Module 3: Sections.1 though.8 Module 4: Sections.9 though.14 1 Table of Contents.1 Electic Chage... -3. Coulomb's Law... -3..1 Van de Gaaff Geneato (link)... -4.3 Pinciple of Supeposition... -5 Example.1: Thee Chages... -5.4 Electic Field... -7.4.1 Electic Field of Point Chages (link)... -8.5 Electic Field Lines... -9.6 Foce on a Chaged Paticle in an Electic Field... -1.7 Electic Dipole... -11.7.1 The Electic Field of a Dipole... -1.7. Electic Dipole Animation (link)... -13.8 Dipole in Electic Field... -13.8.1 Potential Eneg of an Electic Dipole... -14.9 Chage Densit... -16.9.1 Volume Chage Densit... -16.9. Suface Chage Densit... -17.9.3 Line Chage Densit... -17.1 Electic Fields due to Continuous Chage Distibutions... -18 Example.: Electic Field on the Axis of a Rod... -18 Example.3: Electic Field on the Pependicula Bisecto (link)... -19 Example.4: Electic Field on the Axis of a Ring (link)... -1 Example.5: Electic Field Due to a Unifoml Chaged Disk... -3.11 Summa... -5.1 Poblem-Solving Stategies... -7.13 Solved Poblems... -9.13.1 Hdogen Atom... -9.13. Millikan Oil-Dop Expeiment... -3 1 These notes ae excepted Intoduction to Electicit and Magnetism b Sen-Ben Liao, Pete Doumashkin, and John Belche, Copight 4, ISBN -536-817-1. -1
.13.3 Chage Moving Pependiculal to an Electic Field... -3.13.4 Electic Field of a Dipole... -34.13.5 Electic Field of an Ac... -36.13.6 Electic Field Off the Axis of a Finite Rod... -37.14 Conceptual Questions... -39 -
Coulomb s Law.1 Electic 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 = 1.6 1 19 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, q 1 and q, sepaated b a distance in vacuum. The foce exeted b q 1 on q is given b Coulomb's law: F = k 1 e qq 1 ˆ (..1) whee k e is the Coulomb constant, and ˆ = / is a unit vecto diected fom q 1 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 k e is given b -3
k e 1 9 = = 8.9875 1 N m / C (..) whee 1 ε = = 885. 1 1 C N m (..3) 9 4π (8.99 1 N m C ) is known as the pemittivit of fee space. Similal, the foce on q 1 due to q is given b F 1 = F 1, as illustated in Figue..1(b). This is consistent with Newton's thid law. As an example, conside a hdogen atom in which the poton (nucleus) and the electon ae sepaated b a distance = 5.3 1 11 m. The electostatic foce between the two paticles is appoximatel F = k e / = 8. 1 8 N. On the othe hand, one ma show e e that the gavitational foce is onl F g 3.6 1 47 N. Thus, gavitational effect can be neglected when dealing with electostatic foces!..1 Van de Gaaff Geneato (link) 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
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 (link) 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 q 3 due to q 1 and q will be The supeposition pinciple is illustated in the example below. Example.1: Thee Chages F 3 = F 13 + F 3 (.3.1) Thee chages ae aanged as shown in Figue.3.1. Find the foce on the chage q 3 assuming that q 1 = 6. 1 6 C, q = q 1 = 6. 1 6 C, q 3 =+3. 1 6 C and a =. 1 m. -5
Figue.3.1 A sstem of thee chages Solution: Using the supeposition pinciple, the foce on q 3 is 1 qq qq 1 3 3 F 3 = F 13 + F 3 = ˆ + ˆ 13 3 13 3 In this case the second tem will have a negative coefficient, since q is negative. The unit vectos ˆ13 and ˆ3 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 ˆ13 which points fom q 1 to q 3 can be witten as ˆ13 = cosθ î + sinθ ĵ = ( î + ĵ) Similal, the unit vecto ˆ3 = î points fom q to q 3. Theefoe, the total foce is q q F = 1 qq 1 3 3 ˆ13 + q q 3 ˆ3 1 qq = 1 3 (ˆi + ˆj) + ( 1 3 ) î 4πε 13 3 ( a) a 1 qq 1 3 = 1 î + ĵ a 4 4 upon adding the components. The magnitude of the total foce is given b -6
1 qq 1 1 3 F 3 = 1 + 4πε a 4 4 9 (6. 1 6 C)(3. 1 6 C) = (9. 1 N m / C ) (.74) = 3. N (. 1 m) The angle that the foce makes with the positive x -axis is F φ = tan 1 3, = tan 1 /4 F 3,x + 1 / 4 =151.3 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 ĵ components. Fo a sstem of N chages, the net foce expeienced b the jth paticle would be N F j = F ij (.3.) i=1 i j whee F ij denotes the foce between paticles i and 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 Electic 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 E is defined as: F E= lim e (.4.1) q q 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 / m is depicted in Figue.4.1. m -7
Figue.4.1 Analog between the gavitational field g and the electic field E. Fom the field theo point of view, we sa that the chage q ceates an electic field E which exets a foce F e = q E on a test chage q. Using the definition of electic field given in Eq. (.4.1) and the Coulomb s law, the electic field at a distance fom a point chage q is given b 1 q E = 4πε 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: E = E = i 1 q i ˆ (.4.3) 4πε i i i.4.1 Electic Field of Point Chages (link) 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 (link), (b) a moving negative chage (link), when the speed of the chage is small compaed to the speed of light. -8
.5 Electic Field Lines Electic 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 = i Q i. Thus, the lines should be adiall outwad, unless Q =. -9
(4) Null point: This is a point at which E=, and no field lines should pass though it. The popeties of electic field lines ma be summaized as follows: The diection of the electic field vecto E 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 Electic Field Conside a chage +q moving between two paallel plates of opposite chages, as shown in Figue.6.1. Figue.6.1 Chage moving in a constant electic field Let the electic field between the plates be E = E ĵ, with E >. (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 e = qe (.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. Even 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, E is the field that aises fom the souce chages onl. -1
Accoding to Newton s second law, a net foce will cause the chage to acceleate with an acceleation F e qe qe a = = = ˆj (.6.) m m m 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 v = a = qe (.6.3) m whee is the distance between the two plates. The kinetic eneg of the paticle when it stikes the plate is.7 Electic Dipole K = 1 mv = qe (.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 Electic dipole The dipole moment vecto p which points fom q to +q (in the + - diection) is given b p = qa ĵ (.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 -11
i = N p q i i=1 i (.7.) whee is the position vecto of the chage q i i. Examples of dipoles include HCL, CO, H O 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 Electic Field of a Dipole What is the electic field due to the electic dipole? Refeing to Figue.7.1, we see that the x-component of the electic field stength at the point P is q E x = + cosθ + cos θ q = x x + ( a) 3/ x (.7.3) x + ( + a) 3/ whee ± = + a m cos θ = x a + ( m a) (.7.4) Similal, the -component is q sin θ E = + sin θ q 4πε = a 4πε + a 3/ 3/ (.7.5) + x + ( a) x + ( + a) In the point-dipole limit whee a, one ma veif that (see Solved Poblem.13.4) the above expessions educe to and 3p E x = sinθ cos θ (.7.6) 3 p E = (3cos θ 4πε 1) (.7.7) 3 whee sinθ = x / and cosθ = /. With 3 p cos θ = 3p and some algeba, the electic field ma be witten as 1 p 3( p ) E () = + 4πε 3 5 (.7.8) -1
Note that Eq. (.7.8) is valid also in thee dimensions whee = x î + ĵ + zkˆ. The equation indicates that the electic field E due to a dipole deceases with as 1/ 3, 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. Electic field lines fo (a) a finite dipole and (b) a point dipole..7. Electic Dipole Animation (link) 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 Electic Field What happens when we place an electic dipole in a unifom field E = E î, with the dipole moment vecto p making an angle with the x-axis? Fom Figue.8.1, we see that the unit vecto which points in the diection of p is cosθ î + sin θ ĵ. Thus, we have -13
p = qa (cos θ î + sin θ ĵ) (.8.1) Figue.8.1 Electic 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 F net = F + + F =. Even 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 = (a cos θ ˆi + a sin θ ĵ ) ( F ˆ i) + ( a cos θ ˆ i a sinθ ˆ) j ( F ˆi) + + + = a sinθ F + ( kˆ ) + a sin θ F ( kˆ ) (.8.) = af sin θ ( kˆ ) whee we have used F + = F = F. The diection of the toque is kˆ, 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 E. With F = qe, the magnitude of the toque can be ewitten as τ = aqe ( )sin θ = ( aqe ) sin θ = pe sin θ and the geneal expession fo toque becomes τ = p E (.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 Eneg of an Electic Dipole The wok done b the electic field to otate the dipole b an angle dθ is dw = τd θ = pe sinθ d θ (.8.4) -14
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 θ θ W = ( pe sin θ)d θ = pe (cos θ cos θ ) (.8.5) The esult shows that a positive wok is done b the field when cosθ >cosθ. The change in potential eneg ΔU of the dipole is the negative of the wok done b the field: Δ U = U U = W = pe (cosθ cos θ ) (.8.6) whee U = PE cos θ is the potential eneg at a efeence point. We shall choose ou efeence point to be θ = 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 = pe cos θ = pe (.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 E, making U a minimum with U min = pe. On the othe hand, when p and E ae anti-paallel, U max =+pe is a maximum and the sstem is unstable. If the dipole is placed in a non-unifom 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 E + at +q diffes fom the electic field E at q. Figue.8. Foce on a dipole Assuming the dipole to be ve small, we expand the fields about x : ( + a) E x + a de de E+ x ( ), E (x a) E( x ) a (.8.8) dx dx -15
The foce on the dipole then becomes de de F e = q(e + E ) = qa î = p î (.8.9) dx 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 non-unifom due to its iegula shape (Figue.8.3). Figue.8.3 Electostatic 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 Electic field due to a small chage element Δq 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 ΔV i which contains an amount of chage Δq i. The distances between chages within the volume element ΔV i ae much smalle than compaed to, the distance between ΔV i and P. In the limit whee ΔV i becomes infinitesimall small, we ma define a volume chage densit ρ( ) as -16
() = lim Δq i dq ρ = ΔV i ΔV i dv (.9.1) The dimension of ρ( ) is chage/unit volume chage within the entie volume V is 3 (C/m ) in SI units. The total amount of Q () dv = Δq i = ρ (.9.) i The concept of chage densit hee is analogous to mass densit ρ m( ). 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.9. Suface Chage Densit V M = ρm () dv (.9.3) V 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): dq σ () = da (.9.4) The dimension of σ is chage/unit aea suface is:.9.3 Line Chage Densit (C/m ) in SI units. The total chage on the entie Q = σ () da (.9.5) S If the chage is distibuted ove a line of length l, then the linea chage densit λ (lowecase Geek lette lambda) is λ() dq = dl (.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 () dl (.9.7) line -17
If chages ae unifoml distibuted thoughout the egion, the densities ( ρ,σ o λ) then become unifom..1 Electic 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 de = 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 E is the vecto sum (integal) of all these infinitesimal contibutions: 1 E = dq V ˆ (.1.) This is an example of a vecto integal which consists of thee sepaate integations, one fo each component of the electic field. Example.: Electic Field on the Axis of a Rod A non-conducting od of length l with a unifom positive chage densit λ and a total chage Q is ling along the x -axis, as illustated in Figue.1.1. Figue.1.1 Electic 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: -18
The linea chage densit is unifom and is given b λ = Q / l. The amount of chage contained in a small segment of length dx is dq = λ dx. 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 = î. The contibution to the electic field due to dq is 1 dq 1 λdx 1 Qdx de = ˆ = ( î ) = î x l x Integating ove the entie length leads to 1 Q x +l dx 1 Q 1 1 1 Q E = de = l x x î = l x x + l î = î (.1.3) x (l + x ) Notice that when P is ve fa awa fom the od, x becomes l, and the above expession 1 Q E 4πε x î (.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. Example.3: Electic Field on the Pependicula Bisecto (link) A non-conducting od of length l 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. Solution: Figue.1. -19
We follow a simila pocedue as that outlined in Example.. The contibution to the electic field fom a small length element dx caing chage dq = λdx is 1 dq de = = 1 λ dx (.1.5) x + Using smmet agument illustated in Figue.1.3, one ma show that the x component of the electic field vanishes. The -component of de is Figue.1.3 Smmet agument showing that E x =. 1 λ dx 1 λ dx de = de cosθ = = (.1.6) 3/ x + x + (x + ) B integating ove the entie length, the total electic field due to the od is E = de = 1 l / λ dx = λ l / dx l / 3/ (.1.7) l / 3/ (x + ) (x + ) B making the change of vaiable: x = tan θ, which gives dx = sec θ d θ, the above integal becomes l / dx θ sec θ d θ = = 1 θ sec θ d θ = 1 θ sec θ dθ l / (x + ) 3/ θ 3 (sec θ +1) 3/ θ (tan θ +1) 3/ θ sec θ 3 (.1.8) 1 θ dθ 1 θ d = sin θ = θ θ θ secθ = cos θ which gives -
1 λ sin θ 1 λ l / E = = (.1.9) 4πε + ( l /) In the limit whee l, the above expession educes to the point-chage limit: On the othe hand, when l, we have 1 λ l / 1 λl 1 Q E = = (.1.1) 1 λ E (.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 Eq. (.1.11), as we shall show in Chapte 4. The chaacteistic behavio of / = /4 E E (with E Q πε l ) as a function of / l is shown in Figue.1.4. Figue.1.4 Electic field of a non-conducting od as a function of / l. Example.4: Electic Field on the Axis of a Ring (link) A non-conducting 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. -1
Figue.1.5 Electic field at P due to the chage element dq. Solution: Conside a small length element dl on the ing. The amount of chage contained within this element is dq = λ dl = λr d φ. Its contibution to the electic field at P is 1 dq de = ˆ = 1 λr d φ 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. 1 λrdφ z λ Rzd φ de z = de cosθ = = (.1.13) R + z R + z (R + z ) 3/ Upon integating ove the entie ing, we obtain λ Rz λ π Rz 1 Qz E z = (R + z ) dφ = = (.1.14) 3/ (R + z ) 3/ (R + z 3/ ) 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 Electic field along the axis of smmet of a non-conducting ing of adius R, with E = Q / R. Notice that the electic field at the cente of the ing vanishes. This is to be expected fom smmet aguments. Example.5: Electic Field Due to a Unifoml Chaged Disk A unifoml chaged disk of adius R with a total chage Q lies in the x-plane. Find the electic field at a point P, along the z-axis 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 Example.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 Eq. (.1.14), we see that the ing gives a contibution -3
1 zdq 1 z( πσ d ) de z = = ( + z ) 3/ ( + z ) 3/ (.1.15) Integating fom = to = R, the total electic field at P becomes σ z R d σ z R + z 1/ du σ z u R + z E z = de = z ε = ( z ) 3/ 4ε = z + u 3/ 4ε ( 1/) z σ z 1 1 σ z z = = ε R + z z ε z R + z (.1.16) The above equation ma be ewitten as σ z 1, z > ε z + R E z = σ z 1, z < ε z + R (.1.17) The electic field E / z E ( E = σ / ε ) as a function of z/ Ris shown in Figue.1.9. Figue.1.9 Electic field of a non-conducting plane of unifom chage densit. To show that the point-chage limit is ecoveed fo z R, we make use of the Talo-seies expansion: This gives z R 1/ 1 R 1 R 1 =1 1+ =1 1 +L (.1.18) z + R z z z -4
σ R 1 σπ R 1 Q E z = = = (.1.19) ε z 4 πε z z which is indeed the expected point-chage 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 unit-vecto notation, σ ˆ k, z > E = ε (.1.) σ kˆ, z < ε The plot of the electic field in this limit is shown in Figue.1.1. Figue.1.1 Electic field of an infinitel lage non-conducting plane. Notice the discontinuit in electic field as we coss the plane. The discontinuit is given b σ σ σ ΔE z = E z+ E z = = (.1.1) ε ε ε 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 ΔE n = σ / ε..11 Summa The electic foce exeted b a chage q 1 on a second chage q is given b Coulomb s law: -5
whee qq 1 1 qq F = k 1 ˆ 1 e = ˆ is the Coulomb constant. 1 9 k = N m / C e = 8.99 1 4πε The electic field at a point in space is defined as the electic foce acting on a test chage q divided b q : F E= lim e q q The electic field at a distance fom a chage q is 1 q E = ˆ Using the supeposition pinciple, the electic field due to a collection of point chages, each having chage q i and located at a distance i awa is 1 E = 4πε A paticle of mass m and chage q moving in an electic field E has an acceleation i qe a = m q i ˆi i 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 E is τ = p E The potential eneg of an electic dipole in a unifom extenal electic field E is -6
U = pe The electic field at a point in space due to a continuous chage element dq is 1 dq de = ˆ At sufficientl fa awa fom a continuous chage distibution of finite extent, the electic field appoaches the point-chage limit..1 Poblem-Solving 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 E = 4πε i Fo the latte, we must evaluate the vecto integal 1 E = q i ˆi i 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 de = ˆ () Rewite the chage element dq as λ dl dq = σ da ρ dv (length) (aea) (volume) depending on whethe the chage is distibuted ove a length, an aea, o a volume. -7
(3) Substitute dq into the expession fo de. (4) Specif an appopiate coodinate sstem (Catesian, clindical o spheical) and expess the diffeential element ( dl, 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 d ρ, ρd φ, dz d, d θ, sin θ dφ da dx d, d dz, dz dx dρ dz, ρd φ dz, ρd φ dρ dd θ, sin θ dd φ, sin θ d θ dφ dv dx d dz ρ dρdφ dz sinθ d d θ dφ Table.1 Diffeential elements of length, aea and volume in diffeent coodinates (5) Rewite de in tems of the integation vaiable(s), and appl smmet agument to identif non-vanishing component(s) of the electic field. (6) Complete the integation to obtain E. 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 () Expess dq in tems of chage densit (3) Wite down de dq de = k = λ dx dq = λ dl dq = σ da e λ dx λ dl σ da de = k e de = k e -8
(4) Rewite and the diffeential element in tems of the appopiate coodinates dx cos θ = = x + d l = R dφ z cos θ = = R + z da = π ' d ' z cos θ = = + z (5) Appl smmet agument to identif non-vanishing component(s) of de de = de cos θ = k λ dx + e 3/ ( x ) de z = de cos θ = k λ Rz d φ de = de cos θ πσ z d ( + z ) e 3/ e ( R + z ) 3/ z = k (6) Integate to get E E = k λ e + / l l / dx ( x + ) 3/ k e λ l / = ( l / ) + Rλ z E z = k e dφ 3/ ( R + z ) = k = k ( πrλ ) z e 3/ ( R + z ) Qz e 3/ ( R + z ) E = πσ k z z e R d ( + z ) 3/ z z = πσ k e z z + R.13 Solved Poblems.13.1 Hdogen Atom In the classical model of the hdogen atom, the electon evolves aound the poton with a adius of = 53. 1 1 m. The magnitude of the chage of the electon and poton is e = 1.6 1 19 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 -9
1 e F e = 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 19 (9. 1 N m / C )(1.6 1 C) 8 = 8. 1 N 11 (5.3 1 m) (b) The magnitude of the electic field due to the poton is given b E = 9 19 1 q (9. 1 N m / C )(1.6 1 C) = = 5.76 1 11 N / C 1 (.5 1 m) (c) The mass of the electon is m e = 9. 1 1 31 kg and the mass of the poton is m = 1. 7 1 7 p kg. Thus, the atio of the magnitudes of the electic and gavitational foce is given b 1 e 1 4πε e 9 19 4 πε (9. 1 N m / C )(1.6 1 C) 39 γ = = = =. 1 11 7 31 mm p Gm pm e (6.67 1 N m / kg )(1.7 1 kg)(9.1 1 kg) G e 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 of1 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 Oil-Dop Expeiment 6 An oil dop of adius = 1.64 1 m and mass densit ρ oil = 8.51 1 kg m 3 is allowed to fall fom est and then entes into a egion of constant extenal field E applied in the downwad diection. The oil dop has an unknown electic chage q (due to iadiation b busts of X-as). The magnitude of the electic field is adjusted until the -3
gavitational foce F = mg g = mg ĵ on the oil dop is exactl balanced b the electic foce, F e = qe. Suppose this balancing occus when the electic field is ˆ 5 ˆ 5 E = E j = (1.9 1 N C) j, with E =1.9 1 N C. (a) What is the mass of the oil dop? (b) What is the chage on the oil dop in units of electonic chage e =1.6 1 19 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, M = ρ oil V = ρ oil 4 π 3 3 whee the oil dop is assumed to be a sphee of adius with volume V = 4π 3 /3. Now we can substitute ou numeical values into ou smbolic expession fo the mass, 4 3 3 4π 6 3 14 M = ρ oil π = (8.51 1 kg m ) (1.64 1 m) =1.57 1 kg 3 3 (b) The oil dop will be in static equilibium when the gavitational foce exactl balances the electical foce: F g + F e =. Since the gavitational foce points downwad, the electic foce on the oil must be upwad. Using ou foce laws, we have = mg + qe mg = qe 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 (1.57 1 kg)(9.8m / s ) 19 q = = = 8.3 1 C 5 E 1.9 1 N C 19 Since the electon has chage e = 16 1. C, the chage of the oil dop in units of e is q 8. 1 19 C N = = = 5 19 e 1.6 1 C -31
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 Electic Field An electon is injected hoizontall into a unifom field poduced b two oppositel chaged plates, as shown in Figue.13.1. The paticle has an initial velocit v = v î pependicula to E. Figue.13.1 Chage moving pependicula to an electic field (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 L 1 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 t 1 when it leaves the plates? (e) What is the vetical displacement of the electon afte time t 1 when it leaves the plates? (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 L 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 sceen at t? Solutions: -3
(a) Since the electon has a negative chage, q = e, the foce on the electon is u u Fe = qe = e E = ( e )( E )ˆ j = ee whee the electic field is witten as E = E ĵ, with E >. 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 and its diection is upwad. qe qe a = = ĵ = ee ĵ m m m (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 î. Since the acceleation of the electon is in the + -diection, onl the -component of the velocit changes. The velocit at a late time t 1 is given b î + v ĵ = î + a ĵ = v ˆ i ee ee L x 1 + 1 ĵ v ˆ 1 v = v v t t = i+ ˆj m mv (e) Fom the figue, we see that the electon tavels a hoizontal distance L 1 in the time t 1 = L 1 v and then emeges fom the plates with a vetical displacement 1 1 ee L 1 1 = a t 1 = m v (f) When the electon leaves the plates at timet 1, the electon makes an angle θ 1 with the hoizontal given b the atio of the components of its velocit, ˆ j v (ee / m)( L 1 / v ) ee L tan θ = = = v x v mv 1-33
(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 and the total deflection becomes ee L L = L tan θ 1 = mv 1 1 ee L 1 ee LL 1 ee L 1 1 = 1 + = + = L 1 + L mv mv mv.13.4 Electic 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 whee sinθ = x / and cos θ = /. 3p E x = sinθ cos θ, E p 4πε 3 = 4πε 3 (3cos θ 1) (b) Show that the above expession fo the electic field can also be witten in tems of the pola coodinates as E (, θ ) = E ˆ + E θ θˆ whee E = p cos θ 3, Eθ = p sin θ 3 Solutions: (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 q cosθ cos θ q x x E = + = x + x + ( a) 3/ x + ( + a) 3/ -34
whee Similal, the -component is given b q E = 4πε sin ± = + a m a cos θ = x + ( m a) θ + sin θ + q = 4πε a x + ( a) 3/ + a x + ( + a) 3/ We shall make a polnomial expansion fo the electic field using the Talo-seies expansion. We will then collect tems that ae popotional to 1/ 3 and ignoe tems that ae popotional to 1/ 5, whee =+(x + ) 1. We begin with 3/ 3/ 3 a ± a [x + ( ± a) ] = [x + + a ± a ] = 1 + In the limit whee >> a, we use the Talo-seies expansion with s (a ± a )/ : (1 + s) 3/ = 1 3 s + 15 s... 8 and the above equations fo the components of the electic field becomes 3/ and q 6xa E x = +... 5 q a 6 a E = + +... 3 5 whee we have neglected the Os ( )tems. The electic field can then be witten as E = E î + E ĵ = q a 6a ĵ + (x î + ĵ) p 3x = î + 3 1 ĵ x 4πε 3 5 3 whee we have made used of the definition of the magnitude of the electic dipole moment p = aq. -35
In tems of the pola coodinates, with sinθ = x and cosθ = (as seen fom Figue.13.4), we obtain the desied esults: 3p E x = sin θ cos θ, 4πε E p = 4πε 3 3 ( 3cos θ 1) (b) We begin with the expession obtained in (a) fo the electic dipole in Catesian coodinates: E(, ) θ p = 4πε 3sin θ cos θ î + (3cos θ 1) ĵ 3 With a little algeba, the above expession ma be ewitten as p E(,θ ) = 4πε 3 cos θ (sinθ î + cos θ ĵ) + sin θ cos θ î + (cos θ 1) ĵ p = 4πε î sin θ ĵ) 3 cos θ (sinθ î + cos θ ĵ) + sin θ (cos θ whee the tigonometic identit (cos θ 1) = sin θ has been used. Since the unit vectos ˆ and θˆ in pola coodinates can be decomposed as ˆ = sin θ î + cos θ ĵ θˆ = cos θ î sin θ ĵ, the electic field in pola coodinates is given b u and the magnitude of E is p E(, θ ) = 4πε 3 cos θ ˆ + sin θ θˆ E = (E + E θ ) 1/ p = 4πε 3 (3cos θ +1) 1/.13.5 Electic 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 x-axis, as shown in Figue.13.. What is the electic field E at the oigin O? -36
Solution: Conside a diffeential element of length dl = 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 dl = λr d θ. The contibution to the electic field at O is 1 dq 1 dq 1 λdθ de = ˆ = ( cos θî R sin θ ĵ) = ( cos R θî sin θ ĵ) Figue.13. (a) Geomet of chaged souce. (b) Chage element dq Integating ove the angle fom θ to +θ, we have 1 λ θ 1 λ θ = 1 λ sin θ î E = R θ dθ ( cosθî sin θ ĵ) = R ( sin θî+ cos θ ĵ) 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 non-conducting ing is zeo. This is to be expected fom smmet aguments. On the othe hand, fo ve small θ, sinθ θ and we ecove the point-chage limit: 1 λθ E î = 1 λθ R î = 1 Q R R R whee the total chage on the ac is Q = λl = λ(rθ ). î.13.6 Electic Field Off the Axis of a Finite Rod -37
A non-conducting od of length l with a unifom chage densit λ and a total chage Q is ling along the x -axis, as illustated in Figue.13.3. Compute the electic field at a point P, located at a distance off the axis of the od. Figue.13.3 Solution: The poblem can be solved b following the pocedue used in Example.3. Conside a length element dx on the od, as shown in Figue.13.4. The chage caied b the element is dq = λ dx. The electic field at P poduced b this element is Figue.13.4 1 dq 1 λ dx de = ˆ = ( sinθ î + cos θ ĵ) x + whee the unit vecto ˆ has been witten in Catesian coodinates: ˆ = sin θ î + cos θ ĵ. In the absence of smmet, the field at P has both the x- and -components. The x- component of the electic field is -38
1 λ dx 1 λ dx x 1 λx dx de = sinθ = = x + x + x + (x + ) x 3/ Integating fom x = x 1 to x = x, we have E = x λ x xdx λ 1 x + du λ = x1 (x + ) 3/ x 1 + u 3/ λ = (cosθ cos θ 1 ) = u 1/ x + x + λ 1 1 λ = = 4πε x + x 1 + 4πε x + x 1 + 1 Similal, the -component of the electic field due to the chage element is 1 λ dx 1 λ dx 1 λ dx de = cos θ = = x + x + x + (x + ) 3/ Integating ove the entie length of the od, we obtain λ E = 4 dx λ = 4 x θ πε x 1 1 (x + ) 3/ πε θ 1 λ cos θ d θ = 4πε (sin θ sin θ 1 ) whee we have used the esult obtained in Eq. (.1.8) in completing the integation. In the infinite length limit whee x 1 and x +, with x i = tan θ i, the coesponding angles ae θ 1 = π / and θ =+π /. Substituting the values into the expessions above, we have 1 λ E x =, E = in complete ageement with the esult shown in Eq. (.1.11)..14 Conceptual Questions 1. Compae and contast Newton s law of gavitation, F = Gm m /, and g 1 Coulomb s law, F = kqq e 1 /.. Can electic field lines coss each othe? Explain. -39
3. Two opposite chages ae placed on a line as shown in the figue below. 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 positivel-chaged insulating od. 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 non-unifom electic field. Does it expeience a foce and/o a toque? -4
MIT OpenCouseWae http://ocw.mit.edu 8.SC Phsics II: Electicit and Magnetism Fall 1 Fo infomation about citing these mateials o ou Tems of Use, visit: http://ocw.mit.edu/tems.