Chapter 22 The Electric Field II: Continuous Charge Distributions

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1 Chapte The lectic Field II: Continuous Chage Distibutions A ing of adius a has a chage distibution on it that vaies as l(q) l sin q, as shown in Figue -9. (a) What is the diection of the electic field at the cente of the ing? (b) What is the magnitude of the field at the cente of the ing? Pictue the Poblem The following diagam shows a segment of the ing of length ds that has a chage dq lds. We can expess the electic field d at the cente of the ing due to the chage dq and then integate this expession fom q to p to find the magnitude of the field in the cente of the ing. (a) and (b) The field d at the cente of the ing due to the chage dq is: The magnitude d of the field at the cente of the ing is: Because dq lds: d d x + d y -d cosq i - d sinq j kdq d klds d (1) The linea chage density vaies with kl sin q ds d q accoding to l(q) l sin q : Substitute dq fo ds: kl sinq dq kl sinq dq d 89

2 9 Chapte Substitute fo d in equation (1) to obtain: Integate d fom q to p and simplify to obtain: kl sinq cosq dq d - i kl sin q dq - j p p kl k k d i l sin sin d p l - q q - q q j - j p k - l (b) The field at the oigin is in the negative y diection p kl and its magnitude is. 4 A non-conducting solid sphee of adius has a volume chage density that is popotional to the distance fom the cente. That is, fo, whee A is a constant. (a) Find the total chage on the sphee. (b) Find the expessions fo the electic field the sphee ( < ) and outside the sphee ( > ). (c) Sketch the magnitude of the electic field as a function of the distance fom the sphee s cente. Pictue the Poblem We can find the total chage on the sphee by expessing the chage dq in a spheical shell and integating this expession between and. By symmety, the electic fields must be adial. To find the chaged sphee we choose a spheical Gaussian suface of adius <. To find outside the chaged sphee we choose a spheical Gaussian suface of adius >. On each of these sufaces, is constant. Gauss s law then elates to the total chage the suface. j (a) xpess the chage dq in a shell of thickness d and volume 4p d: Integate this expession fom to to find the total chage on the sphee: (b) Apply Gauss s law to a spheical suface of adius > that is concentic with the nonconducting sphee to obtain: dq 4p d 4p 4pA d ( A) d 4 4 [ pa ] p 4p A d A S 1 da Þ 4p

3 The lectic Field II: Continuous Chage Distibutions 91 Solving fo yields: ( > ) 4p 1 k kap 4 4 A 4 Apply Gauss s law to a spheical suface of adius < that is concentic with the nonconducting sphee to obtain: S 1 da Þ 4p Solve fo to obtain: ( < ) 4p A 4 4 pa 4p (c) The following gaph of vesus /, with in units of A/(4 ), was plotted using a speadsheet pogam / emaks: Note that the esults fo (a) and (b) agee at. 4 [SSM] A sphee of adius has volume chage density B/ fo <, whee B is a constant and fo >. (a) Find the total chage on the sphee. (b) Find the expessions fo the electic field and outside the chage distibution (c) Sketch the magnitude of the electic field as a function of the distance fom the sphee s cente. Pictue the Poblem We can find the total chage on the sphee by expessing the chage dq in a spheical shell and integating this expession between and

4 9 Chapte. By symmety, the electic fields must be adial. To find the chaged sphee we choose a spheical Gaussian suface of adius <. To find outside the chaged sphee we choose a spheical Gaussian suface of adius >. On each of these sufaces, is constant. Gauss s law then elates to the total chage the suface. (a) xpess the chage dq in a shell of thickness d and volume 4p d: Integate this expession fom to to find the total chage on the sphee: dq 4p d 4p 4pBd 4pB d pb B [ pb ] d (b) Apply Gauss s law to a spheical suface of adius > that is concentic with the nonconducting sphee to obtain: S 1 da o 4p Solving fo yields: ( > ) 4p 1 k kpb B Apply Gauss s law to a spheical suface of adius < that is concentic with the nonconducting sphee to obtain: S 1 da Þ 4p Solving fo yields: ( < ) 4p B pb 4p

5 The lectic Field II: Continuous Chage Distibutions 9 (c) The following gaph of vesus /, with in units of B/( ), was plotted using a speadsheet pogam / emaks: Note that ou esults fo (a) and (b) agee at. 5 Figue -4 shows a potion of an infinitely long, concentic cable in coss section. The inne conducto has a chage of 6. nc/m and the oute conducto has no net chage. (a) Find the electic field fo all values of, whee is the pependicula distance fom the common axis of the cylindical system. (b) What ae the suface chage densities on the and the outside sufaces of the oute conducto? Pictue the Poblem The electic field is diected adially outwad. We can constuct a Gaussian suface in the shape of a cylinde of adius and length L and apply Gauss s law to find the electic field as a function of the distance fom the centeline of the infinitely long, unifomly chaged cylindical shell. (a) Apply Gauss s law to a cylindical suface of adius and length L that is concentic with the inne conducto: S 1 n da Þ pl whee we ve neglected the end aeas because thee is no flux though them. Solving fo yields: k (1) L Fo < 1.5 cm, and: ( <1.5cm) Letting 1.5 cm, expess fo 1.5 cm < < 4.5 cm: ll psl

6 94 Chapte Substitute in equation (1) to obtain: k ( 1.5 cm < < 4.5 cm) L kl Substitute numeical values and evaluate n (1.5 cm < < 4.5 cm): ( ll) ( ) ( ) ( 6.nC/m ) ( 18N m/c 1.5cm 4.5cm N m /C ) 9 < < xpess fo 4.5 cm < < 6.5 cm: Letting s epesent the chage density on the oute suface, expess fo > 6.5 cm: Substitute in equation (1) to obtain: and ( 4.5cm < < 6.5cm) s A ps L whee 6.5 cm. ( ps L) k s ( > ) L In (b) we show that s 1. nc/m. Substitute numeical values to obtain: ( ) ( 1. nc/m )( 6.5 cm).5cm > 6-156N m/c 1 ( C / N m ) (b) The suface chage densities on the and the outside sufaces of the oute conducto ae given by: s - l p and l s outside p outside Substitute numeical values and - 6. nc/m p evaluate s and s outside : (.45 m) s - 1. nc/m and 6. nc/m s outside p (.65 m) 14.7 nc/m -1. nc/m

7 The lectic Field II: Continuous Chage Distibutions An infinitely long non-conducting solid cylinde of adius a has a nonunifom volume chage density. This density vaies linealy with, the pependicula distance fom its axis, accoding to () b, whee b is a constant. (a) Show that the linea chage density of the cylinde is given by l pba /. (b) Find expessions fo the electic field fo < a and > a. Pictue the Poblem Fom symmety consideations, we can conclude that the field tangent to the suface of the cylinde must vanish. We can constuct a Gaussian suface in the shape of a cylinde of adius and length L and apply Gauss s law to find the electic field as a function of the distance fom the centeline of the infinitely long nonconducting cylinde. (a) Apply Gauss s law to a cylindical suface of adius and length L that is concentic with the infinitely long nonconducting cylinde: S 1 n da o pl n Þ (1) pl whee we ve neglected the end aeas because thee is no flux though them. xpess d fo () a: d ( ) dv a( pl) pa Ld d Integate dfom to to obtain: pal pal é ù d palê ú ë û Divide both sides of this equation by L to obtain an expession fo the chage pe unit length l of the cylinde: L l pa (b) Substitute fo in equation pal < p L (1) and simplify to obtain: ( ) Fo > : pal a

8 96 Chapte Substitute fo in equation (1) and simplify to obtain: ( > ) pal pl a 66 Conside the concentic metal sphee and spheical shells that ae shown in Figue -4. The innemost is a solid sphee that has a adius 1. A spheical shell suounds the sphee and has an inne adius and an oute adius. The sphee and the shell ae both suounded by a second spheical shell that has an inne adius 4 and an oute adius 5. None of these thee objects initially have a net chage. Then, a negative chage is placed on the inne sphee and a positive chage + is placed on the outemost shell. (a) Afte the chages have eached equilibium, what will be the diection of the electic field between the inne sphee and the middle shell? (b) What will be the chage on the inne suface of the middle shell? (c) What will be the chage on the oute suface of the middle shell? (d) What will be the chage on the inne suface of the outemost shell? (e) What will be the chage on the oute suface of the outemost shell? (f) Plot as a function of fo all values of. Detemine the Concept We can detemine the diection of the electic field between sphees I and II by imagining a test chage placed between the sphees and detemining the diection of the foce acting on it. We can detemine the amount and sign of the chage on each sphee by ealizing that the chage on a given suface induces a chage of the same magnitude but opposite sign on the next suface of lage adius. (a) The chage placed on sphee III has no beaing on the electic field between sphees I and II. The field in this egion will be in the diection of the foce exeted on a test chage placed between the sphees. Because the chage at the cente is negative, the field will point towad the cente. (b) The chage on sphee I (- ) will induce a chage of the same magnitude but opposite sign on sphee II: + (c) The induction of chage + on the inne suface of sphee II will leave its oute suface with a chage of the same magnitude but opposite sign: - (d) The pesence of chage - on the oute suface of sphee II will induce a chage of the same magnitude but opposite sign on the inne suface of sphee III: + (e) The pesence of chage + on the inne suface of sphee III will leave the oute suface of sphee III neutal:

9 (f) A gaph of as a function of is shown to the ight: The lectic Field II: Continuous Chage Distibutions [SSM] A thin, non-conducting, unifomly chaged spheical shell of adius (Figue -44a) has a total positive chage of. A small cicula plug is emoved fom the suface. (a) What is the magnitude and diection of the electic field at the cente of the hole? (b) The plug is now put back in the hole (Figue - 44b). Using the esult of Pat (a), find the electic foce acting on the plug. (c) Using the magnitude of the foce, calculate the ²electostatic pessue² (foce/unit aea) that tends to expand the sphee. Pictue the Poblem If the patch is small enough, the field at the cente of the patch comes fom two contibutions. We can view the field in the hole as the sum of the field fom a unifom spheical shell of chage plus the field due to a small patch with suface chage density equal but opposite to that of the patch cut out. (a) xpess the magnitude of the electic field at the cente of the hole: Apply Gauss s law to a spheical gaussian suface just outside the given sphee: spheical + shell ( p ) hole enclosed spheical 4 shell Solve fo spheical shell to obtain: spheical shell 4p The electic field due to the small hole (small enough so that we can teat it as a plane suface) is: hole -s Substitute and simplify to obtain: 4p 4p 8p -s + - ( 4p ) adially outwad

10 98 Chapte (b) xpess the foce on the patch: F q whee q is the chage on the patch. Assuming that the patch has adius a, expess the popotion between its chage and that of the spheical shell: q p a a o q 4p 4 Substitute fo q and in the expession fo F to obtain: æ a F ç è 4 öæ ç øè 8p a p 4 ö ø adially outwad (c) The pessue is the foce exeted a on the patch divided by the aea of 4 p the patch: P 4 pa p 77 An infinite non-conducting plane sheet of chage that has a suface chage density +. µc/m lies in the y.6 m plane. A second infinite non-conducting plane sheet of chage that has a suface chage density of. µc/m lies in the x 1. m plane. Lastly, a non-conducting thin spheical shell of adius of 1. m and that has its cente in the z plane at the intesection of the two chaged planes has a suface chage density of. µc/m. Find the magnitude and diection of the electic field on the x axis at (a) x.4 m and (b) x.5 m. Pictue the Poblem Let the numeal 1 efe to the infinite plane whose chage density is s 1 and the numeal to the infinite plane whose chage density is s. We can find the electic fields at the two points of inteest by adding the electic fields due to the chage distibutions on the infinite planes and the sphee.

11 The lectic Field II: Continuous Chage Distibutions 99 xpess the electic field due to the infinite planes and the sphee at any point in space: (a) Because (.4 m, ) is the sphee: + + (1) sphee sphee (.4m,) 1 Find the field at (.4 m, ) due to plane 1: s1. µ C/m j ( ) 1.4m, j -1 ( 1 C /N m ) Find the field at (.4 m, ) due to plane : s (.4m,) (- i ) -. µ C/m Substitute in equation (1) to obtain: ( kn/c)j ( ) ( - i ) ( 11.9 kn/c 1 C /N m )i 1 (.4 m,) + ( kn/c) j + ( 11.9 kn/c) i ( 11.9 kn/c) i + ( kn/c)j Find the magnitude and diection of (.4 m,) : (.4m,) ( 11.9kN/C) + ( 169.4kN/C).6kN/C 4kN/C and - æ kn/c ö q tan 1 ç è 11.9 kn/c ø (b) Because (.5 m, ) is outside ksphee ( ) sphee.4m, the sphee: whee is a unit vecto pointing fom (1. m, -.6 m) to (.5 m, ). valuate sphee : sphee sa 4ps 4p sphee (-. µ C/m )( 1. m) -7.7 µ C efeing to the diagam above, detemine and : 1.616m and.985 i j

12 1 Chapte Substitute and evaluate (.5 m,) sphee : sphee 9 ( ) ( N m /C )(- 7.7 µ C ).5 m, ( 1.616m) Find the field at (.5 m, ) due to plane 1: (-19.8kN/C)(.985 i j ) (-1.5kN/C) i + (- 48. kn/c)j s1. µ C/m j ( ) 1.5 m, j -1 ( 1 C /N m ) Find the field at (.5 m, ) due to plane : s. C/m i - µ (.5 m,) - -1 ( 1 C /N m ) Substitute in equation (1) to obtain: ( 169.4kN/C)j i ( 11.9 kn/c)i (.4m,) (-1.5kN/C) i + ( kn/c) j + ( 169.4kN/C) j + (-11.9kN/C) (-.5kN/C) i + ( 11.kN/C)j Find the magnitude and diection of (.5 m,) : i and (.5 m,) (-.5kN/C) + ( 11.kN/C) 6kN/C - æ 11.kN/C ö q tan 1 ç 15 è -.5kN/C ø 8 A stationay ing of adius a that lies in the yz plane has a unifomly distibuted positive chage. A small paticle that has mass m and a negative chage q is located at the cente of the ing. (a) Show that if x << a, the electic field along the axis of the ing is popotional to x. (b) Find the foce on the paticle of mass m as a function of x. (c) Show that if the paticle is given a small displacement in the +x diection, it will pefom simple hamonic motion. (d) What is the fequency of that motion? Pictue the Poblem Stating with the equation fo the electic field on the axis of ing chage, we can facto the denominato of the expession to show that, fo

13 The lectic Field II: Continuous Chage Distibutions 11 x << a, x is popotional to x. We can use F x q x to expess the foce acting on the paticle and apply Newton s nd law to show that, fo small displacements fom equilibium, the paticle will execute simple hamonic motion. Finally, we can find the peiod of the motion fom its angula fequency, which we can obtain fom the diffeential equation of motion. (a) xpess the electic field on the kx x x + a axis of the ing of chage: ( ) Facto a fom the denominato of x to obtain: (b) xpess the foce acting on the paticle as a function of its chage and the electic field: x kx é æ x êa ç1 + ë è a kx æ x ö a ç1 + a è ø povided x << a. F q öù ú øû kq x a x x» k x a (c) Because the negatively chaged paticle expeiences a linea estoing foce, we know that its motion will be simple hamonic. Apply Newton s nd law to the negatively chaged paticle to obtain: (d) elate the fequency of the simple hamonic motion to its angula fequency: d x kq m - x dt a o d x kq + x dt ma the diffeential equation of simple hamonic motion. w f (1) p Fom the diffeential equation we kq kq w Þ have: w ma ma Substitute fo w in equation (1) 1 kq and simplify to obtain: f p ma

14 1 Chapte 87 [SSM] Conside a simple but supisingly accuate model fo the hydogen molecule: two positive point chages, each having chage +e, ae placed a unifomly chaged sphee of adius, which has a chage equal to e. The two point chages ae placed symmetically, equidistant fom the cente of the sphee (Figue -48). Find the distance fom the cente, a, whee the net foce on eithe point chage is zeo. Pictue the Poblem We can find the distance fom the cente whee the net foce on eithe chage is zeo by setting the sum of the foces acting on eithe point chage equal to zeo. ach point chage expeiences two foces; one a Coulomb foce of epulsion due to the othe point chage, and the second due to that faction of the sphee s chage that is between the point chage and the cente of the sphee that ceates an electic field at the location of the point chage. Apply åf to eithe of the point chages: xpess the Coulomb foce on the poton: The foce exeted by the field is: Apply Gauss s law to a spheical suface of adius a centeed at the oigin: F - F (1) Coulomb field ke F Coulomb F field e ( 4p a ) ke ( a) 4a enclosed elate the chage density of the e enclosed ea electon sphee to enclosed : Þ 4 4 enclosed p pa Substitute fo enclosed : Solve fo to obtain: Substitute fo F Coulomb and F field in equation (1): ke 4a o ke 4a ea ( 4p a ) ea Þ F field p p e a - p ke a 1 - Þ a 1 8 e a

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