Chapter 21 The Electric Field 1: Discrete Charge Distributions

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Russell Ramsey
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1 hapte The Electic Field : Discete hage Distibutions onceptual Poblems 3 Two point paticles that have chages of + and 3 ae sepaated b distance d. (a) Use field lines to sketch the electic field in the neighbohood of this sstem. (b) Daw the field lines at distances much geate than d fom the chages. Detemine the oncept (a) We can use the ules fo dawing electic field lines to daw the electic field lines fo this sstem. In the field-line sketch we ve assigned field lines to each chage. (b) At distances much geate than the sepaation distance between the two chages, the sstem of two chaged bodies will look like a single chage of and the field patten will be that due to a point chage of. Fou field lines have been assigned to each chage. (a) (b) 7 Two molecules have dipole moments of eual magnitude. The dipole moments ae oiented in vaious configuations as shown in Figue -34. Detemine the electic-field diection at each of the numbeed locations. Explain ou answes. omment [EPM]: Detemine the oncept Figue -3 shows the electic field due to a single dipole, whee the dipole moment is diected towad the ight. The electic field due two a pai of dipoles can be obtained b supeposing the two electic fields. 3 (a) down up up (b) up ight left (c) down up up (d) down up up

2 hapte hage 3 What is the total chage of all of the potons in. kg of cabon? Pictue the Poblem We can find the numbe of coulombs of positive chage thee ae in. kg of cabon fomq 6ne, whee n is the numbe of atoms in. kg of cabon and the facto of 6 is pesent to account fo the pesence of 6 potons in each atom. We can find the numbe of atoms in. kg of cabon b setting up a popotion elating Avogado s numbe, the mass of cabon, and the molecula mass of cabon to n. See Appendix fo the mola mass of cabon. Expess the positive chage in tems of the electonic chage, the numbe of potons pe atom, and the numbe of atoms in. kg of cabon: Q 6n e Using a popotion, elate the numbe of atoms in. kg of cabon n, to Avogado s numbe and the molecula mass M of cabon: n m N M A n N Am M Substitute fo n to obtain: 6N Ame Q M Substitute numeical values and evaluate Q: 6 6. Q 3 atoms 9 (. kg)(.6 ) mol kg. mol oulomb s Law 7 Thee point chages ae on the x-axis: 6. μ is at x 3. m, 4. μ is at the oigin, and 3 6. μ is at x 3. m. Find the electic foce on.

3 The Electic Field : Discete hage Distibutions 3 Pictue the Poblem exets an attactive electic foce F, on point chage and 3 exets a epulsive electic foce F 3, on point chage. We can find the net electic foce on b adding these foces (that is, b using the supeposition pinciple). F 3, F 3, μ 4.μ Expess the net foce acting on : F F, + F3, Expess the foce that exets on : F, k, 3 6.μ Expess the foce that 3 exets on k 3 F ( 3, ) : 3, x, m Substitute and simplif to obtain: F k k 3, 3, 3 k, 3, Substitute numeical values and evaluate F : F 9 4. μ 6. μ ( )( ) N m / 6. μ (.5 N) ( ) ( ) 3.m 6.m 35 Five identical point chages, each having chage Q, ae euall spaced on a semicicle of adius R as shown in Figue -37. Find the foce (in tems of k, Q, and R) on a chage located euidistant fom the five othe chages. Pictue the Poblem B consideing the smmet of the aa of chaged point paticles, we can see that the component of the foce on is zeo. We can appl oulomb s law and the pinciple of supeposition of foces to find the net foce acting on. Expess the net foce acting on the point chage : F F + F Q on x axis, Q at 45,

4 4 hapte Expess the foce on point chage due to the point chage Q on the x axis: Expess the net foce on point chage due to the point chages at 45 : F Q on x axis, F Q at 45, kq R kq cos45 R kq R Substitute fo F F to obtain: Q at 45, Qon xaxis, The Electic Field and F kq kq + R R kq ( + ) R 37 A point chage of 4. μ is at the oigin. What is the magnitude and diection of the electic field on the x axis at (a) x 6. m, and (b) x m? (c) Sketch the function E x vesus x fo both positive and negative values of x. (Remembe that E x is negative when E points in the x diection.) Pictue the Poblem Let epesent the point chage at the oigin and use oulomb s law fo E due to a point chage to find the electic field at x 6. m and m. (a) Expess the electic field at a point P located a distance x fom a point chage : E k x ˆ ( x ) P, Evaluate this expession fo x 6. m: E (.m) N m ( 6.m) ( 4. μ) (. kn/) (b) Evaluate E at x m: E 9 N m ( m) ( m) ( 4. μ) ( ) (.36 kn/)

5 The Electic Field : Discete hage Distibutions 5 (c) The following gaph was plotted using a speadsheet pogam: E x (kn/) E x (N/) x (m) 4 Two point chages and both have a chage eual to +6. n and ae on the axis at +3. cm and 3. cm espectivel. (a) What is the magnitude and diection of the electic field on the x axis at x 4. cm? (b) What is the foce exeted on a thid chage. n when it is placed on the x axis at x 4. cm? Deleted: Pictue the Poblem The diagam shows the locations of the point chages and and the point on the x axis at which we ae to find E. Fom smmet consideations we can conclude that the component of E at an point on the x axis is zeo. We can use oulomb s law fo the electic field due to point chages and the pinciple of supeposition fo fields to find the field at an point on the x axis and F E to find the foce on a point chage placed on the x axis at x 4. cm., cm n θ 4. θ x, cm E n

6 6 hapte (a) Letting, expess the x- component of the electic field due to one point chage as a function of the distance fom eithe point chage to the point of inteest: Expess E x fo both chages: E E x x k cosθ k cos θ Substitute fo cosθ and, substitute numeical values, and evaluate to obtain: E ( 4. cm) x k.4m k (.4 m) 9 ( N m / )( 6.n)(.4 m) 3 [(.3 m) + (.4 m) ] ( 34.5kN/) ( 35kN/) 3 The magnitude and diection of the electic field at x 4. cm is: (b) Appl F E to find the foce on a point chage placed on the x axis at x 4. cm: 35 F (.n)( 34.5kN/) ( 69 μn) 47 Two point paticles, each having a chage, sit on the base of an euilateal tiangle that has sides of length L as shown in Figue -38. A thid point paticle that has a chage eual to sits at the apex of the tiangle. Whee must a fouth point paticle that has a chage eual to be placed in ode that the electic field at the cente of the tiangle be zeo? (The cente is in the plane of the tiangle and euidistant fom the thee vetices.) Pictue the Poblem The electic field of 4 th chaged point paticle must cancel the sum of the electic fields due to the othe thee chaged point paticles. B smmet, the position of the 4 th chaged point paticle must lie on the vetical centeline of the tiangle. Using tigonomet, one can show that the cente of an euilateal tiangle is a distance L 3 fom each vetex, whee L is the length of the side of the tiangle. Note that the x components of the fields due to the base chaged paticles cancel each othe, so we onl need concen ouselves with the components of the fields due to the chaged point paticles at the vetices of the tiangle. hoose a coodinate sstem in which the oigin is at the midpoint of the base of the tiangle, the +x diection is to the ight, and the + diection is upwad. Deleted: Note that the x components of the electic field vectos add up to zeo.

7 The Electic Field : Discete hage Distibutions 7 3 L 3 E 6 E L 3 L 3 6 E 3 x Expess the condition that must be satisfied if the electic field at the cente of the tiangle is to be zeo: E i i to 4 Substituting fo E, E, E 3, and E 4 ields: ( ) k( ) k( ) k L 3 cos6 ˆj + L 3 cos6 ˆj L 3 ˆ k j + ˆj Solving fo ields: ± The positive solution coesponds to the 4 th point paticle being a distance L 3 above the base of the tiangle, whee it poduces the same stength and same diection electic field caused b the thee chages at the cones of the tiangle. So the chaged point paticle must be placed a distance L 3 below the midpoint of the tiangle. Point hages in Electic Fields 5 The acceleation of a paticle in an electic field depends on /m (the chage-to-mass atio of the paticle). (a) ompute /m fo an electon. (b) What is the magnitude and diection of the acceleation of an electon in a unifom electic field that has a magnitude of N/? (c) ompute the time it takes fo an electon placed at est in a unifom electic field that has a magnitude of N/ to each a speed of.c. (When the speed of an electon appoaches the speed of light c, elativistic kinematics must be used to calculate its motion, but at speeds L 3 Deleted: Because the positive solution coesponds to the 4 th chage being at the cente of the tiangle, it follows that:... [] Deleted: The chaged point paticle must be placed a distance eual to the length of the side of the tiangle divided b the suae oot of thee below the midpoint of the base of the tiangle, whee L is the length of a side of the tiangle.

8 8 hapte of.c o less, non-elativistic kinematics is sufficientl accuate fo most puposes.) (d) How fa does the electon tavel in that time? Pictue the Poblem We can use Newton s second law of motion to find the acceleation of the electon in the unifom electic field and constant-acceleation euations to find the time euied fo it to each a speed of.c and the distance it tavels while acuiing this speed. (a) Use data found at the back of ou text to compute e/m fo an electon: e m e kg /kg (b) Appl Newton s second law to elate the acceleation of the electon to the electic field: Substitute numeical values and evaluate a: F ee a net m a e m e 9 (.6 )( N/) m/s m/s kg The diection of the acceleation of an electon is opposite the electic field. (c) Using the definition of acceleation, elate the time euied fo an electon to each.c to its acceleation: Substitute numeical values and evaluate Δt: (d) Use a constant-acceleation euation to expess the distance the electon tavels in a given time inteval: Substitute numeical values and evaluate Δx: v.c Δ t a a 8 ( m/s)..998 t.759 Δ 3.μs Δx v Δt + i a ( Δt) o, because v i, Δx a Δt ( ) m/s.74 μs 3 (.759 m/s )(.74 s) Δx μ.3m Deleted: Page Beak Deleted: Δx v Δt ( v v o, because v i, Δx vf Δt av i + Deleted: (d) The distance the electon tavels in a given time inteval is the poduct of its aveage speed and the elapsed time: Deleted: Substitute numeical values and evaluate Δx:... [] f

9 The Electic Field : Discete hage Distibutions 9 57 An electon stats at the position shown in Figue -39 with an initial speed v 5. 6 m/s at 45º to the x axis. The electic field is in the + diection and has a magnitude of N/. The black lines in the figue ae chaged metal plates. On which plate and at what location will the electon stike? Pictue the Poblem We can use constant-acceleation euations to expess the x and coodinates of the electon in tems of the paamete t and Newton s second law to expess the constant acceleation in tems of the electic field. Eliminating the paamete will ield an euation fo as a function of x,, and m. We can decide whethe the electon will stike the uppe plate b finding the maximum value of its coodinate. Should we find that it does not stike the uppe plate, we can detemine whee it stikes the lowe plate b setting (x). Ignoe an effects of gavitational foces. Expess the x and coodinates of the electon as functions of time: Appl Newton s second law to elate the acceleation of the electon to the net foce acting on it: Substitute in the -coodinate euation to obtain: Eliminate the paamete t between the two euations to obtain: To find max, set d/dx fo extema: Solve fo x to obtain: Substitute x in (x) and simplif to obtain max : x ( v cosθ )t and ( ) v sinθ t a t F a net, me ee m e ( v sin ) t t θ ee m ( x) ( tanθ ) x x d dx x' e ee () m v cos θ ee tanθ x' mev cos θ fo extema mev sin θ (See emak below.) ee max mev sin θ ee e

10 hapte Substitute numeical values and evaluate max : 3 6 ( 9.9 kg)( 5. m/s) sin 9 3 (.6 )( 3.5 N/) 45 max. cm and, because the plates ae sepaated b cm, the electon does not stike the uppe plate. To detemine whee the electon will stike the lowe plate, set in euation () and solve fo x to obtain: mev sin θ x ee Substitute numeical values and evaluate x: x 3 6 ( 9.9 kg)( 5. m/s) 9 3 (.6 )( 3.5 N/) sin 9 4.cm Deleted: Remaks: x is an extemum, that is, eithe a maximum o a minimum. To show that it is a maximum we need to show that d /dx, evaluated at x, is negative. A simple altenative is to use ou gaphing calculato to show that the gaph of (x) is a maximum at x. Yet anothe altenative is to ecognize that, because euation () is uadatic and the coefficient of x is negative, its gaph is a paabola that opens downwad. Geneal Poblems 6 Show that it is onl possible to place one isolated poton in an odina empt coffee cup b consideing the following situation. Assume the fist poton is fixed at the bottom of the cup. Detemine the distance diectl above this poton whee a second poton would be in euilibium. ompae this distance to the depth of an odina coffee cup to complete the agument. Pictue the Poblem Euilibium of the second poton euies that the sum of the electic and gavitational foces acting on it be zeo. Let the upwad diection be the + diection and appl the condition fo euilibium to the second poton. Appl F to the second poton: Fe + Fg o kp mpg h h kp m g p omment [EPM]: DAVID: M coffee mug is moe than 5 in high, so I changed mug to cup. omment [MSOffice3]: At close distances, potons will bind togethe b the stong nuclea foce. This uestion confuses students. omment [EPM4]: To the best of m knowledge, Helium- does not exist. Deleted:

11 Substitute numeical values and evaluate h: The Electic Field : Discete hage Distibutions h N m 9 (.6 ) 7 (.673 kg)( 9.8 m/s ) cm 5 in This sepaation of about 5 in is geate than the height of a tpical coffee cup. Thus the fist poton will epel the second one out of the cup and the maximum numbe of potons in the cup is one. 65 A positive chage Q is to be divided into two positive point chages and. Show that, fo a given sepaation D, the foce exeted b one chage on the othe is geatest if Q. Pictue the Poblem We can use oulomb s law to expess the foce exeted on one chage b the othe and then set the deivative of this expession eual to zeo to find the distibution of the chage that maximizes this foce. Using oulomb s law, expess the foce that eithe chage exets on the othe: k F D Expess in tems of Q and : Q Substitute fo to obtain: k ( Q ) F D Diffeentiate F with espect to and set this deivative eual to zeo fo exteme values: Solve fo to obtain: df d k d D d k D fo extema [ ( Q )] [ ( ) + Q ] Q Q Q

12 hapte To detemine whethe a maximum o a minimum exists at Q, diffeentiate F a second time and evaluate this deivative at Q : d F k d [ Q ] d D d k ( ) D < independentl of. Q maximizes F. 69 A igid.-m-long od is pivoted about its cente (Figue -4). A chage 5. 7 is placed on one end of the od, and a chage is placed a distance d. cm diectl below it. (a) What is the foce exeted b on? (b) What is the toue (measued about the otation axis) due to that foce? (c) To countebalance the attaction between the two chages, we hang a block 5. cm fom the pivot as shown. What value should we choose fo the mass of the block? (d) We now move the block and hang it a distance of 5. cm fom the balance point, on the same side of the balance as the chage. Keeping the same, and d the same, what value should we choose fo to keep this appaatus in balance? Pictue the Poblem We can use oulomb s law, the definition of toue, and the condition fo otational euilibium to find the electostatic foce between the two chaged bodies, the toue this foce poduces about an axis though the cente of the od, and the mass euied to maintain euilibium when it is located eithe 5. cm to the ight o to the left of the mid-point of the od. (a) Using oulomb s law, expess the electic foce between the two chages: k F d Substitute numeical values and evaluate F: F 9 7 ( N m / )( 5. ) (. m).47 N.5 N (b) The toue (measued about the otation axis) due to the foce F is: Substitute numeical values and evaluate τ: τ Fl τ (.47 N)(.5 m).4 N m counteclockwise.. N m, (c) Appl τ cente of to the od: the od τ τ mgl' m gl' Deleted:

13 Substitute numeical values and The Electic Field : Discete hage Distibutions 3 m.4 N m evaluate m: ( 9.8m/s )(.5 m ).4583 kg 45.8g (d) Appl τ cente of to the od: τ + mgl' the od Substitute fo τ to obtain: Fl + mgl' Substituting fo F gives: k ' d l + mg ' d mgl' ' k l whee is the euied chage. Substitute numeical values and evaluate : ' (.m) (.458kg)( 9.8m/s )(.5 m) ( N m / )( 5. )(.5m) 7 Two point chages have a total chage of μ and ae sepaated b.6 m. (a) Find the chage of each paticle if the paticles epel each othe with a foce of N. (b) Find the foce on each paticle if the chage on each paticle is μ. Pictue the Poblem Let the numeal denote one of the point chages and the numeal the othe. Knowing the total chage on the two sphees, we can use oulomb s law to find the chage on each of them. A second application of oulomb s law when the sphees ca the same chage and ae.6 m apat will ield the foce each exets on the othe. (a) Use oulomb s law to expess the epulsive foce each point chage exets on the othe: Expess in tems of the total chage and : k F, Q Substitute fo to obtain: k ( Q ) F,

14 4 hapte Substitute numeical values to obtain: N 9 ( N m / )[( μ) ] (.6m) Simplif to obtain the uadatic euation: Use the uadatic fomula o ou gaphing calculato to obtain: Hence the chages on the paticles ae: (b) Use oulomb s law to expess the epulsive foce each point chage exets on the othe when μ: ( μ) + 486( ) + μ 7.9 and 7μ μ 7.9 μ and 7 μ k F, Substitute numeical values and evaluate F: F ( ) ( N m / ) 9 μ (.6m) 5 N 77 Figue -46 shows a dumbbell consisting of two identical small paticles, each of mass m, attached to the ends of a thin (massless) od of length a that is pivoted at its cente. The paticles ca chages of + and, and the dumbbell is located in a unifom electic field E. Show that fo small values of the angle θ between the diection of the dipole and the diection of the electic field, the sstem displas a otational fom of simple hamonic motion, and obtain an expession fo the peiod of that motion. Pictue the Poblem We can appl Newton s second law in otational fom to obtain the diffeential euation of motion of the dipole and then use the small angle appoximation sinθ θ to show that the dipole expeiences a linea estoing toue and, hence, will expeience simple hamonic motion. Appl τ Iα to the dipole: d θ pe sinθ I dt whee τ is negative because acts in such a diection as to decease θ.

15 The Electic Field : Discete hage Distibutions 5 Fo small values of θ, sinθ θ d θ peθ I and: dt Expess the moment of inetia of the dipole: Relate the dipole moment of the dipole to its chage and the chage sepaation: Substitute fo p and I to obtain: Expess the peiod of a simple hamonic oscillato: I ma p a d θ ma aeθ dt o d θ E θ dt ma the diffeential euation fo a simple hamonic oscillato with angula feuenc ω E ma. π T ω Substitute fo ω and simplif to obtain: T π ma E 79 An electon (chage e, mass m) and a positon (chage +e, mass m) evolve aound thei common cente of mass unde the influence of thei attactive coulomb foce. Find the speed v of each paticle in tems of e, m, k, and thei sepaation distance L. Pictue the Poblem The foces the electon and the poton exet on each othe constitute an action-and-eaction pai. Because the magnitudes of thei chages ae eual and thei masses ae the same, we find the speed of each paticle b finding the speed of eithe one. We ll appl oulomb s foce law fo point chages and Newton s second law to elate v to e, m, k, and thei sepaation distance L. Appl Newton s second law to the positon to obtain: ke L v ke m mv L L

16 6 hapte Solving fo v gives: v ke ml 85 Duing a famous expeiment in 99, Enest Ruthefod shot doubl ionized helium nuclei (also known as alpha paticles) at a gold foil. He discoveed that vituall all of the mass of an atom esides in an extemel compact nucleus. Suppose that duing such an expeiment, an alpha paticle fa fom the foil has an initial kinetic eneg of 5. MeV. If the alpha paticle is aimed diectl at the gold nucleus, and the onl foce acting on it is the electic foce of epulsion exeted on it b the gold nucleus, how close will it appoach the gold nucleus befoe tuning back? That is, what is the minimum cente-to-cente sepaation of the alpha paticle and the gold nucleus? omment [MSOffice5]: What???? Pictue the Poblem The wok done b the electic field of the gold nucleus changes the kinetic eneg of the alpha paticle eventuall binging it to est. We can appl the wok-kinetic eneg theoem to deive an expession fo the distance of closest appoach. Because the epulsive oulomb foce F e vaies with distance, we ll have to evaluate Fe d in ode to find the wok done on the alpha paticles b this foce. Appl the wok-kinetic eneg theoem to the alpha paticle to obtain: min Wnet Fe d ΔK o, because min min k e Fe d K f, min d 58ke K ( )( 79e) i d and Evaluating the integal ields: min 58ke 58ke min K i Solve fo min and simplif to obtain: 58ke min K i

17 The Electic Field : Discete hage Distibutions 7 Substitute numeical values and evaluate min : min 9 N m Mev ev 9 (.6 ) 9 J m 87 In Poblem 86, thee is a desciption of the Millikan expeiment used to detemine the chage on the electon. Duing the expeiment, a switch is used to evese the diection of the electic field without changing its magnitude, so that one can measue the teminal speed of the micosphee both as it is moving upwad and as it is moving downwad. Let v u epesent the teminal speed when the paticle is moving up, and v d the teminal speed when moving down. (a) If we let u v u + v d, show that 3πηu E, whee is the micosphee s net chage. Fo the pupose of detemining, what advantage does measuing both v u and v d have ove measuing onl one teminal speed? (b) Because chage is uantized, u can onl change b steps of magnitude N, whee N is an intege. Using the data fom Poblem 86, calculate Δu. Pictue the Poblem The fee bod diagam shows the foces acting on the micosphee of mass m and having an excess chage of Ne when the electic field is downwad. Unde teminal-speed conditions the sphee is in euilibium unde the influence of the electic foce F e, its weight mg, and the dag foce F d. We can appl Newton s second law, unde teminalspeed conditions, to elate the numbe of excess chages N on the sphee to its mass and, using Stokes law, to its teminal speed. (a) Appl Newton s second law to the micosphee when the electic field is downwad: Substitute fo F e and obtain: F d, teminal to F e mg Fd ma o, because a, F mg F m, Ne e d,teminal F e mg F d E mg 6πη vu o, because Ne, NeE mg 6πη vu

18 8 hapte Solve fo v u to obtain: With the field pointing upwad, the electic foce is downwad and the application of Newton s second law to the micosphee ields: Solve fo v d to obtain: v u F NeE mg () 6πη d, teminal Fe mg o 6πη vd NeE mg v d NeE + mg () 6πη Add euations () and () and simplif to obtain: NeE mg NeE + mg NeE E u vu + vd + 6πη 6πη 3πη 3πη Measuing both v u and v d has the advantage that ou don t need to know the mass of the micosphee. (b) Letting Δu epesent the change in the teminal speed of the micosphee due to a gain (o loss) of one electon we have: Δ + u v N v N Noting that Δv will be the same whethe the micosphee is moving upwad o downwad, expess its teminal speed when it is moving upwad with N electonic chages on it: v N NeE mg 6πη Expess its teminal speed upwad when it has N + electonic chages: v N + ( N + ) ee mg 6πη Substitute and simplif to obtain: ( N + ) Substitute numeical values and evaluate Δu: Δu 6πη ee 6πη ee mg NeE mg 6πη 9 4 (.6 )( 6. N/) 5 π (.8 Pa m)( 5.5 m) Δu 7 6 5μm/s

19 Page 7: [] Deleted Lea Mills 9/3/7 7:5: PM Because the positive solution coesponds to the 4 th chage being at the cente of the tiangle, it follows that: L 3 Page 8: [] Deleted Lea Mills 9/3/7 7:54: PM Substitute numeical values and evaluate Δx: 8 [(.)(.998 m/s) ](.74 s).3m Δx μ

7.2. Coulomb s Law. The Electric Force

7.2. Coulomb s Law. The Electric Force Coulomb s aw Recall that chaged objects attact some objects and epel othes at a distance, without making any contact with those objects Electic foce,, o the foce acting between two chaged objects, is somewhat