ() = lim = (.9.1) Vi V dv i 16
17 The dimension of ρ( 3 ) is chage/unit volume (C/m ) in SI units. The total amount of chage within the entie volume V is i (.9.) i V Q= q = ρ() dv The concept of chage densit hee is analogous to mass densit ρm ( ). When a lage numbe of atoms ae tightl packed within a volume, we can also take the continuum limit and the mass of an object is given b M = ρ () m dv (.9.3) V.9. Suface Chage Densit In a simila manne, the chage can be distibuted ove a suface S of aea A with a suface chage densit σ (lowecase Geek lette sigma): The dimension of σ is chage/unit aea suface is: dq σ () = (.9.4) da (C/m ) in SI units. The total chage on the entie Q= σ () da S (.9.5).9.3 Line Chage Densit If the chage is distibuted ove a line of length, then the linea chage densit λ (lowecase Geek lette lambda) is dq λ () = d (.9.6) whee the dimension of λ is chage/unit length (C/m). The total chage is now an integal ove the entie length: Q= λ( ) d (.9.7) line 17
18 If chages ae unifoml distibuted thoughout the egion, the densities ( ρ, σ o λ ) then become unifom..1 lectic Fields due to Continuous Chage Distibutions The electic field at a point P due to each chage element dq is given b Coulomb s law: 1 dq d= ˆ (.1.1) whee is the distance fom dq to P and ˆ is the coesponding unit vecto. (See Figue.9.1). Using the supeposition pinciple, the total electic field is the vecto sum (integal) of all these infinitesimal contibutions: 1 dq = ˆ V (.1.) This is an example of a vecto integal which consists of thee sepaate integations, one fo each component of the electic field. xample.: lectic Field on the Axis of a Rod A nonconducting od of length with a unifom positive chage densit λ and a total chage Q is ling along the axis, as illustated in Figue.1.1. x Figue.1.1 lectic field of a wie along the axis of the wie Calculate the electic field at a point P located along the axis of the od and a distance x fom one end. Solution: The linea chage densit is unifom and is given b λ = Q /. The amount of chage contained in a small segment of length d x is dq = λ dx. 18
19 Since the souce caies a positive chage Q, the field at P points in the negative x diection, and the unit vecto that points fom the souce to P isˆ = ˆi. The contibution to the electic field due to dq is 1 dq 1 λdx ˆ 1 Qdx d= ˆ = ( i) = ˆi x x Integating ove the entie length leads to 1 Q x + dx ˆ 1 Q 1 1 ˆ 1 Q = d ˆ = = = x x i x x + i i x( + x) (.1.3) Notice that when P is ve fa awa fom the od, becomes 1 Q ˆi x x, and the above expession (.1.4) The esult is to be expected since at sufficientl fa distance awa, the distinction between a continuous chage distibution and a point chage diminishes. xample.3: lectic Field on the Pependicula Bisecto A nonconducting od of length with a unifom chage densit λ and a total chage Q is ling along the x axis, as illustated in Figue.1.. Compute the electic field at a point P, located at a distance fom the cente of the od along its pependicula bisecto. Figue.1. Solution: We follow a simila pocedue as that outlined in xample.. The contibution to the electic field fom a small length element d x caing chage dq = λ dx is 19
20 1 dq 1 λ dx d = = x + (.1.5) Using smmet agument illustated in Figue.1.3, one ma show that the x  component of the electic field vanishes. Figue.1.3 Smmet agument showing that x =. The component of d is d 1 λ dx 1 λ dx = d cosθ = = x + x + 4 πε ( x + ) 3/ (.1.6) B integating ove the entie length, the total electic field due to the od is 1 / λ dx λ / dx = d = 4 πε = ( ) 4 ( / 3/ / 3/ x + πε x + ) (.1.7) B making the change of vaiable: above integal becomes x = tanθ, which gives = θ, the dx sec θ d / dx θ sec θ dθ 1 θ sec θ dθ 1 θ sec θ dθ = = = ( x + ) (sec θ + 1) (tan θ + 1) secθ 1 θ dθ 1 θ sinθ = cosθ dθ = = θ secθ θ / 3/ θ 3 3/ θ 3/ θ 3 (.1.8) which gives 1 λ sinθ 1 λ / = = + 4 πε ( /) (.1.9) 
21 In the limit whee, the above expession educes to the pointchage limit: 1 λ / 1 λ 1 Q = = πε πε πε (.1.1) On the othe hand, when, we have 1 λ (.1.11) In this infinite length limit, the sstem has clindical smmet. In this case, an altenative appoach based on Gauss s law can be used to obtain q. (.1.11), as we shall show in Chapte 4. The chaacteistic behavio of / (with = Q/ ) as a function of / is shown in Figue.1.4. Figue.1.4 lectic field of a nonconducting od as a function of /. xample.4: lectic Field on the Axis of a Ring A nonconducting ing of adius R with a unifom chage densit λ and a total chage Q is ling in the x  plane, as shown in Figue.1.5. Compute the electic field at a point P, located at a distance z fom the cente of the ing along its axis of smmet. Figue.1.5 lectic field at P due to the chage element dq. 1
22 Solution: Conside a small length element d on the ing. The amount of chage contained within this element is dq = λ d = λr dφ. Its contibution to the electic field at P is 1 dq 1 λr dφ d= ˆ = ˆ (.1.1) Figue.1.6 Using the smmet agument illustated in Figue.1.6, we see that the electic field at P must point in the +z diection. d z 1 λrdφ z λ Rzdφ = d cosθ = = R + z R + z 4 πε ( R + z ) 3/ (.1.13) Upon integating ove the entie ing, we obtain z Rz Rz 1 Qz = λ dφ = λ π = 4 πε ( R + z ) 4 πε ( R + z ) 4 πε ( R + z ) 3/ 3/ 3/ (.1.14) whee the total chage is Q= λ( π R). A plot of the electic field as a function of z is given in Figue.1.7. Figue.1.7 lectic field along the axis of smmet of a nonconducting ing of adius R, with = Q/ R. 
23 Notice that the electic field at the cente of the ing vanishes. This is to be expected fom smmet aguments. xample.5: lectic Field Due to a Unifoml Chaged Disk A unifoml chaged disk of adius R with a total chage Q lies in the xplane. Find the electic field at a point P, along the zaxis that passes though the cente of the disk pependicula to its plane. Discuss the limit whee R z. Solution: B teating the disk as a set of concentic unifoml chaged ings, the poblem could be solved b using the esult obtained in xample.4. Conside a ing of adius and thickness d, as shown in Figue.1.8. Figue.1.8 A unifoml chaged disk of adius R. B smmet aguments, the electic field at P points in the + z diection. Since the ing has a chage dq = σ ( π d ), fom q. (.1.14), we see that the ing gives a contibution d z 1 zdq 1 z(πσ d ) = = 4 πε ( + z ) 4 πε ( + z ) 3/ 3/ (.1.15) Integating fom = to = R, the total electic field at P becomes z σ σ σ = dz = ε = = ( + z ) 4ε u 4 ε ( 1/) z 1/ z R d z R + z du z u R + z 3/ z 3/ σz 1 1 σ z z = = ε R + z z ε z R + z (.1.16) 3
24 The above equation ma be ewitten as z σ z 1, z > ε z + R = σ z 1, z < ε z + R (.1.17) The electic field / ( = σ /ε ) as a function of z/ Ris shown in Figue.1.9. z Figue.1.9 lectic field of a nonconducting plane of unifom chage densit. To show that the pointchage limit is ecoveed fo z R, we make use of the Taloseies expansion: 1/ 1 z 1 1 R R = + = + z z z + R 1 R z (.1.18) This gives σ R 1 σπr 1 = = = z ε z z z Q (.1.19) which is indeed the expected pointchage esult. On the othe hand, we ma also conside the limit whee R z. Phsicall this means that the plane is ve lage, o the field point P is extemel close to the suface of the plane. The electic field in this limit becomes, in unitvecto notation, σ ˆ k, z > ε = σ kˆ, z < ε (.1.) 4
25 The plot of the electic field in this limit is shown in Figue.1.1. Figue.1.1 lectic field of an infinitel lage nonconducting plane. Notice the discontinuit in electic field as we coss the plane. The discontinuit is given b σ σ σ z = z+ z = = ε ε ε (.1.1) As we shall see in Chapte 4, if a given suface has a chage densitσ, then the nomal component of the electic field acoss that suface alwas exhibits a discontinuit with = σ / ε. n.11 Summa The electic foce exeted b a chage q1 on a second chage q is given b Coulomb s law: whee is the Coulomb constant. k e qq F = k ˆ = e 1 = 1 = qq ˆ N m / C The electic field at a point in space is defined as the electic foce acting on a test chage divided b q : q F = lim q q e 5
26 The electic field at a distance fom a chage q is 1 = q ˆ Using the supeposition pinciple, the electic field due to a collection of point chages, each having chage q and located at a distance awa is i 1 q = i i i ˆ i i A paticle of mass m and chage q moving in an electic field has an acceleation q a = m An electic dipole consists of two equal but opposite chages. The electic dipole moment vecto p points fom the negative chage to the positive chage, and has a magnitude p = aq The toque acting on an electic dipole places in a unifom electic field is τ = p The potential eneg of an electic dipole in a unifom extenal electic field is U = p The electic field at a point in space due to a continuous chage element dq is 1 d= dq ˆ At sufficientl fa awa fom a continuous chage distibution of finite extent, the electic field appoaches the pointchage limit. 6
27 .1 PoblemSolving Stategies In this chapte, we have discussed how electic field can be calculated fo both the discete and continuous chage distibutions. Fo the fome, we appl the supeposition pinciple: 1 q = i i i ˆ i Fo the latte, we must evaluate the vecto integal 1 = dq ˆ whee is the distance fom dq to the field point P and ˆ is the coesponding unit vecto. To complete the integation, we shall follow the pocedues outlined below: 1 dq (1) Stat with d= ˆ () Rewite the chage element dq as λ d dq = σ da ρ dv (length) (aea) (volume) depending on whethe the chage is distibuted ove a length, an aea, o a volume. (3) Substitute dq into the expession fo d. (4) Specif an appopiate coodinate sstem (Catesian, clindical o spheical) and expess the diffeential element ( d, da o dv ) and in tems of the coodinates (see Table.1 below fo summa.) Catesian (x,, z) Clindical (ρ, φ, z) Spheical (, θ, φ) dl dx, d, dz dρ, ρdφ, dz d, dθ, sinθ dφ da dx d, d dz, dz dx dρ dz, ρdφdz, ρdφdρ ddθ, sin θ ddφ, sinθ dθ dφ dv dx d dz ρ dρdφ dz sinθ ddθ dφ Table.1 Diffeential elements of length, aea and volume in diffeent coodinates 7
28 (5) Rewite d in tems of the integation vaiable(s), and appl smmet agument to identif nonvanishing component(s) of the electic field. (6) Complete the integation to obtain. In the Table below we illustate how the above methodologies can be utilized to compute the electic field fo an infinite line chage, a ing of chage and a unifoml chaged disk. Line chage Ring of chage Unifoml chaged disk Figue () xpess dq in tems of chage densit (3) Wite down d dq = λ dx dq = λ d dq = σ da d = k e λ dx d k λ dl d σ da = e = e k (4) Rewite and the diffeential element in tems of the appopiate coodinates dx cosθ = = x + d = Rdφ z cosθ = = R + z da = π ' d ' z cosθ = = + z (5) Appl smmet agument to identif nonvanishing component(s) of d d = d cosθ λ dx = ke ( x + ) 3/ d z = d cosθ λrz dφ = ke ( R z ) d = d cosθ πσ z d ( + z ) 3/ e + 3/ z = k (6) Integate to get = kλ e + / / dx ( x + ) 3/ keλ / = ( /) + Rλz z = ke dφ 3/ ( R + z ) ( πrλ) z = ke ( R + z ) 3/ Qz = ke ( R + z ) 3/ = πσkz z e R d ( + z ) 3/ z z = πσke z z + R 8
29 .13 Solved Poblems.13.1 Hdogen Atom In the classical model of the hdogen atom, the electon evolves aound the poton with 1 a adius of = m. The magnitude of the chage of the electon and poton is 19 e = C. (a) What is the magnitude of the electic foce between the poton and the electon? (b) What is the magnitude of the electic field due to the poton at? (c) What is atio of the magnitudes of the electical and gavitational foce between electon and poton? Does the esult depend on the distance between the poton and the electon? (d) In light of ou calculation in (b), explain wh electical foces do not influence the motion of planets. Solutions: (a) The magnitude of the foce is given b F e 1 e = 4 πε Now we can substitute ou numeical values and find that the magnitude of the foce between the poton and the electon in the hdogen atom is F e (9. 1 N m / C )(1.6 1 C) = 11 (5.3 1 m) 9 19 = N (b) The magnitude of the electic field due to the poton is given b q (9. 1 N m / C )(1.6 1 C) = = = 1 4 πε (.5 1 m) N / C (c) The mass of the electon is and the mass of the poton is 7 m = kg. Thus, the atio of the magnitudes of the electic and gavitational p foce is given b m =. e kg 9
30 4 (9. 1 N m / C )(1.6 1 C) γ = = = = G 1 e 1 e 9 19 πε mm p e Gmpme ( N m / kg )(1.7 1 kg)(9.1 1 kg) which is independent of, the distance between the poton and the electon. (d) The electic foce is 39 odes of magnitude stonge than the gavitational foce between the electon and the poton. Then wh ae the lage scale motions of planets detemined b the gavitational foce and not the electical foce. The answe is that the magnitudes of the chage of the electon and poton ae equal. The best expeiments show that the diffeence between these magnitudes is a numbe on the ode of 1 4. Since objects like planets have about the same numbe of potons as electons, the ae essentiall electicall neutal. Theefoe the foce between planets is entiel detemined b gavit..13. Millikan OilDop xpeiment 6 3 An oil dop of adius = m and mass densit ρ oil = kg m is allowed to fall fom est and then entes into a egion of constant extenal field applied in the downwad diection. The oil dop has an unknown electic chage q (due to iadiation b busts of Xas). The magnitude of the electic field is adjusted until the gavitational foce Fg = mg= mgˆj on the oil dop is exactl balanced b the electic foce, F e = q. Suppose this balancing occus when the electic field is ˆ 5 = j= (1.9 1 N C) ˆ 5 j, with = N C. (a) What is the mass of the oil dop? 19 (b) What is the chage on the oil dop in units of electonic chage e = C? Solutions: (a) The mass densit ρ oil times the volume of the oil dop will ield the total mass M of the oil dop, 4 3 M = ρoilv = ρoil π 3 3 whee the oil dop is assumed to be a sphee of adius with volume V = 4 π /3. Now we can substitute ou numeical values into ou smbolic expession fo the mass, 3
31 M 4 4π 14 ( kg m ) ( m) kg = ρoil π = = (b) The oil dop will be in static equilibium when the gavitational foce exactl balances the electical foce: F + F =. Since the gavitational foce points downwad, the g e electic foce on the oil must be upwad. Using ou foce laws, we have = mg+ q mg = q With the electical field pointing downwad, we conclude that the chage on the oil dop must be negative. Notice that we have chosen the unit vecto ĵ to point upwad. We can solve this equation fo the chage on the oil dop: 14 mg ( kg)(9.8 m / s ) q = = = N C C Since the electon has chage 19 e = C, the chage of the oil dop in units of e is q N = = e C C = 19 5 You ma at fist be supised that this numbe is an intege, but the Millikan oil dop expeiment was the fist diect expeimental evidence that chage is quantized. Thus, fom the given data we can asset that thee ae five electons on the oil dop!.13.3 Chage Moving Pependiculal to an lectic Field An electon is injected hoizontall into a unifom field poduced b two oppositel chaged plates, as shown in Figue The paticle has an initial velocit v = vˆ i pependicula to. Figue.13.1 Chage moving pependicula to an electic field 31
32 (a) While between the plates, what is the foce on the electon? (b) What is the acceleation of the electon when it is between the plates? (c) The plates have length L1 in the x diection. At what time t 1 will the electon leave the plate? (d) Suppose the electon entes the electic field at time t =. What is the velocit of the electon at time when it leaves the plates? t 1 (e) What is the vetical displacement of the electon afte time plates? t 1 when it leaves the (f) What angle θ 1 does the electon make θ 1 with the hoizontal, when the electon leaves the plates at time t 1? (g) The electon hits the sceen located a distance fom the end of the plates at a time t. What is the total vetical displacement of the electon fom time t = until it hits the t sceen at? L Solutions: (a) Since the electon has a negative chage, q = e, the foce on the electon is Fe = q= e= ( e)( )ˆj= e ˆj whee the electic field is witten as = ˆ j, with >. The foce on the electon is upwad. Note that the motion of the electon is analogous to the motion of a mass that is thown hoizontall in a constant gavitational field. The mass follows a paabolic tajecto downwad. Since the electon is negativel chaged, the constant foce on the electon is upwad and the electon will be deflected upwads on a paabolic path. (b) The acceleation of the electon is q q ˆ e a = = j= m m m ˆj and its diection is upwad. 3
33 (c) The time of passage fo the electon is given b t 1 = L 1 / v. The time t 1 is not affected b the acceleation because v, the hoizontal component of the velocit which detemines the time, is not affected b the field. (d) The electon has an initial hoizontal velocit, v = v ˆ i. Since the acceleation of the electon is in the + diection, onl the component of the velocit changes. The velocit at a late time is given b t 1 e e L v ˆ ˆ ˆ ˆ ˆ 1 = v i ˆ ˆ x + v j = vi + at1j = vi+ t1j = vi+ ˆ j m mv (e) Fom the figue, we see that the electon tavels a hoizontal distance t 1 = L1 v and then emeges fom the plates with a vetical displacement L 1 in the time 1 1 e L 1 1 = a t1 = m v (f) When the electon leaves the plates at time t1, the electon makes an angle θ 1 with the hoizontal given b the atio of the components of its velocit, v ( e / m)( L / v ) e L tanθ = = = v v m 1 1 x v (g) Afte the electon leaves the plate, thee is no longe an foce on the electon so it tavels in a staight path. The deflection is e L L = L = 1 tanθ1 mv and the total deflection becomes e L e LL e L 1 = + = + = L + L mv mv mv.13.4 lectic Field of a Dipole Conside the electic dipole moment shown in Figue.7.1. (a) Show that the electic field of the dipole in the limit whee a is 33
34 x = 3p p sin cos, 3cos 1 θ θ = 4 θ ) 3 3 ( πε whee sin θ = x / and cos θ = /. (b) Show that the above expession fo the electic field can also be witten in tems of the pola coodinates as (, θ ) = ˆ ˆ + θ θ whee Solutions: pcosθ psinθ =, = 4 3 θ 3 πε (a) Let s compute the electic field stength at a distance a due to the dipole. The x  component of the electic field stength at the point P with Catesian coodinates ( x,,) is given b x q cosθ cosθ q x x = = x + a x + + a + 3/ 3/ 4 πε + ( ) ( ) whee ± = + a acos θ = x + ( a) Similal, the component is given b q sinθ sinθ q a + a = = x + a x + + a + 3/ 3/ 4 πε + ( ) ( ) We shall make a polnomial expansion fo the electic field using the Taloseies 3 expansion. We will then collect tems that ae popotional to 1/ and ignoe tems that 5 1 ae popotional to 1/, whee =+ ( x + ). We begin with 34
35 3/ 3/ 3/ 3 a ± a + ± = + + ± = + [ x ( a) ] [ x a a] 1 In the limit whee >> a, we use the Taloseies expansion with s ( a ± a)/ : s = s + s 8 3/ (1 ) 1... and the above equations fo the components of the electic field becomes and x q 6xa = q a 6 a = whee we have neglected the Os ( ) tems. The electic field can then be witten as ˆ ˆ q aˆ 6a ˆ ˆ p 3xˆ 3 x ( x ) 1 ˆ = i + j = j+ i + j = + i j whee we have made used of the definition of the magnitude of the electic dipole moment p = aq. In tems of the pola coodinates, with sinθ = x and cosθ = (as seen fom Figue.13.4), we obtain the desied esults: x = 3p p sin cos 3cos 1 θ θ, = 4 θ ) 3 3 ( πε (b) We begin with the expession obtained in (a) fo the electic dipole in Catesian coodinates: ( p ˆ, θ) = 3sinθ cosθ + 3 ( 3cos θ 1)ˆ i j With a little algeba, the above expession ma be ewitten as 35
36 p (, θ) = cosθ 3 ( sinθ ˆi + cosθ ˆj ) + sinθ cosθ ˆi + ( cos θ 1) ˆj ( ˆi ˆj) ( ˆi ˆj) p = cosθ sinθ + cosθ + sinθ cosθ sinθ 3 whee the tigonometic identit ( θ ) vectos ˆ and cos 1 = sin θ has been used. Since the unit ˆθ in pola coodinates can be decomposed as ˆ = sinθ ˆi + cosθ ˆj θˆ = cosθ ˆi sin θ ˆj, the electic field in pola coodinates is given b ( p, θ ) = cosθ ˆ + sin ˆ 3 θ and the magnitude of is 1/ p = ( + θ ) = 3 ( 3cos θ + 1 ) 1/.13.5 lectic Field of an Ac A thin od with a unifom chage pe unit length λ is bent into the shape of an ac of a cicle of adius R. The ac subtends a total angle θ, smmetic about the xaxis, as shown in Figue.13.. What is the electic field at the oigin O? Solution: Conside a diffeential element of length d = Rdθ, which makes an angle θ with the x  axis, as shown in Figue.13.(b). The amount of chage it caies is dq = λ d = λr dθ. The contibution to the electic field at O is 1 dq 1 dq 1 d d= ˆ = cos i sin = cos sinθ R R ( ˆ ˆ λ θ θ θj) ( θˆi ˆj ) 36
37 Figue.13. (a) Geomet of chaged souce. (b) Chage element dq Integating ove the angle fom θ to + θ, we have θ λ sinθ i j i+ j i ( ˆ ˆ λ cos sin ) ( sin ˆ cos ˆ) = 1 λ θ 1 1 dθ θ θ θ θ ˆ R = θ R θ = R We see that the electic field onl has the x component, as equied b a smmet agument. If we take the limit θ π, the ac becomes a cicula ing. Since sinπ =, the equation above implies that the electic field at the cente of a nonconducting ing is zeo. This is to be expected fom smmet aguments. On the othe hand, fo ve smallθ, sinθ θ and we ecove the pointchage limit: 1 λθ ˆ 1 λθ R ˆ 1 i = i = R R R Q ˆi whee the total chage on the ac is Q= λ = λ( Rθ ) lectic Field Off the Axis of a Finite Rod A nonconducting od of length with a unifom chage densit λ and a total chage Q is ling along the x axis, as illustated in Figue Compute the electic field at a point P, located at a distance off the axis of the od. Figue
38 Solution: The poblem can be solved b following the pocedue used in xample.3. Conside a length element dx on the od, as shown in Figue The chage caied b the element is dq = λ dx. The electic field at P poduced b this element is Figue dq 1 λ dx d= ˆ = sin ˆi+ cos ˆj x + ( θ θ ) whee the unit vecto ˆ has been witten in Catesian coodinates: ˆ = sinθ ˆi+ cosθ ˆj. In the absence of smmet, the field at P has both the x and components. The x component of the electic field is d x 1 λ dx 1 λdx x 1 λx dx = sinθ = = x + x + x + 4 πε ( x + ) 3/ Integating fom x = x1 to x = x, we have x λ xdx λ 1 du λ = = = 4 πε ( ) x x+ 1/ u 3/ x 3/ 1 x1 x + + u x + 1 λ 1 1 λ = = x 4 x1 πε + + x + x1 + λ = ( cosθ cosθ1) x + Similal, the component of the electic field due to the chage element is 38
39 d 1 λ dx 1 λdx 1 λdx = cosθ = = x + x + x + 4 πε ( x + ) 3/ Integating ove the entie length of the od, we obtain λ x dx λ 1 θ λ = = cosθ dθ = sinθ sin 4 πε ( x + ) x 3/ 1 θ1 ( ) θ1 whee we have used the esult obtained in q. (.1.8) in completing the integation. In the infinite length limit whee x1 and x +, with xi = tanθi, the coesponding angles ae θ 1 = π /and θ = + π /. Substituting the values into the expessions above, we have x 1 λ =, = in complete ageement with the esult shown in q. (.1.11)..14 Conceptual Questions 1. Compae and contast Newton s law of gavitation, F = Gmm, and Coulomb s law, F kq q / e = 1.. Can electic field lines coss each othe? xplain. 3. Two opposite chages ae placed on a line as shown in the figue below. g 1 / The chage on the ight is thee times the magnitude of the chage on the left. Besides infinit, whee else can electic field possibl be zeo? 4. A test chage is placed at the point P nea a positivelchaged insulating od. 39
40 How would the magnitude and diection of the electic field change if the magnitude of the test chage wee deceased and its sign changed with evething else emaining the same? 5. An electic dipole, consisting of two equal and opposite point chages at the ends of an insulating od, is fee to otate about a pivot point in the cente. The od is then placed in a nonunifom electic field. Does it expeience a foce and/o a toque?.15 Additional Poblems.15.1 Thee Point Chages Thee point chages ae placed at the cones of an equilateal tiangle, as shown in Figue Figue.15.1 Thee point chages Calculate the net electic foce expeienced b (a) the 9. µ C chage, and (b) the 6. µ C chage..15. Thee Point Chages A ight isosceles tiangle of side a has chages q, +q and q aanged on its vetices, as shown in Figue
41 Figue.15. What is the electic field at point P, midwa between the line connecting the +q and q chages? Give the magnitude and diection of the electic field Fou Point Chages Fou point chages ae placed at the cones of a squae of side a, as shown in Figue Figue.15.3 Fou point chages (a) What is the electic field at the location of chage q? (b) What is the net foce on q?.15.4 Semicicula Wie A positivel chaged wie is bent into a semicicle of adius R, as shown in Figue Figue
42 The total chage on the semicicle is Q. Howeve, the chage pe unit length along the semicicle is nonunifom and given b λ = λ cosθ. (a) What is the elationship between λ, R and Q? (b) If a chage q is placed at the oigin, what is the total foce on the chage?.15.5 lectic Dipole An electic dipole ling in the xplane with a unifom electic field applied in the + x  diection is displaced b a small angle θ fom its equilibium position, as shown in Figue Figue.15.5 The chages ae sepaated b a distance a, and the moment of inetia of the dipole is I. If the dipole is eleased fom this position, show that its angula oientation exhibits simple hamonic motion. What is the fequenc of oscillation?.15.6 Chaged Clindical Shell and Clinde (a) A unifoml chaged cicula clindical shell of adius R and height h has a total chage Q. What is the electic field at a point P a distance z fom the bottom side of the clinde as shown in Figue.15.6? (Hint: Teat the clinde as a set of ing chages.) Figue.15.6 A unifoml chaged clinde 4
43 (b) If the configuation is instead a solid clinde of adius R, height h and has a unifom volume chage densit. What is the electic field at P? (Hint: Teat the solid clinde as a set of disk chages.).15.7 Two Conducting Balls Two tin conducting balls of identical mass m and identical chage q hang fom nonconducting theads of length l. ach ball foms an angle θ with the vetical axis, as shown in Figue Assume that θ is so small that tanθ sinθ. Figue.15.9 (a) Show that, at equilibium, the sepaation between the balls is q = πε mg 13 1 (b) If l = 1. 1 cm, m= 1. 1 g, and x = 5.cm, what is q?.15.8 Toque on an lectic Dipole 19 An electic dipole consists of two chages q 1 = +e and q = e ( e = C ), 9 sepaated b a distance d = 1 m. The electic chages ae placed along the axis as shown in Figue Figue
44 Suppose a constant extenal electic field = (3ˆi+ 3 ˆj)N/C is applied. (a) What is the magnitude and diection of the dipole moment? (b) What is the magnitude and diection of the toque on the dipole? ext (c) Do the electic fields of the chages q1 and q contibute to the toque on the dipole? Biefl explain ou answe. 44
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